当二重积分的积分区域不是圆形但被积函数和圆形有关时，也可以尝试使用极坐标系求解

一、题目

$$I={\int }_0^1\mathrm{d}y{\int }_y^1\sqrt{x^2+y^2}\mathrm{d}x=?$$

难度评级：

本题的计算步骤可以参考 这篇文章 。

二、解析

由题知：

$$x\in (0,1),y=x,x=1$$

于是，我们可以绘制出如下积分区域（阴影部分）：

虽然这个积分区域不像 这道题 一样来自一个圆形，但是本题的被积函数 $\sqrt{x^2+y^2}$ 和圆形有关，且直接在直角坐标系中做积分运算又很复杂，于是，我们可以尝试转换为极坐标系进行积分运算。

又：

$$x=1\Rightarrow r\cos \theta =1\Rightarrow r=\frac{1}{\cos \theta }$$

于是：

$$I={\int }_0^{\frac{\pi }{4}}\mathrm{d}\theta {\int }_0^{\frac{1}{\cos \theta }}\sqrt{(\cos \theta {)}^2+(r\sin \theta {)}^2}\cdot r\mathrm{d}r$$

$$I={\int }_0^{\frac{\pi }{4}}\mathrm{d}\theta {\int }_0^{\frac{1}{\cos \theta }}{r}^2\mathrm{d}r\Rightarrow$$

$$I=\frac{1}{3}{\int }_0^{\frac{\pi }{4}}\left({r^3|}_0^{\frac{1}{\cos \theta }}\right)\mathrm{d}\theta \Rightarrow$$

$$I=\frac{1}{3}{\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^3\theta }\mathrm{d}\theta \Rightarrow$$

解法 1

$$I=\frac{1}{3}{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{d}(\sin \theta )}{{\cos }^4\theta }\Rightarrow$$

$$I=\frac{1}{3}{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{d}(\sin \theta )}{{(1-{\sin }^2\theta )}^2}.$$

令 $t=\sin \theta$, 则：

$$\theta \in (0,\frac{\pi }{4})\Rightarrow t\in (0,\frac{\sqrt{2}}{2})$$

因此：

$$I=\frac{1}{3}{\int }_0^{\frac{\sqrt{2}}{2}}\frac{1}{{(1-t^2)}^2}\mathrm{d}t\Rightarrow$$

$$I=\frac{1}{3}{\int }_0^{\frac{\sqrt{2}}{2}}{\left(\frac{1}{1-t^2}\right)}^2\mathrm{d}t\Rightarrow$$

$$I=\frac{1}{3}\cdot (\frac{1}{2}{)}^2{\int }_0^{\frac{\sqrt{2}}{2}}{(\frac{1}{1-t}+\frac{1}{1+t})}^2\mathrm{d}t\Rightarrow$$

$$I=\frac{1}{3}\cdot \frac{1}{4}{\int }_0^{\frac{\sqrt{2}}{2}}[\frac{1}{(1-t{)}^2}+\frac{1}{(1+t{)}^2}+\frac{2}{1-t^2}]\mathrm{d}t\Rightarrow$$

$$I=\frac{1}{3}\cdot \frac{1}{4}{\int }_0^{\frac{\sqrt{2}}{2}}[\frac{1}{(1-t{)}^2}+\frac{1}{(1+t{)}^2}+\frac{1}{1-t}+\frac{1}{1+t}]\mathrm{d}t\Rightarrow$$

$$I=\frac{1}{3}\cdot \frac{1}{4}{[\frac{1}{1-t}-\frac{1}{1+t}+\ln (1+t)-\ln (1-t)]|}_0^{\frac{\sqrt{2}}{2}}\Rightarrow$$

$$I=\frac{1}{3}\cdot \frac{1}{4}{[\frac{1}{1-t}-\frac{1}{1+t}+\ln \frac{1+t}{1-t}]|}_0^{\frac{\sqrt{2}}{2}}\Rightarrow$$

