Question

Use calculus to find the volume of the following solid S:

The base of S is a circular disk with radius r. Parallel cross-sections perpendicular to the base are squares.

Answer 1

I assume the base of S is the disk centered at the origin, or x² + y² ≤ r². Solving for y gives two solutions corresponding to the upper and lower halves of the circular boundary,

`x^2+y^2 = r^2 \implies y = \pm √(r^2-x^2)`

and the vertical distance between them - i.e. the side length of each cross section - is

`√(r^2-x^2) - \left(-√(r^2-x^2)\right) = 2√(r^2-x^2)`

Then the volume of each square cross section 4 (r² - x²) ∆x, and the volume of S is

`\displaystyle \int_(-r)^r 4(r^2-x^2) \, dx`

By symmetry, this is the same as

`\displaystyle 8 \int_0^r (r^2-x^2) \, dx`

which gives a volume of

`\displaystyle 8 \int_0^r (r^2-x^2) \, dx = 8 \left(r^3 - \frac{r^3}3\right) = \boxed{\frac{16r^3}3}`