Question

How do you find the volume V of the described solid S where the base of S is a circular disk with radius 4r and Parallel cross-sections perpendicular to the base are squares?

Answer 1

`V = 1024/3 r^3`

Explanation:

Place circular base on x-y plane, centred at Origin.

At `z = 0`;

• `x^2 + y^2 = 16r^2`

Considering that part of the solid in the 1st octant, with the square cross-sections running parallel to the x-z axis, the volume of a elemental cross section is:

`dV = x * 2x \ dy = 2 (16r^2 - y^2) dy`

Thus:

`V =2 int_0^(4r) dy qquad (16r^2 - y^2)`

`= 2 [ 16r^2 y - y^3/3 ]_0^(4r) = 256/3 r^3`

Volume in 1st Octant is only `1/4` of the total volume.

So `V_("Tot") = 1024/3 r^3 qquad [ = 341 1/3 r^3]`

Reality check.

• Volume of cube side `8r` is `V_C = 512 r^3`

• Volume of sphere radius `4r` is `V_S = (256 pi R^3)/3 = 268.083 R^3`

• `V_S < V_("Tot") < V_C`