Question

In this problem, we will find the volume of a solid with circular base of radius 2, for which parallel cross-sections perpendicular to the base are squares. To do this, we will assume that the base is the circle x2+y2=4, so that the solid lies between planes parallel to the x-axis at x=2 and x=−2. The cross-sections perpendicular to the x-axis are then squares whose bases run from the semicircle y=√−4−x2 to the semicircle y = √4−x2

Required:
a. What is the area of the cross-section at x?
b. What is the volume of the solid ?

Answer 1

(a) Each cross section is a square whose side length is decided by the distance in the x,y plane between the two curves
`y=√(4-x^2)` and
`y=-√(4-x^2)`, which is
`2√(4-x^2)`. Then each cross section has area

`(2√(4-x^2))^2=4(4-x^2)=\boxed{16-4x^2}`

(b) The volume of the solid is obtained by integrating the cross-sectional area from x = -2 to x = 2.

`\displaystyle\int_(-2)^2(16-4x^2)\,\mathrm dx=16x-\frac43x^3\bigg|_(-2)^2=\boxed{\frac{128}3}`