Let a line BRD be drawn tangent to the curve at B and intersecting the line CA in R and the extended line OC in D. Since the curve KB is always moving away from the line PM in the advance from K to B, therefore the line BD, tangent to the curve in the point B and inclining back toward K, extends to an intersection with the line PM. Therefore, the angle ABR is obtuse, being certainly greater than the right angle MPB, into which it runs when PM is extended. Therefore, the angle ABR is greater than the angle RAB and the side AR is greater than the side BR. By adding the common part, the line AC is more than the lines BR and RC. But BR and RC, touching the curve in the points B and C, respectively, are more than the arc BC. Therefore the line AC is much more than that same arc CB, which it was desired to demonstrate.

By what has been demonstrated up until now the angle ABD is obtuse and likewise, because AP and DO are parallel, angle CDB equals angle ABD. Consequently with the subtending line CB having been drawn, the angle BDC is more than the angle BCD and the line DB less than the line BC. But the line BC is less than the curve BC and therefore the line BD is much less than the curve BC.

Let the tangents AC and EF be extended to L and G, respectively, on the nearest perpendiculars toward K. From the first result, AC is greater than the curve BC, EF is greater than the curve FC, and HK is greater than the curve KF. Consequently, together all of KH, FE, and CA are more than the unbroken curve KB. From the second result, BD (that is, KI) is less than the curve BC, CG (that is, CA) is less than the arc FC, and LF (that is, FE) is less than the arc KF. Consequently all of KI, FE, and CA together are less than the entire curve KB and hence the excess of the lines KH, FE, and CA over the lines KI, FE, and CA is more than their excess over the curve KB. But the lines FE and CA are common to each sum of lines and consequently the excess of the line KH over the line KI is equal to the excess of the sums of the lines KH, FE, and CA over the sum of lines KL, FE, and CA, which demonstrates that this excess is greater than the excess of the lines KH, FE, and CA above the curve KB, which is what ought to have been demonstrated.

If the area of PHδ2 is not equal to the surface of the cylinder mentioned before, there will be a difference between them which will be a planar figure α. Let the planar figure α be applied to the line X and let the side β be the breadth. Let a line 7A be drawn from the point 7 tangent to the curve D7 at the point 7. From the point D let a line Dγ be drawn touching the curve in the point D and let line 7ν be parallel to the line Dγ. From the points 7 and D let perpendiculars 72 and δD be dropped to the line 26 and let the line 2δ be divided into so many equal parts that from N3, the last perpendicular erected, the excess of the tangent 7A beyond the parallel line 7ν is less than the line β. Indeed, it is manifest that this is always able to be done if the tangent 7A is not perpendicular to the line 2δ and the given curve is non-winding. From the points 2, 3, 4, and δ of the partition of 2δ let perpendiculars 2P, 3N, 4L, and δH be erected, cutting the given curve in the points 7, 9, C, and D, the first invented curve in the points R, S, T, and V, and the second invented curve in the points P, N, L, and H. Then, let lines 7A, 9B, CF and Dγ be drawn tangent to the given curve at the points 7, 9, C, and D and terminated at the closest perpendiculars. Let 78, 90, CE, D5, YP, NK, LG, HI, LM, and NQ be lines parallel to and equal to the corresponding parts of 2δ and terminated in the nearest perpendicular.

The angle B96 is right, because 9B is a tangent and 96 is a perpendicular to it. The angle O93 is also right. Consequently, with the common angle O96 being taken away, an angle O9B equal to the angle 693 remains. Therefore, the right triangles O9B and 693 are similar and hence as 93 is to 96, so is 9O to 9B. But as 93 is to 96 so is X to 3N. Consequently, as 9O is to 9B thus is X to 3 N. Therefore, the rectangle formed by 9O and 3N--namely, N4--is equal to the rectangle formed by 9B and X. By this same method it is demonstrated that the rectangle P3 is equal to the rectangle formed by 7A and X and that the rectangle Lδ is equal to the rectangle formed by CF and X. Consequently, the rectangle formed by the line X and the lines 7A, 9B, and CF together is equal to the entire rectilinear figure 2PYNKLGδ. Therefore, the rectangle formed by X and the lines 7A, 9B, and CF together is more the area of the curved figure PNLHδ2.

Let DΠ be the normal to the curve at the point D. Therefore, the angle ΠDγ is right. But the angle δD 5 is also right. Consequently, by taking away the common angle, namely δD γ, it follows the angles δDΠ and γD 5 are equal. Therefore, the right triangles δD Π and γD 5 are similar. Consequently, as δD is to DΠ--that is, as D5 is to Dγ--so is X to δH. Therefore, the rectangle formed by δH and D5--that is, the rectangle H4--is equal to the rectangle formed by X and Dγ--that is, by X and 7ν.

But the rectangle M4 is equal to the rectangle Lδ--that is, to the rectangle formed by X and CF--and the rectangle Q3 is equal to the rectangle N4--that is, to the rectangle formed by X and 9B. Therefore, the rectilinear figure QNMLIHδ2 is equal to the rectangle formed by X and the lines FC, B9, and ν7 together. Therefore, the rectangle formed by X and the lines FC, B9, and ν7 together is less than the curve PNLHδ2. On the other hand, the lines FC, B9, and A7 together are more than the curve 79CD and the lines FC, B9, ν7 together are less than the same curve, each of which is clear from the preceding result. Therefore, the rectangle formed by X and the lines FC, B9, and A7 is more than the surface of the cylinder above the curve 79CD and the rectangle formed by X and the lines FC, B9, and ν7 is less than that same curve.

But it was demonstrated that the rectangle formed by X and the lines FC, B9, and A7 is more than the area of PNLHδ2 and that the rectangle formed by X and the lines FC, B9, and ν7 is less than that same area. Consequently, the difference between these rectangles is greater than the difference between the surface of the cylinder and the area. But the difference of these rectangles is the rectangle formed by X and the difference of the lines 7A and 7 ν. Moreover, the difference of the lines 7A and 7ν is less than the line β by supposition. Consequently, the difference between the rectangle formed by X and FC, B9, and A7 together and the rectangle formed by X and FC, B9, and ν7 together is less than the quantity α--the rectangle formed by X and β, which is, by supposition, the difference between the surface of the cylinder and the curved area. This is absurd. Therefore the surface of the cylinder and the area are equal, which is what ought to have been demonstrated.

Let the curve 79CD be supposed to be extended into a line ΓΛΣΔ equal to itself and let a line Δp equal to the line X be joined so that a rectangle hpΔΓ which is necessarily equal to the surface of the cylinder and to the curved region PNLHδ2 is filled out. Let Γε be equal to the line 27 and from the point ε let a curve ερψΩ be drawn of such a nature that, when any line ΓΛ equal to some small part of the curve, say, 79, has been assumed, the perpendicular to the line ΓΛ at the point Λ on the curve εΩ, namely Λρ, is equal to the perpendicular from the point 9 to the line 2δ--namely, 93. It is manifest that the curved region ερψΩΔΓ is equal to the surface of the trunk, since the inclination of the cutting plane is assumed to be a 45 degree angle.

Therefore, our proof is to demonstrate the equality of the curved regions RSTVδ2 and ερψΩΔΓ. First, we will demonstrate this equality in the latter figures, where we suppose that the curve RV always draws closer to the line 2δ in the progression from R to V and likewise that the curve 7D draws closer to that same line in the progression from 7 to D. Therefore, the curve εΩ draws closer to the line ΓΛ in the progression from ε to Ω , since when 7D draws closer to the line 2δ the curve εΩ draws to the line ΓΛ in the same manner. If the curved surfaces mentioned above are not equal, let α be the difference between them and let the curved region RVδ2 be divided into so many lines parallel to the line R2--namely, S3, T4, Vδ, and R2--that when the perpendiculars Rξ, Sγ, Sλ, Tθ, and Tμ, and Vη have been drawn from the intersections R, S, T, and V to the nearest parallels on either side there are two rectilinear regions, namely, RξSλTμδ2 above the curved region and γSθTηVδ2 within the curved region, the difference of which is less than α. It is evident that this is able to be done. Let the lines S3 and T4 be extended so that they intersect both curves PH and 7D in the points N, L, 9, and C, and let the lines PY, NQ, NK, LM, LG, and HI, parallel to 2δ and terminated by the dividing lines of the curved region, be joined.

Next, let the rectangle hΔ be divided by the lines hΓ, kΛ, mΣ, and pΔ, parallel to the line pΔ into the rectangles hΛ equal to the curved region PN32, kΣ equal to the curved region NL43, and mΔ equal to the curved region LHδ4. From the intersections of the lines dividing the rectangle hΔ with the curve εΩ, namely, ε, ρ, ψ, and Ω, let perpendiculars be drawn in both directions to the nearest dividing lines, namely, εζ, ρπ, ρσ, ψτ, ψω, Ωφ, so that πρτφΩΔΓ is a rectilinear figure within the curved region and εζρσψω ΔΓ is a rectilinear figure circumscribing the curved region. It is apparent from the previous proposition that P2 is to X--or hΓ--as R2 is to 72--or εΣ. By permuting, as P2 is to R2, thus is hΓ to εΓ. Therefore, as P3 is to R3, thus is hΛ to εΛ. But the rectangle hΛ is equal to the curved region PN32. Therefore, by permuting, as P3 is to the region PN32, thus is R3 to εΛ. Since P3 is greater than the curve, R3 will be greater than εΛ. In this same manner it is demonstrated that the rectangle S4 is more than ρΣ and that Tδ is more than ψΔ. Therefore, the rectilinear figure RξS λTμδ2 is more than the rectilinear figure εζρσψωΔΓ. In the same manner, as N3 is to X--or kΛ--thus is S3 to 93--or ρΛ. By permuting, as N3 is to S3 thus is kΛ to ρΛ. That is, as Q3 is to γ3, thus is hΛ--or the region PN32--to πΛ. By permuting, as Q3 is to the region PN32, thus is γ3 to πΛ. But Q3 is less than the curve. Therefore, γ3 is less than πΛ. In the same manner, it is demonstrated the θ4 is less than τΣ and ηδ is less than φΔ. Therefore, the rectilinear figure γS θTηVδ2 is less than the rectilinear figure πρτψφΩΔΓ.

Therefore, both rectilinear figures εζρσψωΔΓ and πρτψφΩΔΓ are between the two rectilinear figures RξSλTμδ2 and γS θ TηVδ2. That is, the greater of the first rectilinear curves is less than the greater of the second curves and the lesser of the first curves is more than the lesser of the second. The curved region ερψΩΔΓ is between the two first rectilinear figures and therefore is likewise between the two latter rectilinear figures, namely RξSλTμδ2 and γS θTηVδ2. But the curved region RSTVδ2 is likewise between these two rectilinear curves. Therefore, the difference between the rectilinear curves RξSλTμδ2 and γSθTηVδ2 is more than the difference of the curved regions RSTVδ2 and ερ ψΩΔΓ. But by supposition, the difference of these rectilinear figures is less than α. Therefore, the difference of the curved regions is much less than α, which is absurd. In fact, equality is shown. Therefore, the curved regions RSTVδ2 and ερψΩΔΓ do not differ. In fact, they are equal, which it was desired to demonstrate.

Second, we will demonstrate the same equality in the first figures, where we suppose the curve V always draws closer to the line 2δ in the progression from R to V and, in contrast, the curve 7D in the progression from 7 toΔ is protracted more from the same line. Therefore, on that account, the curve εΩ is more extended from the line ΓΔ in the progression from ε to Ω, since when 7D is extended from the line ΓΔ in the same manner the curve εΩ is extended from the line ΓΛ. If the curved regions are not equal, let their difference be α. Next, let the curved regions RSTVδ2 and let ερψΩΔΓ be divided by lines perpendicular to their bases 2δ and ΓΔ. It is done entirely as in the preceding demonstration, however, with the stipulation that the difference of the rectilinear figures RξS λTμδ2 and γSθTηVδ2 and also the difference of the rectilinear figures πρτψφ ΩΔΓ and εζρσψωΔ Γ together are less than the quantity α. It is manifest that this is possible, since this division is able to be done infinitely. With the same method which we used in the previous demonstration, it is demonstrated that the rectilinear figure RξS λTμδ2 is more than the rectilinear figure πρτψφΩΔΓ and the rectilinear figure γ SθT ηVδ2 is less than the rectilinear figure εζρσψωΔΓ.

With these things having been understood, if the given curved regions are not equal, let RSTVδ2 be more than the other. It is manifest that the excess of the rectilinear figure RξSλTμδ2 above the rectilinear figure εζρσψω ΔΓ is equal to all the rectangles γξ, θλ, ημ, πζ, τσ, and φω together with the excess of the rectilinear figure πρτψφ ΩΔΓ above the rectilinear figure γSθTηVδ2 having been removed. But the excess of the greater curved region above the lesser curved region is less than the previous-mentioned excess of the rectilinear regions, since the greater curved region is less than the greater rectilinear figure and the lesser curved region is more than the lesser rectilinear figure. Therefore, the excess of the greater curved region above the lesser curved region is less than the previously-mentioned rectangles together with the excess of the rectilinear figure εζρσψωΔΓ above the rectilinear figure γSθTηVδ2. But, by hypothesis, the rectangles together are less than the quantity α. Therefore, the rectangles together with the excess having been removed are much less than the quantity α. Hence, the greater curved region exceeds the lesser by much less excess than α, which is absurd. In fact, it is asserted that there is an excess of the greater curved region about the lesser. Therefore, the curved region RSTVδ2 is not more than the curved region εζρσψωΔΓ.

Let (if it is able to be done) it be less. It is manifest that the excess of the rectilinear region πρτψφΩΔ Γ above the rectilinear curve γSθTηVδ2 is equal to all of the rectangles πζ, πσ, φω, γξ, θλ, and νμ together with the excess of the rectilinear curve RξSλTμδ2 above the rectilinear curve εζρσψωΔΓ having been removed. But the excess of the greater curved region ερψΩΔΓ above the lesser curved region RSTVδ2 is less than the previously-mentioned excess of the rectilinear curves, since the greater curved region is less than the greater rectilinear figure and the lesser curved region is more than the lesser rectilinear figure. Therefore, the excess of the greater curved region above the lesser is less than the previously-mentioned rectangles together with the excess of the rectilinear figure RξSλT μδ2 above the rectilinear curve ερψΩΔΓ having been removed. But together the rectangles are, by hypothesis, less than α. Therefore, the rectangles together with the excess having been removed are much less than α. Hence, the greater curved region exceeds the lesser curved region by much less than α, which is absurd. In fact, it is asserted indeed that α is the excess of the greater curved region above the lesser curved region. Therefore, the curved region ερψΩΔΓ is not more than the curved region RξSλTμδ2. But it was also demonstrated that it is not less. Therefore, the curved regions ερψΩΔΓ and RξSλTμδ2 are equal among themselves, which is what ought to have been demonstrated.

