一类较复杂的二重积分

青青子衿

\[ \int_0^ax^2\int_0^b \sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}y{\rm\,d}x \]
\[ \int_0^b\int_0^ax^2\sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}x{\rm\,d}y \]
\[ {\color{red}{\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\frac{a^5}{10}\operatorname{arsinh}\left(\frac{b}{|a|}\right)-\frac{b^5}{40}\operatorname{arsinh}\left(\frac{a}{|b|}\right)}} \]
\[ \dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\dfrac{a^5}{10}\ln\left(\dfrac{b}{|a|}+\sqrt{1+\dfrac{b^2}{a^2}}\,\right)-\dfrac{b^5}{40}\ln\left(\dfrac{a}{|b|}+\sqrt{\dfrac{a^2}{b^2}+1}\,\right) \]
\[ \iint\limits_{x^\overset{\,}{2}+y^2\le a^2} x^2\sqrt{x^2+y^2}{\rm\,d}x{\rm\,d}y \]
\[ \dfrac{\pi}{5}a^5 \]

1. \frac{3x^3y\sqrt{x^2+y^2}}{20}+\frac{xy^3\sqrt{x^2+y^2}}{40}+\frac{x^5}{10}\operatorname{arcsinh}\left(\frac{y}{\left|x\right|}\right)-\frac{y^5}{40}\operatorname{arcsinh}\left(\frac{x}{\left|y\right|}\right)
2. \frac{3x^3y\sqrt{x^2+y^2}}{20}+\frac{xy^3\sqrt{x^2+y^2}}{40}+\frac{x^5}{10}\sinh^{-1}\left(\frac{y}{\left|x\right|}\right)-\frac{y^5}{40}\sinh^{-1}\left(\frac{x}{\left|y\right|}\right)

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青青子衿

\[ \int_0^ax^2\int_0^b \sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}y{\rm\,d}x \]
...
青青子衿 发表于 2018-10-12 12:56

