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 $\dfrac12bh=\dfrac12b(x)^2$ 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

$y=\pm\sqrt{64-x^2}$

so that

$b(x)=\sqrt{64-x^2}-(-\sqrt{64-x^2})=2\sqrt{64-x^2}$

and so the area of each cross section is

$\dfrac12(2\sqrt{64-x^2})^2=2(64-x^2)=128-2x^2$

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

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