第六题：（15 分）$f(x, y)$ 是 $\left\{(x, y) \mid x^{2}+y^{2} \leq 1\right\}$ 上二次连续可微函数，满足

$$
\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=x^{2} y^{2}
$$

计算积分 $I=\iint_{x^{2}+y^{2} \leq 1}\left(\frac{x}{\sqrt{x^{2}+y^{2}}} \frac{\partial f}{\partial x}+\frac{y}{\sqrt{x^{2}+y^{2}}} \frac{\partial f}{\partial y}\right) \mathrm{d} x \mathrm{~d} y$

【参考解析】：令 $x=r \cos \theta, y=r \sin \theta$ ，则

$$
\begin{aligned}
I & =\int_{0}^{1} d r \int_{0}^{2 \pi}\left(\cos \theta \cdot \frac{\partial f}{\partial x}+\sin \theta \cdot \frac{\partial f}{\partial y}\right) r d \theta=\int_{0}^{1} d r \int_{x^{2}+y^{2}=r^{2}}\left(\frac{\partial f}{\partial x} d y-\frac{\partial f}{\partial y} d x\right) \\
& =\int_{0}^{1} d r \iint_{x^{2}+y^{2} \leq r^{2}}\left(\frac{\partial f}{\partial x^{2}}+\frac{\partial f}{\partial y^{2}}\right) d x d y=\int_{0}^{1} d r \iint_{x^{2}+y^{2} \leq r^{2}}\left(x^{2} y^{2}\right) d x d y \\
& =\int_{0}^{1} d r \int_{0}^{r} \rho^{5} d \rho \int_{0}^{2 \pi} \cos ^{2} \theta \sin ^{2} \theta d \theta=\frac{\pi}{168}
\end{aligned}
$$