If they are not equal, let the difference between them be λ and let the line RV be divided into so many equal parts at the points R, S, T, and V, that, when the lines SC, Tδ, and VZ have been drawn parallel to the line RD and the rectangles RC, RB, Sδ, SY, and TZ have been filled out, the difference of the rectilinear figure Rα BXYT inscribed in the curved region with the rectilinear figure RDCBδYZV circumscribing the curved region is less than the quantity λ. Indeed, it is manifest that this is able to be done, since the line RV is able to be divided into more and more parts ad infinitum. Let the parallel lines DR, CS, δT, and ZV be extended to the points E, G, I, and O on the constructed curve. Let the lines EL, GM, IN, and 04, tangent to the curve at the points E, G, I, and O and which intersect the closest parallels in either direction at the points F, H, N, ω, θ, and μ, be extended to the points L, M, N, and O on the line Lξ. Let EQ, GK, and IP be perpendiculars to the nearest parallel lines.

It is manifest (because the lines GK and SV are parallel) that GK is to GH as SV is to GM--or SB. Consequently, the rectangle Sδ--that is, the rectangle formed by GK and SB--is equal to the rectangle formed by GH and SV. But the rectangle formed by GH and SV is more than the portion of the trunk of the surface above the curve GI, since the line GH is more than the curve GI and--because the plane cuts the base at a 45 degree angle through the line Lξ--the line SV is equal to the highest altitude of the portion of the surface of the cylinder above the curve GI. Therefore, the rectangle Sδ is more than the portion of the trunk of the surface above the curve GI. In the same manner, the rectangle RC is shown to be more than the portion of the trunk of the surface above the curve GE and the rectangle TZ is more than the surface of the trunk above the curve OI. Therefore, the rectilinear figure RDCBδYZV circumscribing the curved region is more than the entire surface of the trunk.

Next, because the lines IP and SV are parallel, as IP is to IN--or GK or ST is to θI--thus is TV to IN--or to TY. Consequently, the rectangle SY--that is, the rectangle formed by ST and TY--is equal to the rectangle formed by θI and TV. But the rectangle formed by θI and TV is smaller than the portion of the surface of the trunk above the curve IG, since the line θI is less than the curve GI and--because the plane cuts the base in a 45 degree angle through the line Lξ--the line TV is equal to the smallest altitude of that same portion trunk of the surface above the curve GI. Therefore the rectangle SY is less than the portion of the trunk of the surface above the curve GI. In the same manner, the rectangle RB is shown to be less than the portion of the trunk of the surface above the curve GE. Consequently, the rectilinear figure TYXBαR inscribed in the curved region is less than the entire surface of the trunk.

But, by supposition, the difference between the inscribed rectilinear figure and the circumscribing figure is less than the quantity λ. Therefore, the difference between the truncated surface and the curved region is much less than λ, since either of the two is demonstrated to be more than the inscribed rectilinear area and less than the circumscribed rectilinear area, which is not able to be done. It is supposed that λ is the difference between the surface of the trunk and the curved region. Therefore, there is no difference between the curved region and the surface of the trunk. Consequently, they are equal, which is what ought to have been demonstrated.

Let G7, cutting the line RV at the point 7, be the line normal to MGμ at the point G. Because of the similarity of the triangles SG7 and GKH, as GS is to G7 thus is GK to GH. And as GK is to GH, thus is SV to GM--or SB--and also SR to Gμ--or B2. Therefore, GS is to G7 as RV is to S2. In this same manner it is able to be demonstrated that IT is to I8 as RV is to T3. And since this holds at all points of the curve EO, it is manifest from Proposition Two that the curved region RD2 3ξV is equal to the surface of the right cylinder above the curve EO and whose altitude is RV. But the surface of the lower trunk is equal to the curved region RVYBD and consequently the surface of the upper trunk is equal to the curved region DBYVξ3 2D, which it was desired to demonstrate.

Let them (if possible) be parallel and let the line AB be drawn parallel and equal to the line ED. Then through the points B and D let a curve agreeing in all parts with the curve AE be drawn, with the provision that the point A is placed above the point B and the point E is placed above the point D. It is manifest that the curve BD cuts the curve AD and that moreover the line CD parallel to the line GE is tangent to the curve AD. But by supposition CD is also tangent to the curve BD, which is absurd, since the curves AD and BD cut each other. Therefore, the lines CD and GE are not parallel, which is what ought to have been demonstrated.

Let the exhibited figure be ABSO and let the inscribed rectangle be ABRO. Let the curve BS be simple or non-winding, but if it is not, the curve ought to be divided into many simple parts and the demonstration carried out separately. Next, let the curve AFLP be of such a nature that when any line IN is drawn normal to the line AO and cutting the curve AFLP in L, the square on IN will equal a sum of the squares of both IL and IM. Next, let the curve AEKQ be drawn of such a nature that when any line IM is drawn perpendicular to the line AO and cutting the curve AEKQ in K and AFLP in L, the rectangle MIK is equal to the curved region IAFL. I say that the figure ABSO is to the rectangle ABRO as the curve AEKQ is to the line AO.

For let K be a point on the curve AEKQ, through which is drawn the line IN perpendicular to the line AO and cutting the curves AFLP, BR, and BHNS in the points L, M, and N. Let IK be to IC as IL is to IM and let KC be drawn. The line KC either cuts or is tangent to the curve AQ in the point K. If possible, let it cut the curve at K and let it will fall within the curve at a point E toward the vertex A. Through the point E let a line DH be drawn parallel to IN and cutting AQ, AP, BR, and BS in the points E, F, G, and H, and the line KC in α. Also, complete the rectangle ILZC, whose side LZ cuts the line DH at X. Since IL is to IM as IK is to IC, the rectangle MIK--or the curved region IAFL--will be equal to the rectangle IZ. Since the rectangle GDE is equal to the curved region DAF, as IK is to DE thus will the curved region IAFL be to the curved region DAF. But IK has a greater ratio to DE than to Dα. Thus, the curved region IAFL has a greater ratio to the curved region DAF than IK has to Dα--or IC has to DC. Therefore, the curved region IAFL has a greater ratio to the curved region DAF than the rectangle IZ has to the rectangle DZ, and, per conversionem rationis, the curved region IAFL has a smaller ratio to the curved region IDFL than the rectangle IZ has to the rectangle IX. By permuting, the curved region IAFL has a smaller ratio to the rectangle IZ than the curved region IDFL has to the rectangle IX. Since the rectangle IZ is equal to the curved region IAFL, the rectangle IX will be less than the curved region IDFL. But it is also more than the curved region IDFL, which is absurd. Consequently, the line KC does not fall within the curve AQ toward the vertex.

If possible, let the line CK fall within the curve toward the base, with the remaining things holding as in the previous situation. As IK is to DE, thus will the curved region IAFL be to the curved region DALF. But IK has a greater ratio to DE than to Dα. Therefore, the curved region IAFL has a greater ratio to the curved region DALF than IK has to Dα--or IC has to DC. Therefore, the curved region IAFL has a greater ratio to the curved region DALF than the rectangle IZ has to the rectangle DZ. Through inverting, conversionem rationis, and inverting in turn, the curved region IAFL has a greater ratio to the curved region IDFL than the rectangle IZ has to the rectangle IX. By permuting, the curved region IAFL has a greater ratio the rectangle IZ than the curved region IDFL has to the rectangle IX. Since the curved region IAFL is equal to the rectangle IZ, the rectangle IX will be more than the curved region IDFL. But it is also less, which is absurd. Therefore, the line CK does not fall within the curve AQ toward the base. Thus, the line KC is tangent to the curve AQ in the point K.

Let the line KT, intersecting the line AO in T, be perpendicular to the line CK. It is manifest that CI is to CK as IK is to KT. But CI is to CK as MI is to NI, since the lines IN, IM, and IL are a right triangle similar to the triangle CIK, with sides IM and IN corresponding to the sides CI and CK. Consequently, as IK is to KT, thus is IM to IN. Since it is done in the same manner for all points of the curve AQ, it is manifest from Proposition Two that the line AO is to the curve AQ as the rectangle OB is to the figure ABSO, which it was desired to demonstrate.

Let the curve BHC be a hyperbola whose diameter is the line AK and whose ordinates EH and KC are of such a nature that the solid formed by the square on BE together with AE is to the solid formed by the square on BK together with AK as the cube on EH is to the cube on KC. Let the given AB be a, let BE be b, and let the ratio of the solid from the square on BE together with AE be to the cube on EH as a³ is to c³. It is desired to find a point F so that the line FH touches the hyperbola in the point H.

From the given lines AB and BE, it is given that AE is a + b and EH is ³√(a b² c³ + b³ c³)/a. Let EF be z and DE be nothing or o. Therefore, BD will be b - o, AD will be a + b - o, and FD will be z - o. Therefore, DG will be

³√(c³ a b² - 2 c³ a b o + c³ a o² + c³ b³ - 3 c³ b² o + 3 c³ b o² - c³ o³)/ a.

Since we suppose that the ordinate DG falls upon the curve in the same point G where the line FH runs into the same curve (if it is able to be done), as EH is to EF thus will DG be to DF. Therefore, the rectangle formed by DF and EH, namely,

³√(b² c³ a z³ - 3 b² c³ a z² o + 3 b² c³ a zo² - b² c³ a o³ + b³ c³ z³ - 3 b³ c³ z² o + 3 b³ c³ z o² - b³ c³ o³)/ a

will be equal to the rectangle formed by EF and DG, namely,

³√(c³ a b² z³ - 2 c³ a b z³ o + c³ a z³ o² + c³ b³ z³ - 3 c³ b² z³ o + 3 c³ b z³ o² - c³ z³ o³)/ a.

By taking away the denominators because they are equal and also by cubing each end of the equation and taking away the equal things, the equation

3 b² c³ a z o² - 3 b² c³ a z² o - b² c³ a o³ - 3 b³ c³ z² o + 3 b³ c³ z o² - b³ c³ o³
= c³ a z³ o² - 2 c³ a b z³ o - 3 c³ b² z³ o + 3 c³ b z³ o² - c³ z³ o³

results. By dividing everything through by o, this is

3 b² c³ a z o - 3 b² c³ a z² - b² c³ a o² - 3 b³ c³ z² + 3 b³ c³ z o - b³ c³ o²
= c³ a z³ o - 2 c³ a b z³ - 3 c³ b² z³ + 3 c³ b z³ o - c³ z³ o².

And by rejecting the quantities in which o or a power of o is found,

-3 b² c³ a z² - 3 b³ c³ z² = -2 c³ a b z³ - 3 c³ b² z³

is left over. By adding the defects and by dividing everything through by c³ b z², the equation is 3 b a + 3 b² = 2 a z + 3 b z. Consequently, the line EF, namely, z = (3 b a + 3 b²)/(2 a + 3 b), which is what ought to have been found.

Let a line BI be drawn tangent to the curve ADIM at the point I (we suppose in fact that this is able to be done) and let the line IB be equal to the curve AI. Let the line BK be drawn. I say this line is tangent to the curve AFKO in the point K.

If it does not touch the curve, let it fall within the curve and let G be a point on the line within the curve in the direction of the vertex. Let a line GFEDC be drawn parallel to KH. It is manifest that IK is to EG as IB is to EB and, per conversionem rationis, IK is to the difference between IK and EG as IB is to IE. By permuting, as IK is to IB (or the curve IA)--that is, as Q is to P--thus the difference between IK and EG is to EI. But as Q is to P, thus DF is to the curve DA. Therefore, as IK is to the curve IA thus DF is to the curve DA. By permuting, as IK is to DF, thus the curve IA is to the curve DA. Per conversionem rationis, as IK is to the difference between IK and DF, thus the curve IA--or the line IB--is to the curve ID. But the difference between IK and EG is more than the difference between IK and DF, since the point G is supposed to lie within the curve. Hence, IK has a smaller ratio to the excess beyond EG than to the excess beyond DF. Therefore, IB is in a smaller ratio to IE than to the curve DI. Therefore, IE is more than the curve DI, which is absurd. (See Proposition One.) Therefore, the line BK does not fall within the curve AFKO toward the vertex.

Let the line fall inside the curve in the direction of the base at the point R (if it is able to be done). IK is to NR as IB is to NB, and IK is to the difference between IK and NR as IB is to IN. By permuting, as IK is to IB (or to the curve IA)--that is, as Q is to P, thus the difference between IK and NR is to IN. But as Q is to P, thus MO is to the curve MA. Therefore, as IK is to the curve IA thus MO is to the curve MA. By permuting, as IK is to MO thus the curve IA is to the curve MA. As IK is to the difference between IK and MO thus the curve IA--or IB--is to the curve IM. But the difference between IK and NR is less than the difference between IK and MO since we suppose that R falls within the curve. Hence, IK has a greater ratio to the difference between IK and NR than to the difference between IK and MO. Therefore, IB is in a greater proportion to IN than to IM. Consequently, IN is less than IM, which is absurd. Hence, the line BK does not fall within the curve toward the base and hence touches the curve in the point K, which it was desired to demonstrate.

Let a line IC be drawn tangent to the curve AEIO at I (we suppose in fact that this is able to be done) and intersecting the line AD parallel to NM at C. Let MZ be to the line IC as Q is to P and let ZD be parallel to the line AR. Let the line DM be drawn. I say this line is tangent to the curve AGMT at the point M.

If it were not tangent, let it fall within the curve and let H be a point within the curve toward the vertex. Let a line be drawn parallel to NM and cutting the remaining lines as in the figure. AE is to BG as P is to Q and AI is to NM as P is to Q. Consequently, as AE is to BG, thus AI is to NM. By permuting, as AE is to AI, thus BG is to NM, and as AE is to EI thus BG is to KM. By permuting, as AE is to BG--that is, as P is to Q--thus EI is to KM. Therefore, as CI is to ZM--that is, as P is to Q--thus FI is to LM, which I demonstrate as follows: The ratio of CI to ZM is compounded of the ratio of CI to DZ and the ratio of DZ to ZM, and the ratio of FI to LM is compounded of the ratio of FI to HL--or CI to DZ--and the ratio of HL to LM--or DZ to ZM. Consequently, as EI is to KM, thus FI is to LM. By permuting, as EI is to FI thus is KM to LM. But since we suppose that H falls within within the curve, KM is less than LM and consequently EI would be less than FI, which is absurd. (See Proposition One.) Therefore, the line DM does not drop within the curve toward the vertex.

Let it fall within the curve in the direction of the base at the point V (if it is able to be done). AI is to NM as P is to Q, and AO is to RT as P is to Q. Therefore, AI is to NM as AO is to RT. By permuting, AI is to AO as NM is to RT, and as AI is to IO thus NM is to XT. By permuting, as AI is to NM--or as P is to Q--thus IO is to XT. Consequently, as CI is to ZM--or P is to Q--thus IS is to XV (which is proved as in the previous case). Therefore, as IO is to XT thus IS is to XV. By permuting, as IO is to IS, thus XT is to XV. But (since we suppose that V falls inside the curve) XT will be more than XV. Therefore IO will be more than IS, which is absurd. (See Proposition One.) Therefore, the line DM does not fall inside the curve in the direction of the base. Consequently, DM is tangent to the curve at the point M, which it was desired to demonstrate.

Let HK be drawn parallel and equal to the line AB. It is manifest that the trapezoid ADEB is equal to the trapezoid ADMK. Also, the trapezoid ADMK is more than the trapezoid ADLO and therefore is much more than the curved region ADLO. Therefore, the proposition is clear, namely, that the trapezoid ADEB is more than the curved region ADLO.

