Volume of a solid with base of circular disk, parallel crosssections perpendicular to base are squares.

Question

Working on a problem of volume using integration:

The problem is this:

The base of is a circular disk with radius . Parallel crosssections perpendicular to the base are squares.

I already have an idea how the solid figure would look like but I am lost at finding a way to define my variables.

Anyone here who can give me, at least a hint?

Thanks

Answer 1

Consider the circle $y=\pm \sqrt{r^2-x^2}$, where $r$ is the radius of the base. Then the cross-sectional area of the volume is $A(x) = (2 y)^2= 4 (r^2-x^2)$. The volume is

$$\int_{-r}^r dx \, A(x) = 8 \int_0^r dx \, (r^2-x^2)$$

I assume you got it from here.