The region bounded by the graph of $y=4-x^2$ and the $x$-axis is revolved about the $x$-axis. What is the volume of the resulting solid?

Core Concept

The curve meets the axis at $x=\pm2$. The volume is $\pi\int_{-2}^2(4-x^2)^2\,dx=\pi[16x-\frac{8x^3}{3}+\frac{x^5}{5}]_{-2}^2=\frac{512\pi}{15}$.