Let the line CF be drawn parallel to the lines AB and DE and let CR be drawn parallel to the lines AQ and DS. It is clear that the trapezoid ABFC is equal to the trapezoid ACRQ and that the trapezoid CFED is equal to the trapezoid CDLI. Consequently, the rectilinear region ABEDC is equal to the rectilinear region ACDLIRQ, which is less than the rectilinear region ACDSQ. Therefore, the rectilinear region ABEDC is much less than the curved region ADSQ, which is what ought to have been demonstrated.

If they are not equal, let X be their difference and let the curvilinear figure BKMY be divided by so many lines CS, GP, and KM, parallel to the line BY, that (with the lines OM, QN, TR, and YV drawn parallel to the line AB) all the parallelograms ON, QR, and TV together are less than X. This is able to be done by the indefinite number of parallels. Let the subtending lines BC, CG, and GK and the tangents, BD, DF, FL, and KL at the points B, C, G, and K be drawn. It is manifest from the preceding proposition that the trapezoid ABCE is more than the curved region BCST, the trapezoid CEHG is more than the curved region CGPQ, and the trapezoid GHIK is more than the curved region GKMO. Therefore, the rectilinear figure ABCGKI is more than the curved region BKMOPQST. It is also clear from the preceding proposition that the rectilinear figure ABDCE is less than the curved region BCVY, the rectilinear figure ECFGH is less than the curved region CGRS, and the rectilinear figure HGKI is less than the curved region GKNP. Consequently, the rectilinear figure ABDFLKI is less than the curved region BKNPRSVY. Therefore, since the curved region BAIK is between the rectilinear figures ABCGKI and ABDFLKI and the curved region BKMY is between the curved regions KMOPQSTB and KNPRSVYB, and likewise the rectilinear figures ABCGKI and ABDFLKI are between the curved regions BKMOPQST and BKNPRSVY, it is manifest that the curved regions ABKI and KMYB differ by a smaller quantity than the curved regions BKMOPQST and KNPRSVYB. But, by supposition, the difference of these is less than X. Therefore, the difference of the areas ABKI and BKMY is much less than X, which is absurd. It is supposed that their difference is more than X. Therefore, there is no difference between the curved regions ABKI and BKMY. Hence they are equal, which is what ought to have been demonstrated.

In fact, as BG is to BH thus the arc EG--or the arc CH--is to the arc OH. But BG is more than BH and therefore CH is more than OH. Thus, the angle CBH is more than the angle EBG, which ought to have been demonstrated.

Let BK be drawn parallel and equal to the line AI, and let EGLI be a rectangle such that the rectilinear figure ABFGLI is inscribed in the figure ABMI. If ABMI is involuted, the angle of involution will be less than the angle of involution of the rectangle ABKI. Now this is noted thusly: The rectilinear figure ABFGLI involuted is the same as the rectangle ABFE involuted together with the rectangle EGLI involuted. The rectangle ABKI involuted is the same as the previous rectangle ABFE involuted together with the rectangle EFKI involuted. But from the preceding result the angle of involution of the rectangle EGLI is less than the angle of involution of the rectangle EFKI. Therefore, the angle of involution of the involuted rectilinear figure ABFGLI is less than the angle of involution of the involuted rectangle ABKI. In exactly the same manner (if the lines QR and SV are drawn parallel to the lines AB and IM and the lines Rθ and VY are drawn parallel to the axis AI so that they might fill up the rectilinear figure ABQRθGSVYI) it will be demonstrated that the angle of involution of this figure is less than the angle of involution of the rectilinear figure ABFGLI. Consequently, it will be much less than the angle of involution of the rectangle ABKI--namely, γNT--for we assumed that the sector γNT is the rectangle ABKI involuted.

Finally, it is always demonstrated in the same manner that the less the inscribed rectilinear figure differs from the figure ABMI, the more will the excess of the angle γNT always be above the angle of involution of the rectilinear figure. Therefore, the angle γNT of the figure ABMI itself exceeds by much more the angle of involution PNX. Consequently, the axis AI of the evolute figure--that is, the arc γT--exceeds the arc PT, which ought to have been demonstrated.

Let the line MD be drawn parallel and equal to the line AI and let AB be extended to D. Let ACGE be a rectangle such that the rectilinear figure ABMI is circumscribed in the figure ACGHMI. If ACGHMI is involuted, the angle of involution will be more than the angle of involution of the rectangle ADMI. Now this is noted thusly: The rectilinear figure ACGHMI involuted is the same as the rectangle EHMI involuted together with the rectangle ACGE involuted, and the rectangle ADMI involuted is the same as the previous rectangle EHMI involuted together with the rectangle ADHE involuted. But from the preceding proposition, the angle of involution of the rectangle ACGE is more than the angle of involution of the rectangle ADHE. Consequently, the angle of involution of the rectilinear figure ACGHMI involuted is more than the angle of involution of the rectangle ADMI. In exactly the same manner (if the lines Rβ and VZ are drawn parallel to the lines AB and IM and the lines Rα and VX are drawn parallel to the axis AI so that they fill up the rectilinear figure AαRβGXVZMI) it will be demonstrated that the angle of involution of this figure is more than the angle of involution of the rectilinear figure ACGHMI. Consequently it will be much more than the angle of involution of the rectangle ADMI--that is, δNX--for we suppose that the sector δNX is the rectangle ADMI involuted. Finally, it is always demonstrated in the same manner that the less the inscribed rectilinear figure differs from the figure ABMI, the more will the excess of the angle of involution of the rectilinear figure always be above the angle δNX. Therefore, the angle of involution of the figure PNX itself exceeds by much more the angle δNX of the figure ABMI itself. Consequently, the axis AI of the evolute figure--that is, the arc δX--is much less than the arc OX, which it was desired to demonstrate.

Let LBK be the involuted figure, the axis of evolution of which it is desired to find. Let MK be the arc of a circle with center L, and let the figure OP78 be contained by the parallel lines OP and 87, by the line 08 cutting those lines normally, and by the curve P7 of such a nature that (with any line LC in the involuted figure extended to N, and likewise the line SV in the figure OP78 drawn perpendicular to the line O8 and cutting it in the ratio of MN to NK) 87 is to SV as LK is the LC. I say that the circumscribed rectangular figure OR78 is to the figure OP78 as the arc MK is to the axis of the figure LBK evoluted.

If it were not thus, let OR78 be to OP78 thus as MK is to α, which differs from the axis of the figure LBK evoluted by the quantity δ. Next, from the center L let similar circular arcs AC, EG, and HK be circumscribed around the involute figure LBK and let just as many similar circular arcs BD, CF, and GI be inscribed within it so that the difference between the inscribed arcs and the circumscribed arcs is less than δ.

Next, let the line 08 be divided into as many equal parts OS, S4, and 48 as the arc MK by the extended lines LC and LG. Let lines SY and 4Z be drawn perpendicular to O8 itself and cutting the curve P7 in the points V and 2. And let PT, QV3, and X26 be joined parallel to the line O8. It is manifest from the description of the figure OP78 that SY is to SV as LN is the LC--that is, as the arc MN is to the arc AC. Therefore, as the rectangle OY is to the rectangle OV thus the arc MN is to the arc AC. In the same manner it is proved that as the rectangle SZ is the rectangle S2 thus the arc NH--that is, the arc MN--is to the arc EG, and as the rectangle 47 is to the rectangle 47 thus the arc HK is to the arc HK.

Since all the firsts are equal among themselves and all the thirds are equal among themselves, as all firsts--that is, the rectangle O7--are to all seconds--that is, the rectilinear figure OQVX2Z78--thus all thirds--that is, the arc MK--will be to all fourths--that is, the arcs AC, EG, and HK. Moreover, as 07 is to the figure OP78 thus MK is to α. But the rectilinear figure OQVX2Z78 is more than the figure OP78. Therefore the arcs AC, EG, and HK together are more than α. But the arcs AC, EG, and HK are also more than the axis of the evolute figure LKB, which is proved thusly: The axis of the entire figure LBK evoluted is equal to the axes of the figures BLC, CLG, and GLK evoluted. But from the preceding proposition the axis of the figure LBC evoluted is less than the arc AC, and the axis of the figure CLG evoluted is less than the arc EG, and likewise the axis of the figure GLK evoluted is less than the arc HK. Therefore, the axes of all the partial figures taken together--that is, the axis of the figure LBK evoluted--is less than all the arcs AC, EG, and HK taken together.

Next, from the description of the figure OP78, as OR is to OP--or OY is to OT--thus LM is to LB--that is, MN to BD. In the same manner it is demonstrated that as SZ is to S3 thus NH is to CF, and as 47 is to 46 thus HK is to GI. Since the firsts are equal among themselves and the thirds are equal among themselves, as all the firsts--namely, O7--are to all the seconds--namely, the rectilinear figure OPTV3268--thus all the thirds--namely, MK--will be to all the fourths--namely, the arcs BD, CF, and GI. Since O7 is to the figure OP78 as MK is to α and the rectilinear figure OPTV3268 is less than the figure OP78, all the arcs BD, CF, and GI will together be less than α. But the arcs BD, CF, and GI are also less than the axis of the evolute figure LBK, which is proved thusly: the axis of the entire figure LBK evoluted is equal to the axes of the figures BLC, CLG, and GLK evoluted. But from the preceding proposition, the axis of the figure LBC evoluted is more than the arc BD, the axis of the figure CLG evoluted is more than more than the arc CF, and the axis of the figure GLK is more than the arc GI. Therefore, the axis of all the partial figures taken together--or the axis of the evolute figure LBK--is more the all the arcs BD, CF, and GI taken together.

Therefore, it is evident that there are four magnitudes--namely, the first all the arcs BD, CF, and GI together, the second the axis of the figure LBK evoluted, the third α, and the fourth all the arcs AC, EG, and HK together--the maximum and minimum of which are the arcs AC, EG, and HK together and all the arcs BD, CF, and GI together. Therefore the difference of these is more than the difference of the two remaining quantities--namely, the axis of the figure LBK and the quantity α, which is absurd, for it is supposed even less. Therefore, there is no difference between α and the axis of the evolute figure LBK. Therefore, they are equal, which it was desired to demonstrate.

Let the figure OP78 be contained by the parallel lines OP and 87, with the line O8 cutting them normally, and let the curve P7 be of such a nature that (with any line LC extended to N in the assumed involute and likewise with the line SV drawn in the figure OP78 perpendicular to the line O8 and cutting it in the ratio of MN to NK) 87 is to SV in the duplicate ratio of the line LK to LC. I say that the circumscribing rectangle OR78 is to the figure OP78 as the sector LMK is to the involute LBK.

If it is not thus, as OR78 is to OP78 thus let MLK be to α, which differs from the involute BLK by the quantity δ. Next, let similar circular sectors LAC, LEG, and LHK be circumscribed around the involuted figure BLK and let as many similar circular sectors LBD, LCF, and LGI be inscribed in the same figure so that the differences between the inscribed figure LBDCFGI and the circumscribed figure LACEGHK is less than δ. Next, let O8 be divided into as many equal parts OS, S4, and 48 as the arc MK by the extended lines LC and LG, and let lines SY and 4Z be drawn perpendicular to O8 itself and cutting the curve P7 in the points V and 2. Let PT, QV3, and X26 be joined parallel to the line O8. It is manifest from the description of the figure OP78 that SY is to SV--that is, OY is to OV--in the duplicate ratio of LN to LC--that is, as the sector LMN is to the sector LAC. It is proved in the same manner that SZ is to S2 as LNH is to LEG, and 47 is to 47 as LHK is to LHK. Since all the firsts are equal among themselves and all the thirds are equal among themselves, as all the firsts--namely, the rectangle O7--are to all the seconds--namely, the rectilinear figure OQVX2Z78--thus all the thirds--namely, the sector MLK--will be to all the fourths--namely, the figure LACEGHK. Moreover, as O7 is to the figure OP78 thus the sector MLK is to α. But the rectilinear figure OQVX2Z78 is more than the figure OP78. Therefore, the figure LACEGHK is more than α.

Next, from the description of the figure OP78, OR is to OP--that is, OY is to OT--in the duplicate ratio of LM to LB--that is, as MLN is to BLD. It is proved in the same manner that SZ is to S3 as NLH is to CLF and 47 is to 46 as HLK is to GLI. Since the firsts and thirds are equal among themselves, all the firsts--namely, 07--will be to all the seconds--namely, the rectilinear figure OPTV3268--thus as all the thirds--namely, MLK--will be to all the fourths--namely, the figure LBDCFGI. But the rectilinear figure OPTV3268 is less than the figure OP78. Therefore, the figure LBDCFGIL is less than α. Therefore, it is evident that there are four magnitudes--namely, the first the figure LBDCFGI, the second the involute LBK, the third α, and the fourth the figure LACEGHK--the maximum and minimum of which are the figures LACEGHK and LBDCFHI. Therefore, the difference of these is more than the difference of the two remaining--namely, α and the involute LBK--which is absurd, for it is supposed less. Therefore, there is no difference between α and the involute BLK. Therefore, they are equal, which it was desired to demonstrate.

Let LBK be the involuted figure, which, when evoluted, makes the figure OP78. I say that the figure OP78 is double the figure LBK.

If it is not thus, let OP78 be the double of the quantity α, which differs from the involute LBK by the quantity δ. Let the figure LBDCFGI be inscribed in involute LBK and let the figure LACEGHK be circumscribed around the same so the difference of those is less than δ. Let B, C, G, and K be points on the evolute relative to P, V, 2, and 7, and let the rectangles OV, OT, S2, S3, 47, and 46 be filled out. The line LC is equal to the line SV and the curve AC is more than the line QV. Therefore, the sector of the circle LAC is more than half the rectangle OV. In the same manner, it is proved that the circular sector LGE is more than half the rectangle S2 and the circular sector LHK is more than half the rectangle 47. Therefore, the figure LACGHK is more than half the figure OQVX2Z78. Consequently, it is much more than half of the evolute OP78--namely, α.

Furthermore, the line OP is equal to the line LB and the curve BD is less than the line PT. Therefore, the sector of the circle LBD is less than half the rectangle OT. In the same manner it is demonstrated that the sector LCF is less than half the rectangle S3 and the sector LGI is less than half the rectangle 46. Therefore, the figure LBDCFGI is less than half the rectilinear figure OPTV3268. Consequently, it is much less than half the evolute--namely α. Therefore, it is evident that there are four quantities--namely, the first the figure LBDCFGI, the second the involute LBK, the third α, and the fourth the figure LACFGHK--the maxima and minima of which are the figures LACEGHK and LBDCFGI. Therefore ,the difference of these is more than the difference of the two remaining--namely, α and the involute LBK--which is absurd, for it is assumed less. Therefore, there is no difference between α and the involute LBK. Therefore, they are equal, which it was desired to demonstrate.

If it is able to be done, let it be less than the axis and let the axis of the evolute of the figure ABC be less than the excess of the axis of the evolute of the figure ABG above the line BK. From the center A let the circular arcs CI and HO be drawn. It is manifest that the arc CI is more than the axis of the evolute of the figure CAH and that HO is more than the axis of the evolute of the figure HAG. But the line DI is more than the arc CI and the line IK is more than the arc HO. Therefore, the line DK is much more than the axis of the evolute of the figure CAH together with the axis of the evolute of the figure HAG.