\begin{align*}
\int_0^b \sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}y&=\left.\left[\dfrac{1}{2}y\sqrt{x^2+y^2}+\dfrac{1}{2}x^2\ln\left(y+\sqrt{x^2+y^2}\,\right)-\dfrac{1}{2}x^2\ln\big|x\big|\right]\right|_0^b\\
&=\dfrac{b}{2}\sqrt{x^\overset{\,}{2}+b^2}+\dfrac{1}{2}x^2\ln\left(b+\sqrt{x^\overset{\,}{2}+b^2}\,\right)-\dfrac{1}{2}x^2\ln\big|x\big|\\
x^2\int_0^b \sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}y&=\frac{b}{2}x^2\sqrt{x^\overset{\,}{2}+b^2}+\frac{1}{2}x^4\ln\left(b+\sqrt{x^\overset{\,}{2}+b^2}\right)-\frac{1}{2}x^4\ln\big|x\big|\\
\end{align*}
\begin{align*}
\int_0^ax^2\int_0^b \sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}y{\rm\,d}x&=\frac{b}{2}\underbrace{\int_0^a x^2\sqrt{x^\overset{\,}{2}+b^2}{\rm\,d}x}_{I_{a,\,y\to\,x}}+\frac{1}{2}\underbrace{\int_0^a x^4\ln\left(b+\sqrt{x^\overset{\,}{2}+b^2}\right){\rm\,d}x}_{I_{b,\,y\to\,x}}-\frac{1}{2}\underbrace{\int_0^a x^4\ln\big|x\big|{\rm\,d}x}_{I_{c,\,y\to\,x}}\\
\end{align*}
\begin{align*}
\frac{b}{2}I_{a,\,y\to\,x}&=\frac{b}{2}\underbrace{\int_0^a x^2\sqrt{x^\overset{\,}{2}+b^2}{\rm\,d}x}_{I_{a,\,y\to\,x}}\\
&=\frac{b}{2}\left.\left[\dfrac{1}{4}x^3\sqrt{x^\overset{\,}{2}+b^2}+\dfrac{b^2}{8}x\sqrt{x^\overset{\,}{2}+b^2}-\dfrac{1}{8}b^4\ln\left(x+\sqrt{x^\overset{\,}{2}+b^2}\,\right)+\dfrac{1}{8}b^4\ln\big|b\big|\right]\right|_0^a\\
&=\frac{b}{2}\left[\dfrac{a^3\sqrt{a^\overset{\,}{2}+b^2}}{4}+\dfrac{ab^2\sqrt{a^\overset{\,}{2}+b^2}}{8}-\dfrac{1}{8}b^4\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)+\dfrac{1}{8}b^4\ln\big|b\big|\right]\\
&=\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{8}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{16}-\dfrac{b^5}{16}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)+\dfrac{b^5}{16}\ln\big|b\big|
\end{align*}
\begin{align*}
\frac{1}{2}I_{b,\,y\to\,x}&=\frac{1}{2}\underbrace{\int_0^a x^4\ln\left(b+\sqrt{x^\overset{\,}{2}+b^2}\right){\rm\,d}x}_{I_{b,\,y\to\,x}}\\
&=\frac{1}{2}\left.\left[
\begin{array}{}
\dfrac{b}{20}x^3\sqrt{x^\overset{\,}{2}+b^2}-\dfrac{3b^3}{40}x\sqrt{x^\overset{\,}{2}+b^2}+\dfrac{1}{5}x^5\ln\left(b+\sqrt{x^\overset{\,}{2}+b^2}\,\right)\\
+\dfrac{3b^5}{40}\ln\left(x+\sqrt{x^\overset{\,}{2}+b^2}\,\right)-\dfrac{1}{25}x^5-\dfrac{3b^5}{40}\ln\big|b\big|\\
\end{array}
\right]\right|_0^a\\
&=\frac{1}{2}\left[
\begin{array}{}
\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}-\dfrac{3ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\dfrac{a^5}{5}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)\\
+\dfrac{3b^5}{40}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)-\dfrac{a^5}{25}-\dfrac{3b^5}{40}\ln\big|b\big|\\
\end{array}
\right]\\
&=\begin{array}{}
\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{40}-\dfrac{3ab^3\sqrt{a^\overset{\,}{2}+b^2}}{80}+\dfrac{a^5}{10}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)\\
+\dfrac{3b^5}{80}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)-\dfrac{a^5}{50}-\dfrac{3b^5}{80}\ln\big|b\big|\\
\end{array} \\
\end{align*}
\begin{align*}
-\frac{1}{2}I_{c,\,y\to\,x}&=-\frac{1}{2}\underbrace{\int_0^a x^4\ln\big|x\big|{\rm\,d}x}_{I_{c,\,y\to\,x}}\\
&=-\frac{1}{2}\left.\left[
-\dfrac{1}{25}x^5+\dfrac{1}{5}x^5\ln\big|x\big|
\right]\right|_0^a\\
&=-\frac{1}{2}\left[
-\dfrac{a^5}{25}+\dfrac{a^5}{5}\ln\big|a\big|\right]\\
&=\dfrac{a^5}{50}-\dfrac{a^5}{10}\ln\big|a\big| \\
\end{align*}

