Question

Find the volume of the solid whose base is the circle x^2+y^2=64 and the cross sections perpendicular to the x-axis are triangles whose height and base are equal.

Answer 1

Because the height and base of each cross section are equal, the area for any given cross section is
`\frac12bh=\frac12b(x)^2` where the base of each section occurring along the line
`x=x_0` is the vertical distance between the upper and lower halves of the circle
`x^2+y^2=64`.

We can write


`y=\pm√(64-x^2)`

so that


`b(x)=√(64-x^2)-(-√(64-x^2))=2√(64-x^2)`

and so the area of each cross section is


`\frac12(2√(64-x^2))^2=2(64-x^2)=128-2x^2`

Over the circular base of the solid, we have
`-8\le x\le8`, so the volume of the solid is given by the integral


`\displaystyle\int_(x=-8)^(x=8)(128-2x^2)\,\mathrm dx=\frac{4096}3`