That is, the line DK is more than the axis of the evolute of the figure CAG. But the axis of the evolute of the figure BAG is above the line BK by a greater excess than the axis of the evolute of the figure BAC. Consequently, the axis of the evolute of the figure BAG minus the axis of the evolute of the figure BAC--that is, the axis of the evolute of the figure CAG--is more than the line BK. But it is also less than the line DK, which is absurd. Therefore the line BK is not less than the axis of the evolute of the figure BAG, which ought to have been demonstrated.

Let GE be a circular arc from the center A. GE is more than the line GF and less than the axis of the evolute of the figure ABC. Therefore, the line GF is much less than the axis of the evolute of the figure ABG, which it was desired to demonstrate.

If it is able to be done, let the curve OP fall outside the curve OQ at the point N: I suppose that the curve OQ (when closer to Q) has a greater distance from the axis IR. Therefore, the line NM perpendicular to the extended line IO falls outside the curve ON. Consequently (from the preceding result) the line NM is not less than the axis of the evolute of the figure ION. In the evolute, let the ordinate VS be equal to the line IN. It is manifest that the figure OVSI is the evolute of ION.

But IN--that is, SV--is more than LT and therefore SI is more than IL. That is, the axis of the evolute figure ION is more than MN, which is absurd. Consequently, OP does not fall outside of OQ. In the same manner it is proved that OP does not coincide with OQ. Therefore, it falls within it, which it was desired to demonstrate.

Second, if it is able to be done, let the curve OB fall outside the curve OA at the point D. I suppose that the curve OA (when closer to A) has a lesser distance from the axis CI. Let the perpendicular DF fall on the line OI within the curve DO. Therefore, DF is less than the axis of the evolute of the figure IDO. In the evolute, let the ordinate EH equal the line DI. It is manifest that the figure OEHI is the evolute of the figure IOD, but ID--that is, HE--is more than the line KG and therefore GI is more than HI. In particular, DF is more than the axis of the evolute figure IDO, which is absurd by the preceding proposition. Therefore, BO does not fall outside of AO. In the same manner it is proved that OB does not coincide with OA. Therefore, it falls within it, which it was desired to demonstrate.

Indeed, since P is to the arithmetic mean between DE and CE, as DC is to NO, therefore the rectangle formed by P and NO is equal to the trapezoid formed by the lines DC, CE, and DE, with right angles at D and C. But such a trapezoid is the common section of the plane above the line EO perpendicular to the base AB with the lower trunk of the cylindrical figure. Since it will always be this way wherever the line EDCNO is drawn, it is manifest from the doctrine of Gregory of Saint Vincent that the right cylindrical figure whose base is MOLN and altitude is P is equal to the lower trunk of the right cylindrical figure of the line above AB cut as stated above, which it was desired to demonstrate.

Since P is to DS as RF is to RT the rectangle from P and RT will be equal to the rectangle from RF and DS. But the rectangle from RF and DS is the common intersection of the lower trunk of the cylinder with the plane above the line RS normal to the base AB.

Therefore, the rectangle formed by P and RT is equal to that same common section. Since it will always be this way wherever the line RS is drawn, it is manifest from the doctrine of Gregory of Saint Vincent that the right cylinder whose base is QTVR and altitude P is equal to the lower trunk of the right cylinder above AB cut as stated above, which it was desired to demonstrate.

Let AD cut the line FG in E and let the line XEZ parallel to the line BC be drawn through E. From the given points F and G, the point E is given, since the figure ABC is assumed to be given. Below XZ let a plane be drawn parallel to the base BOCN and making the lines PQ and RS the common intersection with the planes ABC and FOGN. It is manifest that RS is the radius of the circle in the solid of FOGN passing through R. And on account of the revolution of the solid ABC and the normal section of the plane ABC to the base, the point S is on the semicircle whose diameter is PQ. Therefore, the square on the line RS is equal to the rectangle PRQ, and the circle with radius RS is equal to the circular ring on PRQ. [EDITOR'S NOTE: Use Euclid II.5.]

Since it will always be this way between E and G, the portion of the solid of revolution on FOGN beneath E is equal to all the circular rings between the circle XEZ and the ring on BGC--that is, to the hollowed-out solid equal to the portion of the solid of revolution on XZCB above the removed cone whose vertex is E and whose base is the radius DG.

It is proved in the same manner that the portion of the solid of revolution FOGN between E and F is equal to the hollowed-out solid on FKZX above the removed cone whose vertex is E and whose base radius is IF. Therefore, if from the given portion FKCB of the solid of revolution four times as much toward the vertex of the given cones FEI and DEG were removed (NOTE: his translation.), the sought-after solid of revolution FOGN will be left behind.

Let the center of gravity of the solid of revolution ABC and the center of the removed figure AFK be given. Also, let the center of gravity of the portion on FKCB be given. With these having been given together with the center of gravity of the removed cone, the center of gravity of the hollowed-out portion, which is proportionally analogous with the solid of revolution FOGN, is also given, as is evident from the demonstration. Therefore, the center of gravity of the solid of revolution on FOG is also given.

Let any line HIKLN be drawn. The square on the side IK--that is, the rectangle HKL--is a quarter part of the square on IN. Therefore, the rectangle HKL is to the circle on the diameter IN in a ratio compounded from the ratio of 1 to 4 and the square of the diameter of a circle to the circle. But the point K was assumed arbitrarily. Consequently, the right cylindrical figure with base ALDEK and altitude HK is to the solid of revolution ABC in the ratio compounded from the ratio of 1 to 4 and the square on the diameter of a circle to the circle. Therefore, four times the previously-mentioned cylindrical figure is to the solid of revolution ABC as the square on the diameter of a circle is to the circle--that is, as a rectangular parallelpiped is to the cylinder inscribed in the same altitude. By permutando, four times the original cylindrical figure is to the parallelpiped as the solid of revolution is to the circular cylinder. Consequently, the ratio of four times the cylindrical figure to the parallelpiped is given. Therefore, the cubature of the cylindrical figure and the quadrature of its base ALDEK are given.

Let ABDC be a right cylindrical figure above a figure DCF which is cut by a plane KINM in such a way that the figure RGHQ is the common intersection of the plane with the cylinder.

Let the cutting plane be extended until it cuts the base plane DCF in the line KI and the plane of AB in the line MN. From any point, say, L, on the line IK let a plane OLP be drawn perpendicular to IK and cutting the planes IKMN and DFC normally in the lines OL and LP. Let the line OP be perpendicular to LP. We assume that KI is the axis of rotation. We call the line PL normal to this the radius of rotation. I say that the trunk RQDC of the cylinder is to the solid of revolution arising from the rotation of the figure DEFC around the axis of rotation IK as the altitude OP of the cylinder is to the circumference of the circle whose radius is the radius of rotation LP.

Let EF be any line terminated on both sides of the circumference of the base at E and F and drawn through the base DEFC, which, when extended, intersects the axis of rotation normally in the point V. From the points E and F let EH and FG be sent out perpendicular to the base plane and terminated in the cutting plane at H and G. Since the cylinder is right, these are necessarily on the surface of the trunk. Let the line VH be drawn, which is necessarily in the cutting plane IKMN. It is manifest that the triangles OLP and HVE, which have right angles at the points P and E, are similar (since the angles OLP and HVE are equal to the inclination of the cutting plane IKMN). With the line GV having been drawn, the triangles HEV and GFV are similar for the same reason. Since GF is parallel to the line HE and EF intersects EV, the lines GV and HV lie on one line in the plane IKMN. GHEF will be the common intersection of the plane GFV, parallel to the plane OPL, with the truncated cylinder RQDC. Therefore, it is apparent that OP is to PL as HE is to EV. Therefore, as OP is to the circumference of the circle whose radius is PL, thus HE is to the circumference of the circle whose radius is EV. And as OP is to the circumference of the circle whose radius is PL thus the triangle HEV is to the circle whose radius is EV (by drawing the remaining boundaries to the same altitude, namely, half of the line EV). It is demonstrated in the same manner that as OP is to the circumference of the circle whose radius is PL, thus the triangle GFV is to the circle whose radius is FV. Therefore, the entire triangle GFV is to the entire circle with radius FV thus as the cut-off triangle HEV is to the cut-off circle whose radius is EV. Consequently, the remaining trapezoid GFEH will be in the same ratio to the remaining annulus arising from the revolution of the line FE around the axis of rotation IK, namely, in the ratio of OP to the circumference of the circle whose radius is LP. But this proportion is demonstrated in the same manner for all lines drawn in the base DEFC which, when extended (if necessary), meet the axis of rotation IK normally. But the base DC consists of all such lines, the trunk RQDC consists of all of the trapezoids above these lines, and the solid of revolution arising from the revolution of the base around the axis of revolution IK consists of all of the annuli produced by the revolution of these lines. Consequently, as one antecedent is to one consequent--namely, the altitude OP of the cylinder is to the circumference of the circle whose radius PL is the radius of rotation--thus, all antecedents--namely, all trapezoids--that is, the trunk RQDC--are to all consequents--namely, all annuli--that is, the solid of revolution arising from the rotation of the figure DC around the axis IK, which it was desired to demonstrate.

Let the figure and preparation be just as in the preceding proposition. I say that the surface of the trunk RQDC is to the surface of the solid of revolution arising from the rotation of the figure DEFC around the axis of rotation IK as the altitude OP of the cylinder is to the circumference of the circle whose radius is the radius of rotation LP.

In the previous proposition it was demonstrated that OP is to the circumference of the circle whose radius is PL as HE is to the circumference of the circle whose radius is EV. But this proportion is demonstrated in the same manner for all lines on the surface of the trunk RQDC perpendicular to the base DEFC to all circumferences of circles described by their lowest points in the rotation. But the surface of the trunk itself consists of all these lines. And the surface of the solid of revolution arising from the rotation of the base around the axis IK consists of all the circumferences of the circles out of the bottom points of the lines--that is, from the revolution of the described base. Therefore, as one of the antecedents is to one of the consequents--namely, the altitude OP of the cylinder to the circumferences of the circle whose radius is the radius of rotation LP--thus all the antecedents--that is, the surface of the trunk RQDC--are to all the consequents--that is, the surface of the solid arising from the rotation of the base DEFC around the axis of rotation IK--which ought to have been demonstrated.

Indeed, by the preceding proposition, as the altitude of the cylinder is to the circumference of the circle whose radius is the radius of rotation, thus the surface of the trunk is to the surface of the solid of revolution. Moreover, in this case, because the angle of inclination is semi-right, the altitude of the cylinder is equal to the radius of rotation. Therefore, in our case, as the radius is to its own circumference, thus the surface of the trunk is to the surface of the solid of revolution. Moreover, as the radius is to the circumference, thus half of the square of the radius is to the circle. Therefore, as half of the square of the radius of a circle is to the circle, thus the surface of the trunk is to the surface of the solid of revolution. By convertendo and permutando, a circle is to the surface of the solid of revolution as half of the square on the radius of the circle is to the surface of the trunk. But the circle is assumed equal to the surface of the solid of revolution. Therefore, half of the square of the radius on that circle is equal to the surface of the trunk. Therefore, the square on the radius is double the surface of the trunk, which ought to have been demonstrated.

In fact, the trunk (in this case) is to the solid of revolution as a radius is to its own circumference--that is, as the double of the square on a radius is to the quadruple of the circle--that is, as 2/3 of the cube on a radius is to the sphere. Therefore, the trunk is to the solid of revolution as 2/3 of the cube of the radius is to the sphere. By convertendo and permutando, the solid of revolution is to a sphere as the trunk is to 2/3 of the cube on the radius of the sphere. But the sphere is assumed equal to the solid of revolution. Therefore, 2/3 of the cube on the radius of the sphere is equal to the trunk. But the cube on the radius has the same ratio to 2/3 of itself as 3 has to 2. Therefore, the cube on the radius of the sphere has the same ratio to the trunk, which ought to have been demonstrated.

Let ABCD and NMOYXZ be two right cylindrical figures of equal height with bases DC and XYZ which are cut by planes together into two trunks. Specifically, let ABCD be cut by the plane KHFG into trunks AB43 and 43DC and let NMOYXZ be cut by a plane PSVR into trunks NMO765 and 765ZXY. Let the lines KH and GF be the intersections of the plane HFGK with the planes of the extended (if necessary) parallel bases DC and AB of the cylinder, and let the lines RP and VS be the intersections of the plane RVSR with the planes of the extended (if necessary) parallel bases NMO and ZXY of the cylinders. Let there be planes normal to the lines FG and SV which intersect the cutting planes in the lines IE and QT and the planes of the extended (if necessary) DC and ZXY bases in the lines LE and WT, and let the angles ILE and QWT be right.

It is manifest from Proposition 23, with HFGK and PSVR the assumed cutting planes and GF and VS the axes of rotation, that LE and WT are the radii of rotation. Therefore, I say that the solid of revolution arising from the rotation of the figure DC around GF is to the solid of revolution arising from the rotation of the figure ZXY around VS in the ratio compounded from the proportion of the trunk 34CD to the trunk 567YXZ and from the proportion of the radius of rotation LE to the radius of rotation WT.

The ratio of the solid of revolution arising from the figure DC to the solid of revolution arising from the figure ZXY is compounded from the ratio of the solid of revolution arising from DC to the trunk 34CD, from the ratio of the trunk 34CD to the trunk 567YXZ, and from the ratio of the trunk 567YXZ to the solid of revolution arising from YXZ. But the ratio of the solid of revolution arising from DC to the trunk 34CD is equal to the ratio of the circumference of the circle described by the radius LE to the altitude IL of the cylinder. (See Proposition 23.) The ratio of the trunk 567YXZ to the solid of revolution arising from YXZ is equal to the ratio of the altitude QW--that is, IL--of the cylinder to the circumference of the circle described by the radius TW. Consequently, the ratio of the solid of revolution arising from DC to the solid of revolution arising from ZXY is compounded of the ratio of the trunk 34CD to the trunk 567YXZ, from the ratio of the circumference of the circle described by the radius LE to the line IL, and from the ratio of the line IL to the circumference of the circle described by the radius TW. But these last two ratios compound to the ratio of the circumference described by the radius LE to the circumference of the circle described by the radius TW, which is the same as the ratio of the radius LE to the radius TW. Consequently, the solid of revolution arising from the rotation of the figure DC around the axes FG is to the solid of revolution arising from the rotation of the figure XYZ around the axis VS in a ratio compounded from the proportion of the lower trunk 34CD to the lower trunk 567YXZ and from the proportion of the radius of rotation EL to the radius of rotation TW, which ought to have been demonstrated.

Let the figures and the preparation be the same as in the preceding proposition. I say that the surface of the solid of revolution arising from the rotation of the figure DC around GF is to the surface of the solid of revolution arising from the rotation of the figure ZXY around VS in a ratio compounded from the proportion of the surface of the trunk 34CD to the surface of the trunk 567YXZ and the proportion of the radius of rotation LE to the radius of rotation WT.

