to the one cutting off the section of the cylinder will be By; let the axis of the cylinder and the prism be yd which is bisected at right angles by e and on e let a plane be constructed perpendicular to yd. This will intersect the prism in a square and the cylinder in a circle.

οπρ

στ

Now let the intersection of the prism be the square μv [Fig. 10], that of the cylinder, the circle έorp and let the circle touch the sides of the square at the points έ, o, π and p; let the common line of intersection of the plane cutting off the cylinder-section and that passing through e perpendicular to the axis of the cylinder, be κà; this line is bisected by 0. In the semicircle orp draw a straight line or perpendicular to #x, on or construct a plane perpendicular to έ and produce it to both sides of the plane enclosing the circle Comρ; this will intersect the half-cylinder whose base is the semicircle op and whose altitude is the axis of the prism, in a parallelogram one side of which σT and the other = the vertical boundary of the cylinder, and it will intersect the cylinder-section likewise in a parallelogram of which one side is σr and the other w [Fig. 9]; and accordingly vu will be drawn in the parallelogram de | Bo and will cut off = TX. Now because ey is a parallelogram and || Oy, and ef and ẞy cut the parallels, therefore e0: Oi wy: yv = ẞw: vv. But Bw: wv = parallelogram in the half-cylinder: parallelogram in the cylinder-section, therefore both parallelograms have the same side σr; and 00, 0x0; and since 00g therefore 0: 0x = parallelogram in half-cylinder: parallelogram in the cylinder-section. Imagine the parallelogram in the cylinder-section transferred and so brought to that is its center of gravity, and further imagine

=

=

to be a scale-beam with its center at ; then the parallelogram in the half-cylinder in its present position is in equilibrium at the point with the parallelogram in the cylinder-section when it is transferred and so arranged on the scale-beam at έ that έ is its center of gravity. And since x is the center of gravity in the parallelogram in the half-cylinder, and έ that of the parallelogram in the cylindersection when its position is changed, and έ0:0x= the parallelogram whose center of gravity is x the parallelogram whose center of gravity is έ, then the parallelogram whose center of gravity is x will be in equilibrium at with the parallelogram whose center of gravity is. In this way it can be proved that if another straight line is drawn in the semicircle orp perpendicular to 0 and on this straight line a plane is constructed perpendicular to π and is produced towards both sides of the plane in which the circle έomp lies, then the parallelogram formed in the half-cylinder in its present position will be in equilibrium at the point with the parallelogram formed in the cylinder-section if this is transferred and so arranged on the scale-beam at έ that έ is its center of gravity; therefore also all parallelograms in the half-cylinder in their present positions will be in equilibrium at the point with all parallelograms of the cylinder-section if they are transferred and attached to the scale-beam at the point ; consequently also the half-cylinder in its present position will be in equilibrium at the point with the cylindersection if it is transferred and so arranged on the scale-beam at έ that έ is its center of gravity.

XII.

2

x

ξ

r

Let the parallelogram μv be perpendicular to the axis [of the circle] to [p] [Fig. 11]. Draw 0μ and 6 and erect upon them two planes perpendicular to the plane in which the semicircle op lies and extend these planes on both sides. The result is a prism whose base is a triangle similar to Oun and whose altitude is equal to the axis of the cylinder, and this prism is 4 of the entire prism which contains the cylinder. In the semicircle omp and in the square μv draw two straight lines A and at equal distances from πέ;

Fig. 11.

these will cut the circumference of the semicircle omp at the points

κ and τ, the diameter op at σ and and the straight lines on and oμ at and x. Upon κλ and TV construct two planes perpendicular to op and extend them towards both sides of the plane in which lies the circle corp; they will intersect the half-cylinder whose base is the semicircle оmp and whose altitude is that of the cylinder, in a parallelogram one side of which Ko and the other the axis of the cylinder; and they will intersect the prism On likewise in a parallelogram one side of which is equal to λy and the other equal to the axis, and in the same way the half-cylinder in a parallelogram one side of which = T and the other the axis of the cylinder, and the prism in a parallelogram one side of which up and the other the axis of the cylinder....

