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、题目题目 - 荒原之梦

$$
I=\int_{0}^{1} \mathrm{~d} y \int_{y}^{1} \sqrt{x^{2}+y^{2}} \mathrm{~d} x = ?
$$

难度评级:

本题的计算步骤可以参考 这篇文章

二、解析 解析 - 荒原之梦

由题知:

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

于是,我们可以绘制出如下积分区域(阴影部分):

荒原之梦 | 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
图 01.

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:

$$
I=\int_{0}^{\frac{\pi}{4}} \mathrm{d} \theta \int_{0}^{\frac{1}{\cos \theta}} \sqrt{(\cos \theta)^{2}+(r \sin \theta)^{2}} \cdot r \mathrm{d} r
$$

$$
I=\int_{0}^{\frac{\pi}{4}} \mathrm{d} \theta \int_{0}^{\frac{1}{\cos \theta}} r^{2} \mathrm{d} r \Rightarrow
$$

$$
I=\frac{1}{3} \int_{0}^{\frac{\pi}{4}}\left(\left.r^{3}\right|_{0} ^{\frac{1}{\cos \theta}}\right) \mathrm{d} \theta \Rightarrow
$$

$$
I = \frac{1}{3} \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos ^{3} \theta} \mathrm{d} \theta \Rightarrow
$$

解法 1

$$
I=\frac{1}{3} \int_{0}^{\frac{\pi}{4}} \frac{\mathrm{d} (\sin \theta)}{\cos ^{4} \theta} \Rightarrow
$$

$$
I = \frac{1}{3} \int_{0}^{\frac{\pi}{4}} \frac{\mathrm{d} (\sin \theta)}{\left(1-\sin ^{2} \theta\right)^{2}}.
$$

令 $t=\sin \theta$, 则:

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

因此:

$$
I=\frac{1}{3} \int_{0}^{\frac{\sqrt{2}}{2}} \frac{1}{\left(1-t^{2}\right)^{2}} \mathrm{d} t \Rightarrow
$$

$$
I = \frac{1}{3} \int_{0}^{\frac{\sqrt{2}}{2}}\left(\frac{1}{1-t^{2}}\right)^{2} \mathrm{d} t \Rightarrow
$$

$$
I = \frac{1}{3} \cdot \Big( \frac{1}{2} \Big)^{2} \int_{0}^{\frac{\sqrt{2}}{2}}\left(\frac{1}{1-t}+\frac{1}{1+t}\right)^{2} \mathrm{d} t \Rightarrow
$$

$$
I = \frac{1}{3} \cdot \frac{1}{4} \int_{0}^{\frac{\sqrt{2}}{2}}\left[\frac{1}{(1-t)^{2}}+\frac{1}{(1+t)^{2}}+\frac{2}{1-t^{2}}\right] \mathrm{d} t \Rightarrow
$$

$$
I = \frac{1}{3} \cdot \frac{1}{4} \int_{0}^{\frac{\sqrt{2}}{2}}\left[\frac{1}{(1-t)^{2}}+\frac{1}{(1+t)^{2}}+\frac{1}{1-t}+\frac{1}{1+t}\right] \mathrm{d} t \Rightarrow
$$

$$
I = \frac{1}{3} \cdot \frac{1}{4} {\left.\left[\frac{1}{1-t}-\frac{1}{1+t}+\ln (1+t)-\ln (1-t)\right]\right|_{0} ^{\frac{\sqrt{2}}{2}}} \Rightarrow
$$

$$
I = \frac{1}{3} \cdot \frac{1}{4} {\left.\left[\frac{1}{1-t}-\frac{1}{1+t}+\ln \frac{1+t}{1-t}\right]\right|_{0} ^{\frac{\sqrt{2}}{2}}} \Rightarrow
$$

进而:

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

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

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

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

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

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

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

解法 2

$$
I=\frac{1}{3} \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos ^{3} \theta} \mathrm{d} \theta \Rightarrow
$$

$$
\frac{1}{3} \int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{3} \theta} \mathrm{d} \theta \Rightarrow
$$

$$
\frac{1}{3}\left[\int_{0}^{\frac{\pi}{4}} \frac{\sin ^{2} \theta}{\cos ^{3} \theta} \mathrm{d} \theta+\int_{0}^{\frac{\pi}{4}} \frac{\mathrm{d} \theta}{\cos \theta}\right] \Rightarrow
$$