The ratio of the surface of the solid of revolution arising from the figure DC to the surface of the solid of revolution arising from the figure ZXY is compounded from the ratio of the surface of the solid of revolution arising from DC to the surface of the trunk 34CD, the ratio of the surface of the trunk 34CD to the surface of the trunk 567YXZ, and the ratio of the surface of the trunk 567YXZ to the surface of the solid of revolution arising from ZXY. But the ratio of the surface of the solid of revolution arising from DC to the surface of the trunk 34CD is equal to the ratio of the circumference of the circle described by the radius LE to the altitude IL of the cylinder. (See Proposition 24.) The ratio of the surface of the trunk 567YXZ to the surface of the solid of revolution arising from YXZ is equal to the ratio of QW--that is, of the altitude IL of the cylinder--to the circumference of the circle described by the radius TW. Consequently, the ratio of the surface of the solid of revolution arising from DC to the surface of the solid of revolution arising from ZXY is compounded from the ratio of the surface of the trunk 34CD to the surface of the trunk 567YXZ, from the ratio of the circumference of the circle described by the radius LE to the line IL, and from the ratio of the line IL to the circumference of the circle on the radius TW. But these last two ratios compound the ratio of the circumference of the circle described by the radius LE to the circumference of the circle described by the radius TW, which is the same as the ratio of the radius LE to the radius TW. Consequently, the surface of the solid of revolution arising from the rotation of the figure DC around the axes FG is to the surface of the solid of revolution arising from the rotation of the figure XYZ around the axes VS in a ratio compounded from the proportion of the surface of the lower trunk 34CD to the surface of the lower trunk 567YXZ and from the proportion of the radius of rotation EL to the radius of rotation TW, which ought to have been demonstrated.

Let the figure and the preparation be the same as in the preceding proposition, with this one exception: O and P and are now assumed to be centers of gravity of the perimeters of the opposite bases. I say that the surface of the trunk ABDZY2 is to the surface of the trunk 2YZMKL reciprocally as IO is to OS.

From the midpoints of the lines FI and QS--namely, H and R--let lines HS, RF, and HR be drawn, with HR cutting OP in X. Thus, the lines FI, PO, and QS; and likewise FQ, HR, and IS; and likewise FR and HS will be parallel and equal between themselves. Since FR bisects QS, it will also bisect all the lines in the triangle FQS equidistant to QS itself. Consequently, it will bisect all the diameters of the rectangles in the trunk ABDZY2 which are cut normally by the plane FISQ. Therefore, it will pass through all the centers of gravity of the opposite sides perpendicular from those rectangles to the cylindrical base, since the center of gravity of the opposite sides is in the middle of the diameter. Since the surface of the trunk itself is made up of all of those opposite sides perpendicular to the base of the cylinder, therefore the line FR will also pass through the center of gravity of the surface of the trunk itself. Suppose this is 3. It is demonstrated in the same manner that the center of gravity of the surface of the trunk YZMLK2 is on HS. Therefore, since the midpoint X of the line OP is the center of gravity of the entire surface of the cylinder, if the line 3X4 is drawn through X from 3 until it intersects the line HS in 4, 4 will be the center of gravity of the surface of the trunk YZMLK2. Because the triangles XR3 and XH4 are similar on account of RF and HS being parallel, as X4 is to X3 thus XH will be to XR--that is, IO to OS. But as X4 is to X3 thus the surface of the trunk ABDZY2 will be to the surface of the trunk YZMLK2. Therefore, as the surface of the trunk ABDZY2 is to the surface of the YZMLK2, thus reciprocally IO will be to OS, which ought to have been demonstrated.

Let ABC and HIL be any two figures symmetric around the axes BF and IN, which are rotated around the lines EG and MO cutting the extended (if necessary) axes BF and IN normally at the points F and N. Let D and K be the centers of gravity of the figures ABC and HIL. I say that the ratio of the solid arising from the figure ABC rotated around the line EG to the solid arising from the figure HIL rotating around the line MO, is compounded from the ratio of the figure ABC to the figure HIL and from the ratio of DF to KN.

Above the figures ABC and HIL let right cylindrical figures of equal height be cut by planes passing through the lines EG and MO, each one into two trunks, namely, an upper and a lower trunk. The ratio of the solid of revolution arising from ABC to the solid of revolution arising from HIL is compounded from the ratio of the lower trunk of the cylinder above ABC to the lower trunk of the cylinder above HIL and from the ratio of the radius of rotation of the figure ABC to the radius of rotation of the figure HIL.

But the lower trunk of the cylinder above ABC is to the lower trunk of the cylinder above HIL is in a ratio compounded from the ratio of the lower trunk of the cylinder above ABC to the entire cylinder above ABC, from the ratio of the entire cylinder above ABC to the entire cylinder above HIL, and from the ratio of the entire cylinder above HIL to its lower trunk. But from the convertendo of the Consequence to Proposition 29 the lower trunk of the cylinder above ABC is to the entire cylinder as FD is to the radius of rotation of the figure AB. Also, the cylinder above ABC is to the cylinder above HIL as the figure ABC is to the figure HIL. Similarly, by the Consequence to Proposition 29, the cylinder above HIL is to its own lower trunk as the radius of rotation of the figure HIL is to KN. Consequently, the ratio of the lower trunk of the cylinder above ABC to the lower trunk of the cylinder above HIL is compounded from the ratio of the line DF to the radius of rotation of the figure ABC, from the ratio of the figure ABC to the figure HIL, and from the ratio of the radius of rotation of the figure HIL to the line KN. Therefore, the ratio of the solid arising from the rotation of the figure ABC to the solid arising from the rotation of the figure HIL is compounded from the ratio of the figure ABC to the figure HIL, from the ratio of the line DF to the radius of rotation of the figure ABC, from the ratio of the radius of rotation of the figure ABC to the radius of rotation of the figure HIL, and from the ratio of the radius of rotation of the figure HIL to the line KN. But the last three ratios compound to the ratio of DF to KN. Therefore, the ratio of the solid arising from the rotation of the figure ABC around EG to the solid arising from the rotation of the figure HIL around MO is compounded from the ratio of the figure ABC to the figure HIL and from the ratio of the segment between the center of gravity of the figure ABC and its axis of rotation--namely, DF--to the segment between the center of gravity of the figure HIL and that same axis of rotation--namely, KN--which ought to have been demonstrated.

Let the figure and the preparation be the same as in the preceding proposition, with this one exception: that the points D and K are now assumed to be the centers of gravity of the perimeters of the figures. I say that the ratio of the surface of the solid arising from the figure ABC rotated around the line EG to the surface of the solid arising from the figure HIL rotating around the line MO is compounded from the ratio of the perimeter of ABC to the perimeter of HIL and from the ratio of DF to KN.

Above the figures ABC and HIL are understood right cylinders of equal height cut by planes passing through the lines EG and MO, each one into two trunks, namely, an upper and a lower trunk. The ratio of the surface of the solid arising from ABC to the surface of the solid arising from HIL is compounded from the ratio of the surface of the lower trunk of the cylinder above ABC to the surface of the lower trunk of the cylinder above HIL and from the ratio of the radius of rotation of the figure ABC to the radius of rotation of the figure HIL. But the surface of the lower trunk of the cylinder above ABC is to the surface of the lower trunk of the cylinder about HIL in the ratio compounded from the ratio of the surface of the lower trunk of the cylinder above ABC to the entire surface of the cylinder above ABC, from the ratio of the entire surface of the cylinder above ABC to the entire surface of the cylinder above HIL, and from the ratio of the entire surface of the cylinder above HIL to the surface of its own lower trunk. But by the convertendo of the consequence of Proposition 30 the surface of the lower trunk of the cylinder above ABC is to the entire surface of the cylinder as FD is to the radius of rotation of the figure ABC. The surface of the cylinder above ABC is to the surface of the cylinder above HIL as the perimeter of ABC is to the perimeter of HIL. Likewise, the surface of the cylinder above HIL is to the surface of its own lower trunk as the radius of rotation of the figure HIL is to KN. Consequently, the ratio of the surface of the lower trunk of the cylinder above ABC to the surface of the lower trunk of the cylinder above HIL is compounded from the ratio of the line DF to the radius of the rotation of the figure ABC, from the ratio of the perimeter of ABC to the perimeter of HIL, and from the ratio of the radius of rotation of the figure HIL to the line KN. Therefore, the ratio of the surface of the solid arising from the rotation of the figure ABC to the surface of the solid arising from the rotation of the figure HIL is compounded from the ratio of the perimeter of ABC to the perimeter of HIL, from the ratio of the line DF to the radius of rotation of the figure ABC, from the ratio of radius of rotation of the figure ABC to the radius of rotation of the figure HIL, and from the ratio of the radius of rotation of the figure HIL to the line KN. But the last three ratios compound to the ratio of DF to KN. Therefore, the ratio of the surface of the solid arising from the rotation of the figure ABC around EG to the surface of the solid arising from the rotation of the figure HIL around MO is compounded from the ratio of the perimeter of ABC to the perimeter of HIL and from the ratio of the segment between the center of gravity of the perimeter of ABC and its axis of rotation--namely, DF--to the segment between the center of gravity of the perimeter of HIL and the axis of rotation of the same--namely, KN--which ought to have been demonstrated.

Let ABC and NQP be any two figures which are rotated around the lines EF and Y4, and let their centers of gravity be D and R, which are sent down lines DG and RZ perpendicular to the axes of rotation EF and Y4. I say that the ratio of the solid arising from the figure ABC rotated around the line EF to the solid arising from the figure NQP rotated around the line Y4 is compounded from the ratio of the figure ABC to the figure NQP and from the ratio of DG to RZ.

Let the lines BH and Q2 touching the figures ABC and NQP at B and Q be drawn parallel to the lines DG and RZ. Let the figures ABC and QNP be conceived to be revolved around the lines BH and Q2, like axes, until, attaining the plane on the other part of the axes, they make the figures BLM and QTX, equal and similar to themselves and having exactly the same position toward the lines BH and EF, and Q2 and Y4. Let O and V be the centers of gravity of the figures BLM and QTX. Let the lines OI and V3 be drawn perpendicular to the lines EF and Y4. Also, let the lines DO and RV, intersecting the lines BH and Q2 in the points K and S, be joined.

It is manifest that the point K is the center of gravity of the figure BACBLM symmetric around the axis BH and likewise that the point S is the center of gravity of the entire figure QNPQTX around the axis Q2. It is also apparent that the line DG, KH, OI and also RZ, S2, V3, are equal among themselves. Since the figures BACBLM and QNPQTX are symmetric around the axes BH and Q2, which are normal to the axis of rotation, therefore the solid of revolution arising from the rotation of the figure BACBLM around the line EF is to the solid of revolution arising from the rotation of the figure QNPQTX in the ratio compounded from the ratio of the figure BACBLM to the figure QNPQTX and from the ratio of KH to S2 by Proposition 31. But the solid arising from the figure BACBLM rotated around EF is twice the solid arising from the figure BAC rotated around the same EF. Likewise, the solid arising from the figure QNPQTX rotated around the line Y4 is twice the solid arising from the figure QNP rotated around the same Y4. Also, the figure BACBLM is twice the figure BAC and the figure QNPQTX is twice the figure QNP. Since halves are in the same ratio with their own doubles, the solid of revolution arising from the figure ABC rotated around the line EF will be to the solid of revolution arising from the figure NQP rotated around the line Y4 in the ratio compounded from the ratio of the figure ABC to the figure NQP and from the ratio of KH to S2--or DG to RZ--which it was desired to demonstrate.

Let ABC and NQP be any two figures which are rotated around the lines EF and Y4, and let the centers of gravity of the perimeters be D and R, from which perpendiculars DG and RZ to the axis of rotation EF and Y4 are sent down. I say that the ratio of the surface of the solid of revolution arising from the figure ABC rotated around the line EF to the surface of the solid of revolution arising from the figure NQP rotated around the line Y4 is compounded from the ratio of the perimeter ABC to the perimeter NQP and from the ratio DG to RZ.

Let the lines BH and Q2 be drawn parallel to the lines DG and RZ, touching the figures ABC and NQP in B and Q, Let the figures ABC and QNP be conceived to be revolved around BH and Q2, like axes, until in the other part of the axes on the same plane they make the figures of revolution BLM and QTX , equal to themselves, similar, and having completely the same position to the lines BH and EF, and Q2 and Y4. Let O and V be the centers of gravity for the perimeters BLM and QTX. Let the lines OI and V3 be drawn perpendicular to the lines EF and Y4, and let the lines DO and RV be joined, intersecting the lines BH and Q2 in the points K and S. It is manifest that the point K is the center of gravity of the entire perimeter BACBLM. Also, the figure BACBLM is symmetric around the axes BH. In the same manner, the point S is the center of gravity of the entire perimeter of the figure QNPQTX symmetric around the axis Q2. It is also apparent that the lines DG, KH, OI, and likewise RZ, S2, V3 are equal between themselves. Since the figures BACBLM and QNPQTX, are symmetric around the axes BH and Q2 normal to the axes of rotation EF and Y4, therefore the surface of the solid of revolution arising from the rotation of the figure BACBLM around the line EF is to the surface of the solid of revolution arising from the rotation of the figure QNPQTX in the ratio compounded from the ratio of the perimeter of BACBLM to the perimeter of QNPQTX and from the ratio of KH to S2. But the surface of the solid arising from the figure BACBLM rotated around EF is double the surface of the solid arising from the figure BAC rotated around the same EF. Likewise, the surface of the solid of revolution arising from the figure QNPQTX rotated around the line Y4 is double the surface of the solid arising from the figure QNP rotated around the same line Y4. Also the perimeter of BACBLM is double the perimeter of BAC and the perimeter of QNPQTX is double the perimeter of QNP. (That is, if the figures themselves in turn touch in one point. But if they touch in curves, it desirable that there be imagined a little distance between the unions of the figures so that the demonstrated is general.) Since the halves are in the same ratio with their doubles, the surface of the solid of revolution arising from the figure ABC rotated around the line EF will be to the surface of the solid of revolution arising from the figure NQP rotated around the line Y4 in the ratio compounded from the ratio of the perimeter of ABC to the perimeter of NQP and from the ratio of KQ to S2--or DG to RZ--which it was desired to demonstrate.

Let AB be a figure whose center of gravity is C. Let a solid of revolution be made from the rotation of the figure AB around the line DF. I say this solid of revolution is equal to the cylinder whose base is the figure AB and whose altitude is the circumference of the circle in which the center of gravity C is revolved.

Let HGKI be a rectangle whose center of gravity is L.