\begin{align*}
\frac{b}{2}I_{a,\,y\to\,x}&=\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{8}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{16}-\dfrac{b^5}{16}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)+\dfrac{b^5}{16}\ln\big|b\big| \\
\frac{1}{2}I_{b,\,y\to\,x}&=
\begin{array}{r}
\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{40}-\dfrac{3ab^3\sqrt{a^\overset{\,}{2}+b^2}}{80}+\dfrac{a^5}{10}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right) \\
+\dfrac{3b^5}{80}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)\\
-\dfrac{a^5}{50}-\dfrac{3b^5}{80}\ln\big|b\big|
\end{array}\\
-\frac{1}{2}I_{c,\,y\to\,x}&=\dfrac{a^5}{50}-\dfrac{a^5}{10}\ln\big|a\big|\\
\int_0^ax^2\int_0^b \sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}y{\rm\,d}x
&=\begin{array}{r}   
\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\dfrac{a^5}{10}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)-\dfrac{a^5}{10}\ln\big|a\big|\\   
-\dfrac{b^5}{40}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)+\dfrac{b^5}{40}\ln\big|b\big|\\   
\end{array} \\
&={\color{red}{\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\dfrac{a^5}{10}\ln\left(\dfrac{b}{|a|}+\sqrt{1+\dfrac{b^2}{a^2}}\,\right)-\dfrac{b^5}{40}\ln\left(\dfrac{a}{|b|}+\sqrt{\dfrac{a^2}{b^2}+1}\,\right)}}\\
&={\color{red}{\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\frac{a^5}{10}{\color{blue}{\operatorname{arsinh}}}\left(\frac{b}{|a|}\right)-\frac{b^5}{40}{\color{blue}{\operatorname{arsinh}}}\left(\frac{a}{|b|}\right)}}\\
\end{align*}

青青子衿

\[ \int_0^b\int_0^ax^2\sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}x{\rm\,d}y \]
...
青青子衿 发表于 2018-10-12 12:56