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:

$$
\frac{1}{3}\left[\int_{0}^{\frac{\pi}{4}} \frac{1}{2} \sin \theta \mathrm{d} \left(\frac{1}{\cos ^{2} \theta}\right)+\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \theta} \mathrm{d} \theta\right] \Rightarrow
$$

$$
\frac{1}{3} \cdot\left[\left.\frac{1}{2} \frac{\sin \theta}{\cos ^{2} \theta}\right|_{0} ^{\frac{\pi}{4}}-\frac{1}{2} \int _{0}^{\frac{\pi}{4}} \frac{\mathrm{d} (\sin \theta)}{\cos ^{2} \theta}+\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \theta} \mathrm{d} \theta\right] \Rightarrow
$$

$$
\frac{1}{3} \cdot\left[\frac{1}{2} \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}}-\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \theta} \mathrm{d} \theta+\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \theta} \mathrm{d} \theta\right] \Rightarrow
$$

$$
\frac{1}{3}\left[\frac{\sqrt{2}}{2}+\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{\mathrm{d} (\sin \theta)}{\cos ^{2} \theta}\right] \Rightarrow
$$

$$
\frac{1}{3}\left[\frac{\sqrt{2}}{2}+\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{\mathrm{d} (\sin \theta)}{1-\sin ^{2} \theta}\right] \Rightarrow
$$

接着:

$$
\frac{1}{3}\left[\frac{\sqrt{2}}{2}+ \int_{0}^{\frac{\sqrt{2}}{2}} \frac{\mathrm{d} t}{1-t^{2}}\right] \Rightarrow
$$

$$
\frac{\sqrt{2}}{6}+\left.\frac{1}{6} \cdot \frac{1}{2}\left(\frac{1}{1+t}+\frac{1}{1-t}\right)\right|_{0} ^{\frac{\sqrt{2}}{2}} \Rightarrow
$$

$$
\frac{\sqrt{2}}{6}+\left.\frac{1}{6} \cdot \frac{1}{2}[\ln (1+t)-\ln (1-t)]\right|_{0} ^{\frac{\sqrt{2}}{2}} \Rightarrow
$$

$$
\frac{\sqrt{2}}{6}+\left.\frac{1}{6} \cdot \frac{1}{2} \ln \frac{1+t}{1-t}\right|_{0} ^{\frac{\sqrt{2}}{2}}=\frac{\sqrt{2}}{6}+\frac{1}{6} \cdot \frac{1}{2} \ln (\sqrt{2}+1)^{2} \Rightarrow
$$

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


高等数学箭头 - 荒原之梦

涵盖高等数学基础概念、解题技巧等内容,图文并茂,计算过程清晰严谨。

线性代数箭头 - 荒原之梦

以独特的视角解析线性代数,让繁复的知识变得直观明了。

特别专题箭头 - 荒原之梦

通过专题的形式对数学知识结构做必要的补充,使所学知识更加连贯坚实。

荒原之梦考研数学网 | 让考场上没有难做的数学题!
荒原之梦考研网-圆梦优选-学生党专属优惠好物-学-阿里巴巴普惠体 3.0
荒原之梦考研网-圆梦优选-学生党专属优惠好物-生-阿里巴巴普惠体 3.0
荒原之梦考研网-圆梦优选-学生党专属优惠好物-党-阿里巴巴普惠体 3.0
荒原之梦考研网-圆梦优选-学生党专属优惠好物-专-阿里巴巴普惠体 3.0
荒原之梦考研网-圆梦优选-学生党专属优惠好物-属-阿里巴巴普惠体 3.0
荒原之梦考研网-圆梦优选-学生党专属优惠好物-优-阿里巴巴普惠体 3.0
荒原之梦考研网-圆梦优选-学生党专属优惠好物-惠-阿里巴巴普惠体 3.0
荒原之梦考研网-圆梦优选-学生党专属优惠好物-好-阿里巴巴普惠体 3.0
荒原之梦考研网-圆梦优选-学生党专属优惠好物-物-阿里巴巴普惠体 3.0