Let a cylinder be conceived from the rotation of the rectangle HGKI around the side HI. From the centers of gravity C and L let perpendicular lines CE and LO, which are the radii of rotation, be sent down to the axes of rotation DF and HI. It is manifest that the cylinder arising from the rotation of the rectangle HK around HI is equal to the solid whose base is a circle with radius IK and whose altitude is HI--that is, to the solid whose base is the rectangle on IK and the semicircumference of the circle with radius IK, or the entire circumference of the circle with radius OL, and the altitude HI. This is the same as the solid whose base is the rectangle HK and whose altitude is the circumference of the circle on the radius OL. But the cylinder from the rotation of HK around HI is to the solid of revolution from the rotation of AB around DF in a ratio compounded from the proportion of the figure HK to the figure AB and from the line OL to the line EC--or the circumference on the radius OL to the circumference on the radius EC. (See Proposition 33.) But the solid whose base is HK and whose altitude is the circumference on the radius OL--or the cylinder generated by the rotation of the figure HK around HI--is in the same ratio to the solid whose base is AB and whose altitude is the circumference of the radius EC. Therefore, the cylinder from the rotation of the figure HK around HI is to the solid of revolution from the rotation of the figure AB around DF as the same cylinder is to the solid whose base is AB and whose altitude is the circumference of the circle on the radius EC. Consequently, the solid of revolution from the rotation of the figure AB around DF is equal to the cylinder whose base is AB and whose altitude is the circumference of the radius EC, which it was desired to demonstrate.

With the same things having been assumed as in the preceding proposition, let the centers of gravity of the perimeters of the figures HK and AB be the points L and C. I say that the surface of the solid of revolution arising from the rotation of the figure AB and DF is equal to the rectangle whose base is the perimeter of the figure AB and whose altitude is the circumference of the radius EC.

It is manifest that the surface of the cylinder arising from the rotation of the rectangle HK around HI is equal to the rectangle whose base is the circumference on the radius IK and whose altitude is GK together with IK--that is, to the rectangle whose base is the circumference on the radius OL and whose entire altitude is the boundary HG+ GK+ KI + IH of the figure. This is the same as rectangle whose base is the border of the rectangle HK and whose altitude is the circumference on the radius OL. But the surface of the cylinder arising from the rotation of the rectangle HK around HI is to the surface of the solid of revolution generated by the rotation of the figure AB around DF in the ratio compounded from the proportion of the boundary of the figure HK to the boundary of the figure AB and from the proportion of the line OL to the line EC--that is, the circumference on the radius OL to the circumference on the radius EC. (See Proposition 34.) But the rectangle whose base is the boundary of the figure HK and whose altitude is the circumference on the radius OL--that is, the surface of the cylinder arising from the rotation of the rectangle HK around HI--is in the same ratio to the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference of the radius EC. Therefore, the surface of the cylinder arising from the rotation of the rectangle HK around HI is to the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference of the radius EC, as the same surface is to the surface of the solid of revolution arising from the rotation of the figure AB around the axis of rotation DF. Consequently, the surface of the solid of revolution arising from the rotation of the figure AB around DF is equal to the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference on the radius EC, which it was desired to demonstrate.

Let the figure be the same as in the preceding proposition. Above the figure AB (whose center of gravity is C) let a right cylinder be conceived cut by a plane cutting the base AB seminormally in the line DF. I say that its lower trunk is equal to the cylinder above the base AB having altitude EC.

The solid of revolution arising from the rotation of the figure AB around the line DF is equal to the cylinder whose base is AB and whose altitude is the circumference on the radius EC. (See Proposition 35.) But that same solid of revolution is to the lower trunk of the right cylinder above AB which has been sectioned by a plane intersecting the base plane AB seminormally in the line DF as the circumference on the radius on CE is to the radius CE. (See Proposition 23.) But the cylinder whose base is AB and whose altitude is the circumference on the radius CE--or the previously-mentioned solid of revolution--is to the cylinder whose base is AB and whose altitude is EC as the circumference on the radius EC is to EC. Consequently, the lower section of the trunk is equal to the cylinder whose base is AB and whose altitude is EC, which it was desired to demonstrate.

With the same things being assumed as in the preceding proposition, let C be the center of gravity of the perimeter of the base AB. I say that the surface of the lower trunk is equal to the rectangle whose base is the boundary of the figure AB and whose altitude is the line EC.

The surface of the solid of revolution arising from the rotation of the figure AB around DF is equal to the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference of the radius EC. (See Proposition 36.) But that same surface of the solid of revolution is to the surface of the lower trunk of the right cylinder above AB which has been cut by the plane intersecting the base AB seminormally in the line DF as the circumference on the radius CE is to CE. (See Proposition 24.) But the rectangle whose base is the boundary of the figure AB and whose altitude is the circumference on the radius EC--or the surface of the solid of revolution arising from the rotation of the figure AB around DF--is to the rectangle whose base is the boundary of the figure AB and whose altitude is EC as the circumference on the radius EC is to EC. Consequently, the surface of the lower section of the trunk is equal to the rectangle whose base is the boundary of the figure AB and whose altitude is EC, which it was desired to demonstrate.

Above the figure ABC, terminated in one part by the line BC, let there be a right cylinder which, cut by a plane passing through the line BC, results in a lower trunk BCAG whose center of equilibrium on the base is E. From the rotation of the figure ABC around the line BC let there be a portion of a solid of revolution. I say that the center of gravity of that portion is the same as the center of gravity of the arc of the circle in which the point E is revolved.

If they are not the same, let the line β be the difference between them. Let a solid polyhedron be imagined circumscribing that portion of the solid of revolution and consisting of 2, or 4, or 16, or 32, or however many (as long as it is a number in the progression of doubles from two) equal lower trunks cut by a plane having the line BC as the common intersection with the base. In that same portion of the solid of revolution let there be inscribed another polyhedron similar to the first in such a way that their centers of gravity differ by an interval less than the line β. Indeed, this is able to be done since the more sides the polyhedra have the closer they are to the portion of the solid of revolution ad infinitum. Consequently, through the center of equilibrium E let a plane be drawn normal to the line BC and: making a common intersection with the trunk the triangle HFD, which is right at F; likewise making a common intersection with the portion of the solid of revolution the sector of the circle IVLO, with the circumscribed polyhedron a regular polygon circumscribed around the sector; likewise with the inscribed polyhedron a similar portion of a regular polygon inscribed in the same sector.

It is manifest that the radius IV is equal to the line DF. Let IY9 be the sector of the circle with radius IY equal to the line DE, and let the polygon TS8 similar to the polygon KMN circumscribe the sector IY9. Likewise let the inscribed polygon Y79 be similar to the circumscribed polygon. Let the line IL be drawn cutting the four sides of the similar polygons 7Y, ST, PV and MK in half in the points γ, R, Q, and L. It is manifest that the triangle ILK, right at L, is the common section of the cutting plane with one of the equal trunks out of which the polyhedron circumscribing the portion of the solid of revolution consists.

Let us suppose that this trunk is the same as the trunk ABCG (which we are able to do without absurdity, since all of the lower trunks whose cutting plane cuts the base in BC, have the same center of equilibrium in the base). Consequently, it is apparent that the center of equilibrium of the trunk having common intersection LIK with the cutting plane is the point R. In the same manner it is proved that the point R is the center of equilibrium in the base for the trunk whose intersection with the cutting plane is MIL.

Therefore, the center of gravity of these trunks joined together is the point R. In the same manner it is proved that, in any two trunks on the entire circumscribed polyhedron joined at one surface, the center of gravity of both together is in the point where the line from the vertex I cutting the base of the triangle (which is the common section of cutting plane with the two joined trunks) intersects the arc of the circle 9Y, cutting it in half. It is also proved easily that the point γ is the center of gravity of the two joined trunks whose common intersection with the cutting plane is the triangle PIV. Likewise, the point X is the center of gravity of the two joined trunks whose common intersection with the cutting plane is the triangle φIP. Let the line Xγ, which is cut in half in the point 5, be joined. It is manifest that the point 5 is the center of gravity of the four joined together trunks whose common intersection with the cutting plane is the trapezoid IVPφ. But the point 5 is also the center of gravity of the lines Z7 and 7Y; together. In the same manner it is proved that α is the center of gravity of the four joined together trunks whose common intersection with the cutting plane is the trapezoid IφδO and also of the lines 9ν and νZ together. Let the centers of gravity be joined in the line α5 which is cut in half by the line IZ in the point 3. It is manifest that the center of gravity 3 of the inscribed polyhedron is also the center of gravity of all the inscribed lines of the arc 9Y. In the same manner it is proved that the center of gravity 1 of the circumscribed polyhedron is the same as the center of gravity of all of the circumscribed lines together on the arc 9Y. Let the center of gravity of the portion of the solid of revolution be 4 and the center of gravity of the arc 9Y be 2, which is necessarily on the line IZ between the points 1 and 3. Let the lines 14, 24, and 34 be joined. By construction, 13 is less than 24. Of the angles 124 and 423, let 124 be the greater or certainly not less. Therefore, the side 14 is more than the side 24. But 24 is more than 13. Consequently, 14 is more than 13. That is, the center of gravity of the portion of the solid of revolution is at a distance from the center of gravity of the polyhedron circumscribed in it in an interval more than the center of gravity of of its circumscribing polyhedron is distant from the center of gravity of its inscribing polyhedron, which is absurd. Indeed, the centers of gravity of the inscribed and circumscribed polyhedra are at a greater distance than the centers of gravity of the portion of the solid of revolution and the polyhedron of the circumscribed. Consequently, the centers of gravity of the portion of the solid of revolution and the arc of the circle 9Y are distant by no interval. Therefore, their center of gravity is the same, which it was desired to demonstrate.

With the same things be supposed as in the preceding, let the center of equilibrium of the surface of the trunk BCAG in the base be E. I say that the center of gravity of the surface of portion of the solid of revolution is the same as the center of gravity of the circular arc in which the point E is revolved.

If they were not the same, there would be a line β which is the difference between them. Let it be conceived that a solid polyhedron, consisting of 2, or 4, or 8, or 16, or however many(as long as the number is in the progression of doubles from two) lower trunks equal to the proposed right cylinder cut by a plane having common intersection with the base line BC. And in this same portion of the solid of revolution let be inscribed another polyhedron similar to the previous one in such a way that the centers of gravity of those surfaces are at a distance of an interval less than β. Indeed, this is possible since the more sides such polyhedra have the less is the discrepancy between them ad infinitum, thus so that their distance is able to be less than any proposed quantity. Consequently, through the center of equilibrium E let a plane normal to the line BC be drawn making with the trunk ABCG the common intersection the triangle HFD, right at F, likewise with the portion of the solid of revolution a common intersection the sector IVLO of the circle, with the polyhedron circumscribed to the portion a regular polygon circumscribing the sector, likewise with the inscribed polyhedron a similar portion of a regular polygon inscribed in the same sector. It is manifest that the radius IV is equal to the line DF. Let the sector of the circle Iγ9 with radius Iγ be equal to the line DE. Let the polygon TS8I similar to the polygon IKMN be circumscribed around the sector Iγ9, likewise let the polygon Iγ79 be inscribed similar to the circumscribed. Let the line IL be drawn dividing the four parallel sides of the similar polygons 7γ, ST, PV, and MK in half in the points Y, R, Q, and L. It is manifest that the triangle ILK, right at L, is the common section of the cutting plane with one of the equal trunks from which the polyhedron circumscribing the portion of the solid of revolution consists. Let us suppose that this trunk is the same as the trunk ABCG (which we are able to do without absurdity since all of the surfaces of the lower trunks whose cutting planes cut the base in the line BC have the same center of equilibrium E in the base). Consequently, it is apparent that the center of equilibrium of the base of the trunk, having common intersection LIK with the cutting plane, is the point R. In the same manner it is proved that the point R is the center of equilibrium in the base of the surface of the trunk whose intersection with the cutting plane is MIL. Therefore, it is proved in the same manner that the center of gravity of the surface of those trunks joined together in any two trunks of the entire polyhedron circumscribed to one surface is at the point where the line from the vertex I cutting the base of the triangle (which is the common section of the cutting plane with the two joined trunks) in half intersects the circular arc 9γ. It is also prove easily without a problem that the point Y is the center of gravity of the surface of two joined trunks whose common intersection with the cutting plane is IPV. Likewise, the point is the center of gravity of the surface of the two joined trunks whose common intersection with the cutting plane is the triangle φIP. Let the line XY, which is cut in half by the point 5, be joined. It is manifest that the point 5 is the center of gravity of the surface of the four trunks joined together, whose common intersection with the cutting plane is the trapezium IVPφ. The point 5 is also the center of gravity of the lines Z7 and 7φ together. In the same manner it is proved that the α center of gravity of the lines 9ν and νZ is the center of gravity of the four surfaces of the trunk joined together, whose common intersection is the trapezium IφδA. Let the centers of gravity α and 5 be joined by the line α5, which is cut in half at 3. It is manifest that the center of gravity 3 of the inscribed polyhedron of the surface is also the center of gravity of all the lines inscribed in the arc 9γ. In the same manner it is proved that the center of gravity 1 of the surface of the circumscribed polyhedron is the same as the center of gravity of all of the lines circumscribed around the arc 9γ. Let the center of gravity of the surface of the portion of the solid of revolution be 4 and the center of gravity of the arc 9γ be 2, which is necessarily in the line IZ between the point 1 and 3. Let the lines 14, 24, and 34 be joined. By construction 13 is less than 24. Of the angles 124 and 423, let 124 be more or not even less. Therefore, the side 14 is more than the side 24, but 24 is more than 13. Consequently, 14 is more than 13. That is, the center of gravity of the surface of the portion of the solid of revolution is at a distance from the center of gravity of the surface of the polyhedron circumscribing it in a greater interval than the center of gravity of the surface of its circumscribing polyhedron is from the center of gravity of the surface of the similar inscribed polyhedron, which is absurd. Indeed, the centers of gravity of the inscribing and circumscribing polyhedrons are at a greater distance than the center of gravity of the surface of the portion of the solid of revolution is from the center of gravity of the surface of the polyhedron either inscribed or circumscribed in it. Consequently, the centers of gravity of the surface of the portion of the solid of revolution and of the arc 9δ are distant by no interval. Therefore, their center of gravity is the same, which ought to have been demonstrated.

Let a right cylindrical figure conceived above any base KECD be cut by a plane having the line AB as the common intersection with the base. Let the center of equilibrium in the base of the lower trunk be denoted as the point F. From the rotation of the figure KECD around the line AB let there be a portion of a solid of revolution. I say the center of gravity of this is the same as the center of gravity of the circular arc in which the point F is revolved.

Let the lines CA and KB be joined as you please. Suppose that the right cylinder above the base ACDKB is cut by the above-mentioned plane through the line AB. Suppose also that the center of equilibrium of the lower trunk in the base is the point H, and, likewise, that the center of equilibrium in the base of the trunk above the figure ACEKB is the point G. From the rotation of the figure ACDKB around the line AB (in such a way that its extreme part--namely, the figure CDKE--describes the portion of the previously-mentioned solid of revolution) let there be a portion of the solid of revolution whose center of gravity is M. It is manifest that the portion of the solid of revolution arising from the rotation of the figure ACDKB is equal to the portion of the solid of revolution arising from the rotation of the figure CDKE (whose center of gravity is L) and the solid of revolution arising from the rotation of the figure ACEKB (whose center of gravity is N)). Likewise, the trunk above the figure ACDKB is equal to the trunk above the figures CDKE and ACEKB. Consequently, it is clear that the points F, H, and G are in one line and, likewise, FH is to HG as the trunk above ACEKB is to the trunk above CDKE. Likewise, the points L, M, and N are in one line, and LM is to MN as the portion of the solid of revolution arising from the figure ACEKB is to the portion of the solid of revolution arising from the figure CDKE. But the portions of the solids of revolution between themselves are in a direct ratio to the trunks, as is easily gathered from Proposition 23. Therefore, as FH is to GH, thus LM is to MN. Through the points F and L let the line FLO be drawn, and through the points H and M let the line HMP be drawn. Likewise, through the points G and N let the line GNQ be drawn. It is sufficiently clear that lines HMP and GNQ (since they are radii of circles in whose circumferences the points H and G are revolved) and, therefore, also the line FLO are parallel among themselves and normal to the line AB. Therefore, as PM is to PH--or QN is to QG--thus OL is to OF. But M and N are centers of gravity of similar circular arcs in which the points H and G are revolved around the centers P and Q. Therefore, L is also the center of gravity of the similar circular arc in which the point F is rotated around the center O. But L is also the center of gravity of the portion of the solid of revolution arising from the rotation of the figure CDKE around the line AB, which it was desired to demonstrate.