进而：

$$I=\frac{1}{3}\cdot \frac{1}{4}[\frac{1}{1-\frac{\sqrt{2}}{2}}-\frac{1}{1+\frac{\sqrt{2}}{2}}+\ln \frac{1+\frac{\sqrt{2}}{2}}{1-\frac{\sqrt{2}}{2}}]\Rightarrow$$

$$I=\frac{1}{12}[\frac{2}{2-\sqrt{2}}-\frac{2}{2+\sqrt{2}}+\ln \frac{2+\sqrt{2}}{2-\sqrt{2}}]\Rightarrow$$

$$I=\frac{1}{12}[2\sqrt{2}+\ln \frac{2\sqrt{2}+2}{2\sqrt{2}-2}]\Rightarrow$$

$$I=\frac{1}{12}[2\sqrt{2}+\ln \frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}-1)(\sqrt{2}+1)}]\Rightarrow$$

$$I=\frac{1}{12}[2\sqrt{2}+\ln (\sqrt{2}+1{)}^2]\Rightarrow$$

$$I=\frac{1}{12}[2\sqrt{2}+2\ln (\sqrt{2}+1)]\Rightarrow$$

$$I=\frac{1}{6}[\sqrt{2}+\ln (\sqrt{2}+1)]=\frac{\sqrt{2}}{6}+\frac{1}{6}\ln (\sqrt{2}+1).$$

解法 2

$$I=\frac{1}{3}{\int }_0^{\frac{\pi }{4}}\frac{1}{{\cos }^3\theta }\mathrm{d}\theta \Rightarrow$$

$$\frac{1}{3}{\int }_0^{\frac{\pi }{4}}\frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^3\theta }\mathrm{d}\theta \Rightarrow$$

$$\frac{1}{3}[{\int }_0^{\frac{\pi }{4}}\frac{{\sin }^2\theta }{{\cos }^3\theta }\mathrm{d}\theta +{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{d}\theta }{\cos \theta }]\Rightarrow$$

又：

$${\left(\frac{1}{{\cos }^2\theta }\right)}_{\theta }^{\prime }=\frac{2\cos \theta \sin \theta }{{\cos }^4\theta }=\frac{2\sin \theta }{{\cos }^3\theta }\Rightarrow$$

于是：

$$\frac{1}{3}[{\int }_0^{\frac{\pi }{4}}\frac{1}{2}\sin \theta \mathrm{d}\left(\frac{1}{{\cos }^2\theta }\right)+{\int }_0^{\frac{\pi }{4}}\frac{1}{\cos \theta }\mathrm{d}\theta ]\Rightarrow$$

$$\frac{1}{3}\cdot [{\frac{1}{2}\frac{\sin \theta }{{\cos }^2\theta }|}_0^{\frac{\pi }{4}}-\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{d}(\sin \theta )}{{\cos }^2\theta }+{\int }_0^{\frac{\pi }{4}}\frac{1}{\cos \theta }\mathrm{d}\theta ]\Rightarrow$$

$$\frac{1}{3}\cdot [\frac{1}{2}\frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}}-\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{1}{\cos \theta }\mathrm{d}\theta +{\int }_0^{\frac{\pi }{4}}\frac{1}{\cos \theta }\mathrm{d}\theta ]\Rightarrow$$

$$\frac{1}{3}[\frac{\sqrt{2}}{2}+\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{d}(\sin \theta )}{{\cos }^2\theta }]\Rightarrow$$

$$\frac{1}{3}[\frac{\sqrt{2}}{2}+\frac{1}{2}{\int }_0^{\frac{\pi }{4}}\frac{\mathrm{d}(\sin \theta )}{1-{\sin }^2\theta }]\Rightarrow$$

接着：

$$\frac{1}{3}[\frac{\sqrt{2}}{2}+{\int }_0^{\frac{\sqrt{2}}{2}}\frac{\mathrm{d}t}{1-t^2}]\Rightarrow$$

$$\frac{\sqrt{2}}{6}+{\frac{1}{6}\cdot \frac{1}{2}(\frac{1}{1+t}+\frac{1}{1-t})|}_0^{\frac{\sqrt{2}}{2}}\Rightarrow$$

$$\frac{\sqrt{2}}{6}+{\frac{1}{6}\cdot \frac{1}{2}[\ln (1+t)-\ln (1-t)]|}_0^{\frac{\sqrt{2}}{2}}\Rightarrow$$

$$\frac{\sqrt{2}}{6}+{\frac{1}{6}\cdot \frac{1}{2}\ln \frac{1+t}{1-t}|}_0^{\frac{\sqrt{2}}{2}}=\frac{\sqrt{2}}{6}+\frac{1}{6}\cdot \frac{1}{2}\ln (\sqrt{2}+1{)}^2\Rightarrow$$

$$\frac{\sqrt{2}}{6}+\frac{1}{6}\ln (\sqrt{2}+1).$$

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