Volume of a sphere by revolution

Text

The semicircle $y=\sqrt{9-x^{2}}$ is revolved about the $x$-axis. Compute the volume of the sphere obtained and compare with $V=\tfrac43\pi R^{3}$.

Solution

$V=\pi\displaystyle\int_{-3}^{3}(9-x^{2})dx=\pi\Big[9x-\tfrac{x^{3}}{3}\Big]_{-3}^{3}=36\pi\approx113.097$. Compare: $R=3$, $V=\tfrac43\pi\cdot27=36\pi$. ✓ $\boxed{V=36\pi\approx113.097}$