Question

Find the area cut out of the cylinder x2 z2=9 by the cylinder x2 y2=9.

Answer 1

The cylinder is symmetric across the plane $y=0$, so we need only consider half of the cut-out region.

This region can be parameterized by

$\mathbf r(r,\theta)=(r\cos\theta,\sqrt{9-r^2\cos^2t},r\sin\theta)$

with $0\le\theta\le2\pi$ and $0\le r\le3$. Then the area of this half is given by the surface integral

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\left\|\mathbf r_r\times\mathbf r_\theta\right\|\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\int_0^{2\pi}\int_0^3\frac{3r}{\sqrt{9-r^2\cos^2\theta}}\,\mathrm dr\,\mathrm d\theta$
$=36$

Doubling this gives a total area of 72.