Let ABO be a curve or curves which are assumed to be revolved around the axis IK and let these curves be cut in two points by the radius of rotation. Above the curves of ABO imagine the surface of a cylinder perpendicular to the base, which is cut by a plane also cutting the base in the line IK. Let the center of equilibrium in the base of the surface of the lower trunk be E. From the rotation of the curves ABO, let there be a portion of a surface of revolution whose center of gravity is F. I say that F is the same as the center of gravity of the circular arc in which the point E is revolved.

Let B be the point in which the radius of rotation touches the curves ABO. Let the center of equilibrium in the base of the portion of the surface of the trunk above the curve or curves AB be C and the center of equilibrium in the base of the portion of the surface of revolution arising from AB be D. Let the center of equilibrium in the base of the surface of the trunk above OB be G and the center of equilibrium of the portion of the surface of revolution arising from the same line be H. It is manifest that the points C, E, and G are in the same line. Likewise, CE is to EG as the surface of the trunk above OB is to the surface of the trunk above AB. In the same manner it is clear that the points D, F, and H are in the same line. Likewise, DF is to FH as the portion of the surface of revolution arising from OB is to the portion of the surface of revolution arising from AB. But the portions of the surfaces of revolution are between themselves in the direct proportion of the surfaces of the trunks, as is gathered from Proposition 24. Therefore, as CE is to EG, thus DF is to FH. Let the lines CD, EF, and GH be extended until they cut the line IK in the points L, M, and N. Since the points D and H are centers of gravity of the circular arcs in which the points C and G are revolved, it is manifest that the lines CDL and GHN are normal to the axis of rotation IK and parallel among themselves. Therefore, the line EFM is normal to the same IK, since as CE is to EG thus DF is to FH. (See Proposition 40.) Consequently, as LD is to LC--or NH is to NG--thus MF is to ME. But D and H are centers of gravity of the similar circular arcs in which the points C and G are rotated. Therefore, the point F is also the center of gravity of the similar circular arc in which the point E is rotated, which ought to have been demonstrated.

The figure ABSO is analogous proportionally to the curve AQ. (See Proposition 2.) Therefore, the centers of gravity of the figure ABSO and the curve AQ are distant by the same interval from the line AB. Let this interval be K. The centers of gravity of the line AO and of the rectangle ABRO are also apart by the same interval from the line AB, namely BR/2. But the figure ABSO is to the rectangle ABRO as the curve AQ is to the line AO. (See Proposition 2.) Therefore, by drawing the first and third in the circumference whose radius is K and the second and forth in the circumference whose radius is BR/2, the cylinder whose base is ABSO and whose altitude is the circumference on the radius K--namely, the solid of rotation arising from the rotation of the figure ABSO around AB--is to the cylinder whose base is ABRO and whose altitude is the circumference on the radius BR/2--namely, the cylinder arising form the rotation of ABRO around AB--just as the rectangle whose base is AQ and whose altitude is the circumference of the radius K--namely, the surface generated by the rotation of AQ around AB--is to the rectangle whose base is AO and whose altitude is the circumference of the radius BR/2--namely, the circumference from the radius AO--which ought to have been demonstrated. (See Propositions 35 and 36.)

Let FH be any line parallel and equal to AB and cutting the figure BCE in G. Let the lines FH and GH be bisected at the points O and P. It is manifest that the moment of the line FH is to the moment of the line GH in a ratio compounded from the ratio of FH to GH and from the ratio of OH to PH. That is, in the duplicate ratio of FH to GH--or as the circle on the radius FH is to the circle on the radius GH. Since this always happens thus, both the first and third are always the same.

As the moments of all FH's--namely, the moments of the entire rectangle ABED--are to the moments of all GH's--namely, the moment of the figure BCE--thus all of the circles on the radius FH--namely, the cylinder from the rotation of the rectangle ABED--are BE are to all the circles on the radius GH--namely, the solid of revolution arising from the rotation of the figure BCE around the same BE--which it was desired to demonstrate.

Let the centers of gravity of the semicircles on FH and GH be O and P. It is manifest that the moment of the semicircle FH is to the moment of the semicircle GH in the ratio compounded from the ratio of the semicircle FH to the semicircle GH--namely, from the duplicated ratio of FH to GH--and from the ratio of OH to PH--or FH to GH. Therefore, the moment of the semicircle FH is to the moment of the semicircle GH in the triple ratio of FH to GH--or as the cube from FH is to the cube on GH. Since it always turns out thus, both the first and the third are always the same. As the moments of all of the semicircles FH--namely, the moment of the semisolid BADE itself--are to the moments of all of the semicircles GH--namely, the moment of the semisolid BCE itself--thus are all of the cubes from FH to all of the cubes from GH, which it was desired to demonstrate.

Let a parabolic conoid be generated by the revolution of a semiparabola ABC around the axis AB. It is desired to find a circle equal to this surface. Let AB be extended to D so that AD is a forth part of the latus rectum, and let the line AE equal to half of the latus rectum be tangent to the parabola at the vertex A. From the axis DB let a parabola DEK be drawn with the ordinate BC extended to K and let the line X be the diagonal of a square equal to the segment AEKB of the parabola. I say that the line X is the radius of a circle equal to the surface of the parabolic conoid generated from the rotation of the semiparabola ACB around the axis AB.

From any point on the parabolic curve AC--namely, M--let the ordinate MG, which is extended until it intersects the exterior parabolic curve DEK in N, be dropped down to the axis. Let GH be a line equal to half of the latus rectum. It is apparent from the demonstrations of many that the line MH touching the curve in the point M cuts it normally. Consequently, from the doctrine of asymptotes of parabolas passed down by Gregory of St. Vincent and others, it is manifest that the parabolas ABC and DBK are asymptotic.

Therefore, (from this same doctrine) the square of the line GN is equal to the squares of the lines GM and AE--that is, to the squares of the lines GM and GH. Moreover, the square of the line MH is equal to these squares. Therefore, the lines HM and GN are equals. Since this will always be the case in all points of the parabolic curve AMC, it is manifest from Proposition 3 that the parabolic segment AEKB is equal to the surface of the trunk above the curve AMC cut by a plane passing through the line AB and inclined in a semi-right angle with the parabola--or the base of the trunk. Therefore, since the square with sides the line X is the double of the surface of the trunk, X will be the radius of a circle equal to the surface of the conoid generated from the rotation of the parabolic curve AMC around the axis AB, which ought to have been demonstrated. (See Proposition 25.)

Let there be an oblong spheroid arising from the revolution of a semiellipse EFT around the longer axis ET. It is desired to find a circle equal to the surface of this spheroid. Let the lines TR and ED equal to half of the latus rectum touch the ellipse in the vertices T and E. Next, let an ellipse DFR with center G and vertex F be drawn. Let the square whose diagonal is X be equal to the elliptical segment DFRTE. I say that X is the radius of a circle equal to the surface of the proposed oblong spheroid.

Let the ellipse DFR be extended until it intersects the extended axis ET in the points Z and B. It is easily apparent that the lines GZ and FG are the conjugate semiaxes of the semiellipse BFZ. Let the lines BA, EC, TQ, and ZY, all equal to the line FG, be drawn touching the semiellipses in the vertices B, E, T, and Z. Let the line ACFQY be joined. Also, let GQ, GR, and GY, which cuts QT in S, be drawn. From any point N of the given semiellipse let NP be perpendicular to the axis ET. This should be extended so that it cuts the line AY in I and the lines GY, GR, and GQ in the points O, M, and L, and the ellipse BFZ in K. Let the line PH be equal to the line PM and let HN be joined. It is manifest (from the van Schooten commentaries on Descartes, page 214) that HN cuts the tangent to the ellipse normally in the point N.

Because FRZ is an ellipse and FGZY is a rectangle on the semiaxis whose diameter is GY, the square of the line QT is equal to the squares of the lines RT and TS. Since the lines LP, MP, and OP are in the same ratio as the lines QT, RT, and ST, the square of the line LP will be equal to the squares of the lines MP and OP. But the square of the line IP is equal to the squares of the lines LP and NP on account of the above-mentioned property of ellipses. Therefore, the square of the line IP is equal to the squares of the lines MP, NP, and OP. But the square of the line IP is also equal to the squares of the lines OP and KP. Consequently, the squares of the lines MP, NP, and OP are equal to the squares of the lines KP and OP and, by taking away equals, the square of the line KP is equal to the squares of the lines MP and NP--or HP and NP . But the square of the line HN is equal to those same squares. Therefore, the lines HN and KP are equals. Since this will always be the case in all points of the elliptical curve EFT, it is manifest from Proposition 3 that the elliptical segment EDFRT is equal to the surface of the trunk above the curve EFT cut by a plane passing through the line ET and inclined in a semi-right angle toward the base. Therefore, since the square with sides equal to the line X is double the surface of the trunk, X will be the radius of a circle equal to the surface of a spheroid generated by the rotation of the curve EFT around the axis ET, which it was desired to demonstrate.

Let there be an oblate spheroid generated from the rotation of the semi-ellipse MCR around the smaller axis MR. It is desired to find a circle equal to the surface of this. Let the lines MA and RD, equal to half of the latus rectum, touch the ellipse in the vertices M and R. Next, let a hyperbola ACD with vertex C and center P be drawn through the points A and D. Let the square whose diagonal is the line X be equal to the hyperbolic segment ACDRM. I say that X is the radius of a circle equal to the surface of the proposed oblate spheroid.

Let PF be the hyperbolic asymptote cutting the line DR in F, let BCG be a line touching the hyperbola at the vertex, and let the lines PD and PG be joined. From any point I on the ellipse let IQ, cutting the lines CG, PD, PF and PG in the points H, K, O, and L and likewise intersecting the hyperbola in E, be perpendicular to the axis MR. Let the line QN be equal to the line QK. It is manifest (from the commentary of van Schooten on Descartes, page 214) that the line NI, touching the ellipse in the point I, cuts it normally. Since PF is a hyperbolic asymptote and CG is a line touching the hyperbola in the vertex and since GR is equal to the half-axis CP, the square on the line DR will be equal to the squares on the lines RF and RG. And since the lines KQ, OQ, and LQ are proportional to the lines DR, FR, and GR, the square of the line KQ will be equal to the squares on the lines OQ and LQ.

But the square on the line EQ is equal to the squares on the lines HQ and OQ, and the square on the line HQ is equal to the squares on the lines IQ and LQ. Therefore, the square of the line EQ is equal to the squares on the lines IQ, LQ and OQ--that is, to the squares on the lines IQ and KQ--or IQ and QN. But the square on the line IN is equal to those same squares. Therefore, the lines EQ and IN are equal. And since this will be the case at all the points of the elliptical curve MCR, it is manifest from Proposition 3 that the segment of the hyperbola ACDRM is equal to the surface of the trunk above the elliptical curve MCR cut by a plane passing through the line MR and inclined in a semi-right angle with the base of the trunk. Therefore, since the square with sides equal the line X is the double of the surface of the trunk, X will be the radius of the circle equal to the surface of the oblate spheroid generated by the rotation of the elliptical curve MCR around the axis MR, which ought to have been demonstrated.

Let there be a hyperbolic conoid generated by the revolution of the semi-hyperbola FST around the axis FS. It is desired to find a circle equal to this surface. Let the line FH equal to half of the latus rectum touch the hyperbola in the vertex F. Next let the hyperbola, passing through the point H and having the same center--namely D--and the same conjugate semiaxis--namely DE--as the proposed hyperbola FST, be described. Let the described hyperbola be BHR, cutting the extended ordinate ST in R. Let the square whose diagonal is X be equal to the hyperbolic segment FHRS. I say that X is the radius of a circle equal to the surface of the proposed hyperbolic conoid.

Let the asymptote of the proposed hyperbola be DV, cutting the extended line FH in K, and let DY be the asymptote of the created hyperbola, cutting the extended FH in C. Let the line DH be extended. From any point Q of the proposed hyperbola let the line QG, which intersects the extended lines DH, DV, and DY in the points Z, O, and L, and the created hyperbola in the point P, be an ordinate from the axis. Let GA be equal to the line GZ. It is manifest (from the van Schooten commentaries on Descartes, page 216) that the line QA touching the hyperbola in the point Q cuts it normally. Since the line FK, touching the hyperbola in the vertex, is extended to its asymptote in the point K, FK will be equal to the conjugate semiaxis DE. Because BHR is a hyperbola and DY its asymptote, the square on the line FC is equal to the squares on the lines FH and DE--that is, to FH and FK.

And since the lines GZ, GO, and GL are in the same ratio with the lines FH, FK, and FC, the square on the line GL will be equal to the squares on the lines GO and GZ. But because FQT is a hyperbola and DV its asymptote, the square on the line GO is equal to the squares on the lines GQ and DE. Therefore, the square on the line GL will be equal to the squares on the lines GZ, GQ, and DE. But the square on the line GL is also equal to the squares on the lines GP and DE. Therefore, by taking away equals, the square on the line GP is equal to the squares on the lines GZ and GQ--that is, to the squares on the lines GA and GQ. Moreover, the square on the line AQ is equal to these squares. Therefore, the lines GP and AQ are equal. And since this will always come about at all the points in the hyperbolic curve FQT, it is manifest from Proposition 3 that the hyperbolic segment FHRS is equal to the surface of the trunk above the hyperbolic curve FQT cut by a plane passing through the line FS and inclined in a semi-right angle with the base of the trunk. Therefore, since the square with side the line X is the double of the surface of the trunk, X will be the radius of the circle equal to the surface of the hyperbolic conoid generated by the rotation of the hyperbolic curve FQT around the axis FS, which ought to have been demonstrated.

Let αC be a hyperbolic curve which generates a surface when rotated around its own conjugate axis FZ. It is desired to find a circle equal to this surface. Let Z be the center, α the vertex, and ZD an asymptote of the hyperbolic curve. Let the line αK, equal to half of the latus rectum of the same hyperbola, touch the hyperbola in the vertex α. Let HK be a line parallel and equal to the line Zα which, extended indefinitely, cuts the hyperbola in R and its asymptote ZD in M. Suppose HV is a a segment of this line whose square is equal to the squares on the lines HR and HK. Next, let αVA be a hyperbola with center Z and vertex α which is extended until it intersects the line FCA parallel to the line Zα at the point A. Let the square whose diagonal is X be equal to the hyperbolic segment FAVαZ. I say that X is the radius of the sought after circle.

