$$\iint\left|\sqrt{x^2+y^2}-1\right| d x d y .$$
【答案】 本题目先利用奇偶对称性化简, 再切割积分区域, 把积分区 域分为三块, 分别采用极坐标进行计算:

\begin{aligned} & \iint_D\left|\sqrt{x^2+y^2}-1\right| d \sigma=2 \iiint_{D_1+D_2+D_3}\left|\sqrt{x^2+y^2}-1\right| d \sigma \\ & =2 \iint_{D_1} 1-\sqrt{x^2+y^2} d \sigma+2 \iint_{D_2} 1-\sqrt{x^2+y^2} d \sigma+2 \iint_{D_3} \sqrt{x^2+y^2}-1 d \sigma \end{aligned}

\begin{aligned} & \iint_{D_1} 1-\sqrt{x^2+y^2} d \sigma=\int_0^{\frac{\pi}{3}} d \theta \int_0^1 r(1-r) d r=\frac{\pi}{3} \cdot \frac{1}{6}=\frac{\pi}{18} \\ & \iint_{D_2} 1-\sqrt{x^2+y^2} d \sigma=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} d \theta \int_0^{2 \cos \theta} r(1-r) d r=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} 2 \cos ^2 \theta-\frac{8}{3} \cos ^3 \theta d \theta=\frac{\pi}{6}-\frac{16}{9}+\frac{3}{4} \sqrt{3} \\ & \iint_{D_3} \sqrt{x^2+y^2}-1 d \sigma=\int_0^{\frac{\pi}{3}} d \theta \int_1^{2 \cos \theta} r(r-1) d r=\int_0^{\frac{\pi}{3}} \frac{8}{3} \cos ^3 \theta-2 \cos ^2 \theta+\frac{1}{6} d \theta=\frac{3}{4} \sqrt{3}-\frac{\pi}{3}+\frac{\pi}{18} . \end{aligned}

\begin{aligned} & \iint_D\left|\sqrt{x^2+y^2}-1\right| d \sigma=2 \iint_{D_1} 1-\sqrt{x^2+y^2} d \sigma+2 \iint_{D_2} 1-\sqrt{x^2+y^2} d \sigma+2 \iint_{D_1} \sqrt{x^2+y^2}-1 d \sigma \\ & =-\frac{\pi+32}{9}+3 \sqrt{3} \end{aligned}