\begin{align*}
\int_0^ax^2\sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}x&=\left.\left[\dfrac{1}{4}x^3\sqrt{x^\overset{\,}{2}+y^2}+\dfrac{1}{8}xy^2\sqrt{x^\overset{\,}{2}+y^2}-\dfrac{1}{8}y^4\ln\left(x+\sqrt{x^2+y^2}\,\right)+\dfrac{1}{8}y^4\ln\big|y\big|\right]\right|_0^a\\
&=\dfrac{a^3}{4}\sqrt{a^\overset{\,}{2}+y^2}+\dfrac{a}{8}y^2\sqrt{a^\overset{\,}{2}+y^2}-\dfrac{1}{8}y^4\ln\left(a+\sqrt{a^2+y^2}\,\right)+\dfrac{1}{8}y^4\ln\big|y\big|\\
\end{align*}
\begin{align*}
\int_0^b\int_0^a x^2\sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}x{\rm\,d}y&=\dfrac{a^3}{4}\underbrace{\int_0^b \sqrt{a^\overset{\,}{2}+y^2}{\rm\,d}y}_{I_{a,\,x\to\,y}}+\frac{a}{8}\underbrace{\int_0^b y^2\sqrt{a^\overset{\,}{2}+y^2}{\rm\,d}y}_{I_{b,\,x\to\,y}}-\frac{1}{8}\underbrace{\int_0^b y^4\ln\left(a+\sqrt{a^\overset{\,}{2}+y^2}\right){\rm\,d}y}_{I_{c,\,x\to\,y}}+\frac{1}{8}\underbrace{\int_0^b y^4\ln\big|y\big|{\rm\,d}y}_{I_{d,\,x\to\,y}}\\
\end{align*}
\begin{align*}
\frac{a^3}{4}I_{a,\,x\to\,y}&=\dfrac{a^3}{4}\underbrace{\int_0^b \sqrt{a^\overset{\,}{2}+y^2}{\rm\,d}y}_{I_{a,\,x\to\,y}}\\
&=\frac{a^3}{4}\left.\left[\dfrac{1}{2}y\sqrt{a^\overset{\,}{2}+y^2}+\dfrac{a^2}{2}\ln\left(y+\sqrt{a^\overset{\,}{2}+y^2}\,\right)-\dfrac{a^2}{2}\ln\big|a\big|\right]\right|_0^b\\
&=\frac{a^3}{4}\left[\dfrac{b\sqrt{a^\overset{\,}{2}+b^2}}{2}+\dfrac{a^2}{2}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)-\dfrac{a^2}{2}\ln\big|a\big|\right]\\
&=\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{8}+\dfrac{a^5}{8}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)-\dfrac{a^5}{8}\ln\big|a\big|\\
\end{align*}
\begin{align*}  
\frac{a}{8}I_{b,\,x\to\,y}&=\dfrac{a}{8}\underbrace{\int_0^b y^2\sqrt{a^\overset{\,}{2}+y^2}{\rm\,d}y}_{I_{b,\,x\to\,y}}\\  
&=\frac{a}{8}\left.\left[\dfrac{a^2}{8}y\sqrt{a^\overset{\,}{2}+y^2}+\dfrac{1}{4}y^3\sqrt{a^\overset{\,}{2}+y^2}-\dfrac{a^4}{8}\ln\left(y+\sqrt{a^\overset{\,}{2}+y^2}\,\right)+\dfrac{a^4}{8}\ln\big|a\big|\right]\right|_0^b\\  
&=\frac{a}{8}\left[\dfrac{a^2b\sqrt{a^\overset{\,}{2}+b^2}}{8}+\dfrac{b^3\sqrt{a^\overset{\,}{2}+b^2}}{4}-\dfrac{a^4}{8}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)+\dfrac{a^4}{8}\ln\big|a\big|\right]\\  
&=\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{64}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{32}-\dfrac{a^5}{64}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)+\dfrac{a^5}{64}\ln\big|a\big|\\
\end{align*}
\begin{align*}  
-\frac{1}{8}I_{c,\,x\to\,y}&=-\frac{1}{8}\underbrace{\int_0^b y^4\ln\left(a+\sqrt{a^\overset{\,}{2}+y^2}\right){\rm\,d}y}_{I_{c,\,x\to\,y}}\\  
&=-\frac{1}{8}\left.\left[  
\begin{array}{}  
-\dfrac{3a^3}{40}y\sqrt{a^\overset{\,}{2}+y^2}+\dfrac{a}{20}y^3\sqrt{a^\overset{\,}{2}+y^2}+\dfrac{3a^5}{40}\ln\left(y+\sqrt{a^\overset{\,}{2}+y^2}\,\right)\\  
+\dfrac{1}{5}y^5\ln\left(a+\sqrt{a^\overset{\,}{2}+y^2}\,\right)-\dfrac{1}{25}y^5-\dfrac{3a^5}{40}\ln\big|a\big|\\  
\end{array}  
\right]\right|_0^b\\  
&=-\frac{1}{8}\left[  
\begin{array}{}  
-\dfrac{3a^3b}{40}\sqrt{a^\overset{\,}{2}+b^2}+\dfrac{ab^3}{20}\sqrt{a^\overset{\,}{2}+b^2}+\dfrac{3a^5}{40}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)\\  
+\dfrac{b^5}{5}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)-\dfrac{b^5}{25}-\dfrac{3a^5}{40}\ln\big|a\big|\\  
\end{array}  
\right]\\  
&=\begin{array}{}  
\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{320}-\,\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{160}-\dfrac{3a^5}{320}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)\\  
-\dfrac{b^5}{40}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)+\dfrac{b^5}{200}+\dfrac{3a^5}{320}\ln\big|a\big|\\  
\end{array} \\  
\end{align*}
\begin{align*}  
\frac{1}{8}I_{d,\,x\to\,y}&=\frac{1}{8}\underbrace{\int_0^b y^4\ln\big|y\big|{\rm\,d}y}_{I_{d,\,x\to\,y}}\\  
&=\frac{1}{8}\left.\left[  
-\dfrac{1}{25}y^5+\dfrac{1}{5}y^5\ln\big|y\big|
\right]\right|_0^b\\  
&=\frac{1}{8}\left[  
-\dfrac{b^5}{25}+\dfrac{b^5}{5}\ln\big|b\big|\right]\\  
&=-\dfrac{b^5}{200}+\dfrac{b^5}{40}\ln\big|b\big| \\  
\end{align*}

