Chapter 14 - Multiple Integrals - 14.3 Double Integrals In Polar Coordinates - Exercises Set 14.3 - Page 1024: 16

\[ \int_{0}^{2 \pi} \int_{1}^{3} \mathrm{d} \mathrm{rd} \theta \]

$\left(y^{2}+x^{2}\right)^{-1 / 2}=z$ $\frac{1}{r}=z$ $x^{2}+y^{2}=1$ $1=r^{2}$ $r=1$ $9=x^{2}+y^{2}$ $9=r^{2}$ $r=3$ Thus, the integral is: \[ \int_{0}^{2 \pi} \int_{1}^{3} \mathrm{d} \mathrm{rd} \theta \]