Question

The base of a solid is a circle with radius 6. Each cross section perpendicular to a fixed diameter of the base is semicircular.

If the circle is centered at the origin and the fixed diameter lies on the x-axis, find the cross-section area A(x).

A(x) =


Find the volume V of the solid.

V =

Answer 1

A(x) = π(6^2 - x^2)

V = (1/2)π∫x=-6x=6(6^2 - x^2)dx

V = (1/2)π[6x(6^2 - x^2) - (1/3)(6^2 - x^2)^3]|x=-6 to x=6

V = (1/2)π[6(36 - 36) - (1/3)(36 - 36)^3]

V = 0

Answer 2

V=0?
Surely that cannot be true!
Of course that garbage you wrote prolly didn't help ...
A(x) = π/2 (36-x^2)
using symmetry,
V = 2∫[0,6] π/2 (36-x^2) dx
= ∫[0,6] π (36-x^2) dx
= 144π