Use trig identities to change
\[ \frac{sec^3~\theta}{tan~\theta} \]
to
\[ sec~\theta~tan~\theta~+~csc~\theta \]

For every corresponding pair of cross sections, the area of the cross section of a sphere with radius r is equal to the area of the cross section of a cylinder with radius and height 2r minus the volume of two cones, each with a radius and height of r. a cross section of the sphere is and a cross section of the cylinder minus the cones, taken parallel to the base of cylinder, is the volume of the cylinder with radius r and height 2r is and the volume of each cone with radius r and height r is 1/3 pie r^3. so the volume of the cylinder minus the two cones is therefore, the volume of the cylinder is 4/3pie r^3 by cavalieri's principle. (fill in options are: r/2- r- 2r- an annulus- a circle -1/3pier^3- 2/3pier^3- 4/3pier^3- 5/3pier^3- 2pier^3- 4pier^3)