Double integral exercise

Exercise:

Integrate, $\int_{0}^{1}{\int_{0}^{1}{\frac{xydxdy}{\sqrt{1+{{x}^{2}}+{{y}^{2}}}}}}$

Solution: First we will integrate w.r.t $x$

So \[\int_{0}^{1}{\int_{0}^{1}{\frac{xy}{\sqrt{1+{{x}^{2}}+{{y}^{2}}}}dxdy}=\int_{0}^{1}{\left( y\int_{0}^{1}{\frac{x}{\sqrt{1+{{x}^{2}}+{{y}^{2}}}}dx} \right)dy}}\]

Let $u=1+{{x}^{2}}+{{y}^{2}}\Rightarrow du=2xdx$

If $x=0\Leftrightarrow u=1+{{y}^{2}}$  & If $x=1\Leftrightarrow u=2+{{y}^{2}}$

So $y\int_{0}^{1}{\frac{x}{\sqrt{1+{{x}^{2}}+{{y}^{2}}}}dx}=y\int_{1+{{y}^{2}}}^{2+{{y}^{2}}}{\frac{du}{2\sqrt{u}}}=\frac{y}{2}\int_{1+{{y}^{2}}}^{2+{{y}^{2}}}{{{u}^{-1/2}}du}=\frac{y}{2}\left( 2\sqrt{u} \right)_{1+{{y}^{2}}}^{2+{{y}^{2}}}=y\left( \sqrt{2+{{y}^{2}}}-\sqrt{1+{{y}^{2}}} \right)$

Hence $\int_{0}^{1}{\int_{0}^{1}{\frac{xy}{\sqrt{1+{{x}^{2}}+{{y}^{2}}}}dxdy}}=\int_{0}^{1}{y\left( \sqrt{2+{{y}^{2}}}-\sqrt{1+{{y}^{2}}} \right)dy}$

$=\int_{0}^{1}{y\sqrt{2+{{y}^{2}}}}dy-\int_{0}^{1}{y\sqrt{1+{{y}^{2}}}dy}$ Take ${{t}^{2}}=2+{{y}^{2}}\Leftrightarrow 2tdt=2ydy\Leftrightarrow 2ydy=tdt$

$\Rightarrow \int_{0}^{1}{y\sqrt{2+{{y}^{2}}}dy}=\int_{t\left( 0 \right)}^{t\left( 1 \right)}{{{t}^{2}}dt}=\left[ \frac{{{t}^{3}}}{3} \right]_{\sqrt{2}}^{\sqrt{3}}=\frac{1}{3}\left( \sqrt{{{3}^{3}}}-\sqrt{{{2}^{3}}} \right)=\frac{1}{3}\left( 3\sqrt{3}-2\sqrt{2} \right)$

Also take ${{v}^{2}}=1+{{y}^{2}}\Leftrightarrow 2vdv=2ydy\Leftrightarrow vdv=ydy$

Hence $\int_{0}^{1}{y\sqrt{1+{{y}^{2}}}}dy=\int_{v\left( 0 \right)}^{v\left( 1 \right)}{{{v}^{2}}dv}=\left[ \frac{1}{3}{{v}^{3}} \right]_{1}^{\sqrt{2}}=\frac{1}{3}\left( 2\sqrt{2}-1 \right)$

Thus $\int_{0}^{1}{\int_{0}^{1}{\frac{xy}{\sqrt{1+{{x}^{2}}+{{y}^{2}}}}dxdy}}=\frac{1}{3}\left( 3\sqrt{3}-2\sqrt{2} \right)-\frac{1}{3}\left( 2\sqrt{2}-1 \right)=\frac{1}{3}\left( 3\sqrt{3}-4\sqrt{2}+1 \right)$