=

=

=

β

α

E

Fig. 12.

Y

б

XIII.

Let the square aßyd [Fig. 12] be the base of a perpendicular prism with square bases and let a cylinder be inscribed in the prism whose base is the circle en which touches the sides of the parallelogram aßyd at e, L, 7 and 0. Pass a plane through its center and the side in the square opposite the square aẞyd corresponding to the side yd; this will cut off from the whole prism a second prism which is 14 the size of the whole prism and which will be bounded by three parallelograms and two opposite triangles. In the semicircle en describe a parabola whose origin is ŋe and whose axis is ×, and in the parallelogram & draw μv | K; this will cut the circumference of the semicircle at έ, the parabola at A, and μv × vλ = v¿2 (for this is evident [Apollonios, Con. I, 11]). Therefore μv : vλ = ky2 : λơ2. Upon μv construct a plane parallel to en; this will intersect the prism cut off from the whole prism in a right-angled triangle one side of which is uv and the other a straight line in the plane upon y perpendicular to y8 at v and equal to the axis of the cylinder, but whose hypotenuse is in the intersecting plane. It will intersect the portion which is cut off from the cylinder by the plane passed through en and the side of the square opposite the side y in a right-angled triangle one side of which is μg and the other a straight line drawn in the surface of the cylinder perpendicular

to the plane, and the hypotenuse..

=

and all the triangles in the prism: all the triangles in the cylindersection all the straight lines in the parallelogram &: all the straight lines between the parabola and the straight line en. And the prism consists of the triangles in the prism, the cylinder-section of those in the cylinder-section, the parallelogram & of the straight lines in the parallelogram & and the segment of the parabola of the straight lines cut off by the parabola and the straight line en; hence prism cylinder-section parallelogram nd segment that is en bounded by the parabola and the straight line en. But the parallelogram on the segment bounded by the parabola and the straight. line en as indeed has been shown in the previously published work, hence also the prism is equal to one and one half times the cylindersection. Therefore when the cylinder-section = 2, the prism = 3 and the whole prism containing the cylinder equals 12, because it is four times the size of the other prism; hence the cylinder-section is equal to % of the prism, Q. E. D.

XIV.

[Inscribe a cylinder in] a perpendicular prism with square bases [and let it be cut by a plane passed through the center of the base of the cylinder and one side of the opposite square.] Then this plane will cut off a prism from the whole prism and a portion of the cylinder from the cylinder. It may be proved that the portion cut off from the cylinder by the plane is one-sixth of the whole prism. But first we will prove that it is possible to inscribe a solid figure in the cylinder-section and to circumscribe another composed of prisms of equal altitude and with similar triangles as bases, so that the circumscribed figure exceeds the inscribed less than any given magnitude......

=

But it has been shown that the prism cut off by the inclined plane <1⁄2 the body inscribed in the cylinder-section. Now the prism. cut off by the inclined plane: the body inscribed in the cylindersection parallelogram on the parallelograms which are inscribed in the segment bounded by the parabola and the straight line en. Hence the parallelogram <%1⁄2 the parallelograms in the segment bounded by the parabola and the straight line en. But this is impossible because we have shown elsewhere that the parallelogram ŋ is one and one half times the segment bounded by the parabola and the straight line en, consequently is.

not greater

And all prisms in the prism cut off by the inclined plane: all prisms in the figure described around the cylinder-section = all parallelograms in the parallelogram & all parallelograms in the figure which is described around the segment bounded by the parabola and the straight line en, i. e., the prism cut off by the inclined plane the figure described around the cylinder-section parallelogram on the figure bounded by the parabola and the straight line en. But the prism cut off by the inclined plane is greater than one and one half times the solid figure circumscribed around the cylinder-section

=