设 $D=\left\{(x, y) \mid-2+y\right.$ 叕 $\mid x \sqrt{4-y^{2}}$, 腏 2$\}$, 求二重积分 $I=\iint_{D} \frac{(x-y)^{2}}{x^{2}+y^{2}} d x d y$.
$I=\iint_D \frac{x^2-2 x y+y^2}{x^2+y^2} d x d y=\iint_D\left[1-\frac{2 x y}{x^2+y^2}\right] d x d y=\iint_D d x d y-\iint_D \frac{2 x y}{x^2+y^2} d x d y$ 

$=\pi+2-\int_0^{\frac{\pi}{2}} d \theta \int_{\frac{2}{\cos \theta+\sin \theta}}^2 \frac{2 \rho^2 \cos \theta \sin \theta}{\rho^2} \rho d \rho$

$\begin{aligned} & =\pi+2-4 \int_0^{\frac{\pi}{2}} \cos \theta \sin \theta d \theta+4 \int_0^{\frac{\pi}{2}} \frac{\cos \theta \sin \theta}{(\cos \theta+\sin \theta)^2} d \theta \\ & =\pi+4 \int_0^{\frac{\pi}{2}} \frac{\cos \theta \sin \theta}{1+2 \cos \theta \sin \theta} d \theta=\pi+4 \int_0^{\frac{\pi}{2}} \frac{\tan \theta}{\tan ^2 \theta+2 \tan \theta+1} d \theta \\ & =\pi+4 \int_0^{+\infty} \frac{t}{(t+1)^2} \cdot \frac{1}{1+t^2} d t=\pi+4 \cdot \frac{1}{2} \int_0^{+\infty}\left[\frac{1}{1+t^2}-\frac{1}{(t+1)^2}\right] d t \\ & =\pi+\left.2 \arctan t\right|_0 ^{+\infty}+\left.2 \frac{1}{1+t}\right|_0 ^{+\infty}=\pi+\pi-2=2 \pi-2 .\end{aligned}$