Double integral exercise solved by using polar coordinate system.

Exercise:

Compute, $\iint\limits_{D}{\frac{\ln \left( {{x}^{2}}+{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}dx}dy$ where $D=\left\{ \left( x,y \right)\in {{\mathbb{R}}^{2}}:1\le {{x}^{2}}+{{y}^{2}}\le {{e}^{2}} \right\}$

Solution: Let $x=r\cos \theta \,\,\And \,\,y=r\sin \theta \Leftrightarrow {{x}^{2}}+{{y}^{2}}={{r}^{2}}$

So $1\le {{x}^{2}}+{{y}^{2}}\le {{e}^{2}}\Leftrightarrow 1\le {{r}^{2}}\le {{e}^{2}}\Leftrightarrow 1\le r\le e$

So $\iint\limits_{D}{\frac{\ln \left( {{x}^{2}}+{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}dx}=\int_{0}^{2\pi }{\int_{1}^{e}{\frac{\ln {{r}^{2}}}{{{r}^{2}}}rdrd\theta }}$

$=\int_{0}^{2\pi }{\int_{1}^{e}{\frac{\ln {{r}^{2}}}{r}drd\theta }}=\int_{0}^{2\pi }{\left( \int_{1}^{e}{\frac{1}{r}\ln {{r}^{2}}dr} \right)d\theta }$

Let $u={{r}^{2}}\Leftrightarrow du=2rdr\Leftrightarrow dr=\frac{du}{2r}$

So $\int{\frac{\ln {{r}^{2}}}{r}dr}=\int{\frac{\ln u}{r}}\times \frac{du}{2r}=\int{\frac{\ln u}{2{{r}^{2}}}du}=\int{\frac{\ln u}{2u}du}$

Now put $w=\ln u\Leftrightarrow dw=\frac{1}{u}du$

So $\int{\frac{\ln u}{2u}du}=\frac{1}{2}\int{wdw=\frac{1}{4}{{w}^{2}}+c=\frac{1}{4}{{\ln }^{2}}u+c}$

Thus $\int{\frac{\ln {{r}^{2}}}{r}dr}=\frac{1}{4}{{\ln }^{2}}{{r}^{2}}+c$ hence $\int_{1}^{e}{\frac{\ln {{r}^{2}}}{r}dr}=\frac{1}{4}\left( {{\ln }^{2}}{{e}^{2}}-{{\ln }^{2}}1 \right)=\frac{1}{4}\left( 4\ln e \right)=1$

Therefore $\iint\limits_{D}{\frac{\ln \left( {{x}^{2}}+{{y}^{2}} \right)}{{{x}^{2}}+{{y}^{2}}}dxdy}=\int_{0}^{2\pi }{\int_{1}^{e}{\frac{\ln {{r}^{2}}}{r}drd\theta }=\int_{0}^{2\pi }{d\theta =2\pi }}$