Let ZB, cutting the line HV in L, be the asymptote of the created hyperbola. Let TI and Tβ, perpendicular to each axis (extended if necessary), be dropped down out of a given point T in the original hyperbola. Let Gγ be a line cutting the hyperbolic curve normally in the point T and cutting each axis (extended if necessary) in the points G and γ. Let the lines Tβ and TI be extended (if necessary) and intersect the line ZKY in the points Y and O. It is manifest from the place cited in van Schooten that the lines βY and βγ are equal.

Moreover, as βγ is to βT--or EO--thus IT--or βZ--is to IG. Therefore, as Yβ is to OE, thus Zβ is to IG. But as Yβ is to OE, thus Zβ is to ZE--or IO. Therefore, the lines GI and IO are equal.

Let the line IT be extended so that it intersects the created hyperbola in Q and its asymptote in P, and let its intersection with the asymptote of the original hyperbola be the point N. By construction, the square on the line HV (that is, the square on the lines HL and Zα) is equal to the squares on the lines HR and HK--or the squares on the lines HM, HK, and Zα. Therefore, by taking away the same on each side, the square on the line HL equal to the squares on the lines HK and HM remains. And since the lines IP, IN, and IO are in the same ratio with the lines HL, HM, and HK, the square on the line IP will also be equal to the squares on the lines IN and IO. Therefore, by adding the square on Zα to each, the squares on the lines IP and Zα (that is, the square on the line IQ) is equal to the squares on the lines IO, IN and Zα--that is, to the squares on the lines IO and IT (or IG and IT)--that is, to the square on the line TG. Therefore, the lines QI and TG are equal. And since this always occurs in all the points of the hyperbola αTC, it is manifest from Proposition 3 that the hyperbolic segment FAαZ is equal to the surface of the trunk above the hyperbolic curve αTC cut by a plane passing through the line FZ and inclined in a semi-right angle with the base of the trunk. Therefore, since the square on the line X is double the surface of the trunk, X will be the radius of a circle equal to the surface arising from the rotation of the hyperbolic curve αTC around the conjugate axis ZF, which ought to have been demonstrated. (See Proposition 25.)

Let GN be a parabolic curve having the vertex G and the axis GO. It is desired to find a line equal to this. Let the axis GO be extended to A so that GA is equal to half of the latus rectum, and let GK be a line touching the parabola in the vertex and intersecting the line NK parallel to the axis in K. Next, let GD be a line dividing the right angle AGK in half and let AC be a hyperbola having vertex A, center G, and asymptote GD. Let the line NK be extended until it cuts the hyperbola in the point C, and let AE be parallel to the line GK. Finally, as the rectangle AEKG is to the segment AGKC of the hyperbola, thus let the line GK be to the line X. I say that the line X is equal to the parabolic curve GN.

From any point M on the parabolic curve let ML and MH be dropped down perpendicular to the lines GO and GK, and let OI be a line cutting the parabolic curve normally in the point M and intersecting the lines GO and HK in the points O and I. It is manifest that the line LO is equal to half of the latus rectum, or GA. Let the line MH be extended so that it cuts the hyperbola in the point B and the asymptote GD in F.

It is also manifest, because GD is an asymptote, that the square on the line HB is equal to the squares on the lines AG and HF or, because the angles HGF and HFG are equal, to the squares on the lines LO and GH--or LO and LM. But the square ob the line OM is equal to the squares ob the lines OL and LM. Therefore, the lines OM and HB are equal. Next, because of the similarity of the triangles HMI and LOM, as HM is to MI, thus OL is to OM--or GA to HB. And since this will always be the case in all the points of the parabolic curve GN, by Proposition 2 the rectangle AEKG will be to the segment of the hyperbola ACKG as the line GK is to the parabolic curve GN. Moreover, as the rectangle AEKG was to the hyperbolic segment ACKG, thus the line GK was to the line X. Consequently, the line X is equal to the curve GN, which it was desired to demonstrate.

Let the parabolic curve IED, whose vertex is D, be rotated around the line BD touching it in its vertex so that a surface of revolution is generated. It is desired to find a circle equal to this surface. Let the latus rectum of the parabolic curve be quadruple the line DR, and let its axis be DK. From the point I let IK be a line perpendicular to the axis DK and let DGL, cutting the extended line IK in L, be a hyperbola with transverse axis RD (to which the conjugate axis is also equal). Next, let a square whose diagonal is X be equal to the semi-hyperbola DKL. I say that X is the radius of the sought after circle.

From any point E on the parabolic curve let a line EF, which intersects the extended hyperbola in the point G, be dropped down perpendicular to the axis. Let EA be a line tangent to the parabola at the point E and intersecting the tangent DB in C and the extended axis at A.

Let CH be equal and parallel to the line DF. It is manifest that FA is double the line DA. Therefore, EF is double the line CD. Therefore, EH is equal to the line HF. The rectangle formed by RF and FD is equal to the square on the line DF together with the rectangle RDF. But the rectangle RDF is equal to the square on EH, since RD is a quarter part of the latus rectum. Therefore, the square on the line CE (which is equal to the squares on the lines EH and HC = DF) is equal to the rectangle formed by RF and FD. But the square on the line FG is equal to the same rectangle. Consequently, the lines EC and FG are equals between themselves. And since it is always this way, it is apparent from Proposition 4 that the semi-hyperbola DKL is equal to the surface of the trunk above the parabolic curve DEI cut by a plane passing through the line DB and inclined at a semi-right angle with the base of the trunk. Therefore, since the square on the line X is double the surface of the trunk, X will be the radius of a circle equal to the surface generated by the rotation of the parabolic curve around the line DB, which ought to have been demonstrated.

Let the base of the cylinder be a circle BGQG whose diameter is BQ, and let the line BA touching the circle at A be an altitude of the semi-cylinder equal to the line BQ. In the same way indeed it is as all of the semicylinders above the same base are analogous, which are extended indefinitely to a part B. Let the line PQ be indefinitely long and parallel to it. Also, let AP, CR, and DT be parallel to the diameter BQ and let the line AQ be drawn with the remaining things having themselves as in the figure. Let the figure CRM be of such a nature that (with any line EYKM drawn parallel to the line PQ) EY is to FY as GY is to KM. The figure CRM is proportional in magnitude and gravity to the semi-cylinder. But the center of equilibrium of this figure--or of the semi-cylinder--is found thus: Let CR and CS be equal with the remaining things having themselves as in the figure, and let the figure DOT be of such a nature that (with any line EYKMNO having been drawn) EY is to KL as KM is to NO. And let the rectangle DV circumscribe the figure DOT. Next, as the figure CRM is to the figure DOT thus EY--or RC--is to CZ. It is manifest from Proposition 37 that Z is the center of equilibrium of the figure CMR. Moreover, the figure DOT becomes known in the same manner. The rectangle DV is given, since DT is equal to BQ and DX--or TV--is the fourth part in a continuing proportion whose first is EY and whose second is half the line BQ.

Consequently, NO is always the fourth in a continuing proportion whose first is EY and second is GY, which I prove thus: From the preceding things the following proportions are manifest:

EY: FY = QY :: GY : KM
EY: KL = BY :: KM . NO
EY2 : BY x YQ = GY2:: KM x GY : KM x NO

and therefore,

EY² : GY² :: GY : NO.

and therefore,

GY³ = EY² x NO. Therefore, with EY supposed first and GY supposed second, NO will be the fourth in the continuing proportion. Therefore, the figure DOT is to the rectangle DV as the parabolic conoid (whose base is the circle BGQG) is to the parabolic cylinder circumscribing the conoid. And since the proportion of the parabolic conoid to the cylinder circumscribing it and the rectangle DV are given, the figure DOT will also be given. Therefore the point Z will also be given. Therefore, the center of equilibrium in the base of the semi-cylinder is given. Consequently, the center of gravity of the semi-cylinder will be on the given perpendicular to the base erected from the center of equilibrium. But the same center of gravity is on the line Qα, drawn between the point Q and the midpoint α of the line AB from the point perpendicular to the base BGQ, with it having been supposed of course that the cylinder is cut by a plane cutting the base plane in the line PG. Therefore, if as RC is to CZ, thus Qα is to αδ, δ will be the center of gravity of the semi-cylinder, which ought to have been found.

Let the line IB be extended to G and (with a line CF drawn tangent to the curve ACI at a given point C and with the parallels EC and FH having been drawn) let the curve AHB be drawn of such a nature that the line CH is equal to and parallel to the line EF. It is manifest from Proposition 7 that the line AE is to FE--or LC is to HC--as P is to Q. And since it will always be this way, it is evident that the figure ACIB is to the figure ACIGH as P is to Q. But from Proposition 11 the figure ACIGH is equal to the figure ACIK. Therefore, the figure ACIB is to the figure ACIK as P is to Q. By componendo, the parallelogram ABIK is to the figure ACIk as P + Q is to Q, which ought to have been demonstrated.

If P is less than Q, ACIK will be one of the infinite parabolas. If indeed P is more than Q, ACIK will be one of the infinite curves.

From this proposition it is also manifest that the parallelogram ABIK is to the figure ABIC as P + Q is to P, which is also understood thusly: since the ratio of AB to EC is multiplied with the ratio of AK to AE in the ratio of P to Q--that is, the ratio AB to AL is multiplied with the product of the ratio BI to LC in the same ratio P to Q--by convertendo, BI to LC will be multiplied with the ratio of AB to AL in the ratio of Q to P. Therefore, from the proposition itself, the parallelogram ABIK is to the figure ABIC as P + Q is to P, which ought to have been demonstrated.

From the preceding proposition, ABIK is to ACIK as P + Q is to Q. In the same manner, ALCE is to ACE as P + Q is to Q. Therefore, ABIK is to ACIK as ALCE is to ACE. By permutando, ABIK is to ALCE as ACIK is to ACE. But the ratio of ABIK to ALCE is compounded from the ratio of AK to AE and the ratio of IK to CE. Consequently, the ratio of ACIK to ACE is compounded from the same ratios. But the ratio of IK to CE is multiplied with the ratio of AK to AE in the ratio of P to Q. By convertendo, the ratio AK to AE is multiplied with the ratio IK to CE in the ratio Q to P. By componendo, the ratio compounded from the ratios of AK to AE and IK to CE (namely, the ratio of the figure ACIK to the figure ACE) is multiplied with the ratio IK to CE in the ratio P + Q to P, which it was desired to demonstrate.

Suppose that the plane cutting the cylinder above ACIK is extended so that it also cuts the parallelpiped. It is manifest that its lower trunk is the triangular prism half of the parallelpiped. Next, let AMTK be a rectangle with an inscribed curve AT with the property that (with any point E having been assumed and the line DEO drawn perpendicular to AK and cutting the curves ART and ACI in the points R and C and likewise with a plane normal to AK having been drawn through the point E and cutting the lower cylinder of the trunk in the isosceles right triangles whose bases are DE and CE) OE is to RE just as the triangle above DE is to the triangle above CE. It is apparent that the triangle above DE is to the triangle above CE--or OE is to RE--in the duplicate ratio of DE to CE. But the ratio of DE to CE is multiplied with the ratio AK to AE in the ratio of P to Q. Therefore, the ratio of OE to RE is multiplied with the ratio of AK to AE in the ratio of 2 P to Q. And since this will always be the case wherever the point E is assumed, it is clear from Proposition 54 that the rectangle AMTK is to the figure ARTK as 2 P + Q is to Q. But OE is always to RE as the triangle above DE--namely, the common intersection of the plane normal to AK with the prismatic triangle--is to the triangle above CE--namely, the common intersection of the preceding plane with the lower trunk of the cylinder above ACIK. Also, the preceding quantities--namely, all the lines OE among themselves and all the triangles above DE among themselves--are equal among themselves. Therefore, as all of the lines OE--namely, the rectangle AMTK--are to all the lines RE--namely, the figure ARTK--thus all of the triangles above DE--namely, the prismatic triangle--are to all the triangles above CE--namely, the lower trunk of the cylinder above ACIK. Consequently, the prism is to the trunk as 2 P + Q is to Q. Therefore, the double of the prism--namely, the parallelpiped above ABIK--is to the lower trunk of the right cylinder ACIK as 4 P + 2 Q is to Q, which ought to have been demonstrated.

Let AMTK be a square with an inscribed curve ART of such a nature that (with any point E having been assumed and the line DEO drawn perpendicular to AK and cutting the line ART in the point R and likewise with a plane drawn through E normal to AK whose intersection with the lower trunk is the rectangle LAEC) AE is to RE just as BA is to CE. It is clear from the construction that the ratio of AE to RE is multiplied with the ratio of AK to AE in the ratio of P to Q. By componendo, the ratio of AK to RE--that is, the ratio of KT to RE--is the product of the ratio of AK to AE with the ratio of P + Q to Q. Therefore, the square AMTK is to the figure ARTK as P + 2Q is to Q. (See Proposition 25.) But as ABIK is to AMTK, thus IK--or AB--is to TK--or AK. Therefore, the rectangle ABIK is to the figure ARTK in a ratio compounded from the ratio of P + 2Q to Q and from the ratio of AB to AK. Since AE is to RE as BA is to CE, the rectangle from AE and CE--namely, AECL--will be equal to the rectangle from AB and RE. And since this will always be the case, the cylinder above ARTK having altitude AB will be equal to the lower trunk of the cylinder above ACIK. Therefore, the parallelpiped above ABIK with altitude AB is to the cylinder above ARTK--or the lower trunk of the cylinder above ACIK--as the base ABIK is to the base ARTK--that is, in the ratio compounded from the ratio P + 2 Q to Q and the ratio of AB to AK. But the parallelpiped above ABIK with altitude AB is to the parallelpiped above ABIK with altitude AK as AB is to AK. Therefore, the parallelpiped above ABIK with altitude AK is to the lower trunk of the cylinder above ACIK as P + 2 Q is to Q, which ought to have been demonstrated.

For instance, let the ratio of EF to CD be multiplied in the ratio of AE to AG with the ratio of 5 to 4. Using Proposition 2, as the line AE is to the curve AF thus let any rectangle AGNE be to the region AGLE. Let the latus rectum (namely, the line drawn 4 in the fourth power of the line CD always making the fifth root of the line AC) of the curve AF be l. Also, let AG be b, GI be c, and IL be d, and any increasing length GH be a. Let HO--or CA--be z.

It is manifest from Proposition 7 that BC is 4 z/5 and CD is the fourth root of z⁵/l, and BD is z5 /l + 16 z^2/25). Therefore, BC will be to BD as CK is to CO. That is . . . . . therefore, it is manifest that the trilineum GIL is compounded from the addition of the triline, squared, cubed, and raised to the power 4, of who all of the axes are GI and the bases the trilines squared . . . cubed and fourth-powered . Thereupon, the triline GOLI is . ... Therefore the line AE is to the curve AF as bd or the rectangle AGNE is to bd . . . or the curved region AGLe. Therefore, the line is given equal to the curve AF, which ought to have been found. How is the given surface found from the rotation of the curve AF around the line AG. It is evident from Proposition ?3 since from Propostion 57 it is easy to find the center of gravity of the curved region AGLE. But if . . . .