Consider the solid S described below. The base of S is a circular disk with radius 3r. Parallel cross-sections perpendicular to the base are squares. Find the volume V of this solid. I tried this.... A(x) = pi*r^2 = pi*(3r)^2 = pi*9r^2 V(x) = integral from 0 to 3r(pi*9r^2 dx) = pi*27r^3 Am I approaching this problem the wrong way? Thanks for the help!
The square cross-sections are built upon the circular base, so each cross-section has a side length given by the length of a chord of the circle; these "slices" are standing upright on the circle. (Making a picture of this will be helpful to you.) The solid is symmetric about a diameter of the circle, so you can integrate half the solid and multiply your result by two. So what you need is to find the length of a chord of the circle which has its midpoint at a distance x from the circle's center. That length gives you the area A(x) of each cross-section. You would then integrate A(x) from x = 0 to x = 3r and multiply the resulting volume by 2.