# 转为极坐标系后，怎么确定新的积分上下限？

## 二、解析

\begin{aligned} x^{2}+y^{2} \leqslant y & \rightleftarrows \\ x^{2}+y^{2} – y \leqslant 0 & \rightleftarrows \\ & x^{2}+\left(y-\frac{1}{2}\right)^{2} \leqslant\left(\frac{1}{2}\right)^{2} \end{aligned}

$$\left\{\begin{array}{l}x=r \cos \theta \\ y=r \sin \theta\end{array} \Rightarrow x^{2}+y^{2}=r^{2} \right.$$

\begin{aligned} I & = \iint_{D} \sqrt{1-\left(x^{2}+y^{2}\right)} \mathrm{~d} \sigma \\ & = \iint_{D} \sqrt{1-r^{2}} \cdot r \cdot \mathrm{~d} \sigma \\ & = 2 \int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{\textcolor{orangered}{0}}^{\textcolor{orangered}{\sin \theta}} r \sqrt{1-r^{2}} \mathrm{~d} r \end{aligned}

$$\theta \in\left(0, \frac{\pi}{2}\right)$$

\begin{aligned} \textcolor{orangered}{x^{2}+y^{2} \leqslant y} & \Rightarrow \\ r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=r \sin \theta & \Rightarrow \\ r \cos ^{2} \theta+r \sin ^{2} \theta=\sin \theta & \Rightarrow \\ r\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=\sin \theta & \Rightarrow \\ & \textcolor{orangered}{r=\sin \theta} \end{aligned}

$$\textcolor{orangered}{r \in (0, \ \sin \theta)}$$

\begin{aligned} I & = -\int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\sin \theta}\left(1-r^{2}\right)^{\frac{1}{2}} d\left(1-r^{2}\right) \\ \\ & = -\left.\int_{0}^{\frac{\pi}{2}} \frac{2}{3}\left(1-r^{2}\right)^{\frac{3}{2}}\right|_{0} ^{\sin \theta} \mathrm{~d} \theta \\ \\ & = -\frac{2}{3} \int_{0}^{\frac{\pi}{2}}\left(\cos ^{3} \theta-1\right) \mathrm{~d} \theta \\ \\ & = -\frac{2}{3}\left[\frac{2}{3} \cdot 1-\frac{\pi}{2}\right] \mathrm{~d} \theta \\ \\ & = \frac{\pi}{3}-\frac{4}{9} \end{aligned}