# How will you prove the following?

## $\int {\int}_{R}$ $\frac{{x}^{2} {y}^{2}}{{x}^{2} + {y}^{2}} \mathrm{dx} \mathrm{dy} = \frac{65 \pi}{16}$, where, R is the annulus between ${x}^{2} + {y}^{2} = 4$ and ${x}^{2} + {y}^{2} = 9$

May 17, 2018

See below

#### Explanation:

Write it in polar.

$\int {\int}_{R} \setminus \frac{{x}^{2} {y}^{2}}{{x}^{2} + {y}^{2}} \setminus \textcolor{b l u e}{\mathrm{dx} \setminus \mathrm{dy}}$

$\equiv {\int}_{0}^{2 \pi} {\int}_{2}^{3} \frac{{r}^{2} {\cos}^{2} \theta \setminus {r}^{2} {\sin}^{2} \theta}{{r}^{2}} \textcolor{red}{\setminus r \setminus \mathrm{dr} \setminus d \theta}$

$= \frac{1}{4} {\int}_{0}^{2 \pi} {\int}_{2}^{3} \setminus {r}^{3} {\sin}^{2} \left(2 \theta\right) \setminus \mathrm{dr} \setminus d \theta$

$= \frac{1}{4} {\int}_{0}^{2 \pi} {\left(\setminus {r}^{4} / 4\right)}_{2}^{3} {\sin}^{2} \left(2 \theta\right) \setminus d \theta$

$= \frac{1}{4} {\int}_{0}^{2 \pi} {\left(\setminus {r}^{4} / 4\right)}_{2}^{3} {\sin}^{2} \left(2 \theta\right) \setminus d \theta$

$= \frac{65}{16} {\int}_{0}^{2 \pi} \frac{1 - \cos \left(4 \theta\right)}{2} \setminus d \theta$

$= \frac{65}{32} {\left(\theta - \frac{1}{4} \sin \left(4 \theta\right)\right)}_{0}^{2 \pi}$

$= \frac{65}{16} \pi$