一、题目
设 $D=\left\{(x, y) \mid x^{2}+y^{2} \leqslant x\right \}$, 则 $I=\iint_{D}\left(x+y^{2}\right) \mathrm{~d} \sigma=?$
难度评级：
二、解析 
由题可知：
$$
x^{2}+y^{2} \leq x \Rightarrow x^{2}+y^{2}-x \leq 0 \Rightarrow
$$
$$
\left(x-\frac{1}{2}\right)^{2}+y^{2} \leq \frac{1}{4} \Rightarrow
$$
$$
\left\{\begin{array}{l}x=r \cos \theta \\ y=r \sin \theta \end{array}\right.\Rightarrow
$$
$$
\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \ r \in[0, \cos \theta]
$$
根据积分区域的图象，可以很容易的判断出 $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, 同时，由 $\cos (-\frac{\pi}{2}) = \cos (\frac{\pi}{2}) = 0$, $\cos 0 = 1$ 这几个特殊点的取值，可以判断出 $r \in[0, \cos \theta]$.
于是：
$$
I=\iint_{D}\left(r \cos \theta+r^{2} \sin ^{2} \theta\right) r d \theta=
$$
$$
I=\iint_{D} r^{2} \cos \theta d \theta+\iint_{D}^{3} \sin ^{2} \theta d r=
$$
$$
I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^{2} \theta \int_{0}^{\cos \theta} r^{2} d r+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} \theta \int_{0}^{\cos ^{2} \theta} r^{3} d r=
$$
$$
I=\frac{2}{3} \int_{0}^{\frac{\pi}{2}} \cos ^{4} \theta d \theta+\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(1-\cos ^{2} \theta\right) \cos ^{4} \theta d \theta
$$
$$
I=\frac{2}{3}(\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2})+\frac{1}{2} \Big[(\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2})-(\frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2})\Big] \Rightarrow
$$
$$
I=\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2} \cdot \frac{5}{6}\right)\left(\frac{3}{4} \cdot \frac{\pi}{4}\right) \Rightarrow
$$
$$
I=\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{\pi}{4}=\frac{9 \pi}{64}
$$