\begin{align*}  
\frac{a^3}{4}I_{a,\,x\to\,y}&=\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{8}+\dfrac{a^5}{8}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)-\dfrac{a^5}{8}\ln\big|a\big| \\  
\frac{a}{8}I_{b,\,x\to\,y}&=
\begin{array}{r}   
\dfrac{a^3b\sqrt{a^\overset{\,}{2}+b^2}}{64}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{32}-\dfrac{a^5}{64}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)\\
+\dfrac{a^5}{64}\ln\big|a\big|\\   
\end{array}\\  
-\frac{1}{8}I_{c,\,x\to\,y}&=
\begin{array}{r}   
\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{320}-\,\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{160}-\dfrac{3a^5}{320}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)\\   
-\dfrac{b^5}{40}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)\\
+\dfrac{b^5}{200}+\dfrac{3a^5}{320}\ln\big|a\big|\\   
\end{array}\\  
\frac{1}{8}I_{d,\,x\to\,y}&=-\dfrac{b^5}{200}+\dfrac{b^5}{40}\ln\big|b\big|\\
\int_0^b\int_0^a x^2\sqrt{x^\overset{\,}{2}+y^2}{\rm\,d}x{\rm\,d}y  
&=\begin{array}{r}   
\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\dfrac{a^5}{10}\ln\left(b+\sqrt{a^\overset{\,}{2}+b^2}\,\right)-\dfrac{a^5}{10}\ln\big|a\big|\\   
-\dfrac{b^5}{40}\ln\left(a+\sqrt{a^\overset{\,}{2}+b^2}\,\right)+\dfrac{b^5}{40}\ln\big|b\big|\\   
\end{array} \\
&={\color{red}{\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\dfrac{a^5}{10}\ln\left(\dfrac{b}{|a|}+\sqrt{1+\dfrac{b^2}{a^2}}\,\right)-\dfrac{b^5}{40}\ln\left(\dfrac{a}{|b|}+\sqrt{\dfrac{a^2}{b^2}+1}\,\right)}}\\
&={\color{red}{\dfrac{3a^3b\sqrt{a^\overset{\,}{2}+b^2}}{20}+\dfrac{ab^3\sqrt{a^\overset{\,}{2}+b^2}}{40}+\frac{a^5}{10}{\color{blue}{\operatorname{arsinh}}}\left(\frac{b}{|a|}\right)-\frac{b^5}{40}{\color{blue}{\operatorname{arsinh}}}\left(\frac{a}{|b|}\right)}}\\
\end{align*}

kuing

代码 \$ x^\overset{\,}{2} \$ 这样放在 latex 里会报错，但在 mathjax 里竟然没问题。
$\sqrt{x^2+y^2}\text{ vs }\sqrt{x^\overset{\,}{2}+y^2}$
\[\sqrt{x^2+y^2}\text{ vs }\sqrt{x^\overset{\,}{2}+y^2}\]
KaTeX 对代码的要求更高，所以 KaTeX 不解析，必须再加一个 { } 才行，而且效果也更接近 latex：
`\sqrt{x^2+y^2}\text{ vs }\sqrt{x^{\overset{\,}{2}}+y^2}`
\[\sqrt{x^2+y^2}\text{ vs }\sqrt{x^{\overset{\,}{2}}+y^2}\]

player1703

第3个用极坐标很简单:
\begin{align*}
&\iint\limits_{x^2+y^2\leq a^2}x^2\sqrt{x^2+y^2}\mathrm{d}x\mathrm{d}y \\
=&\int_0^{2\pi}\mathrm{d}\theta\int_0^a\rho^2\cos^2\theta\rho\rho \mathrm{d}\rho \\
=&\int_0^a\rho^4\mathrm{d}\rho\int_0^{2\pi}\cos^2\theta\mathrm{d}\theta \\
=&\;\frac{a^5}{5}\pi
\end{align*}

player1703

回复 5# 青青子衿
其实第1个用极坐标也可以:
记$I=\int_0^{a}\int_0^bx^2\sqrt{x^2+y^2}\rmd y\rmd x$
首先假设$a,b\gt 0$
\begin{align*}
I=&\int_0^{a}\int_0^bx^2\sqrt{x^2+y^2}\rmd y\rmd x \\
=&\int_0^{\arctan\frac{b}{a}}\int_0^{a\sec\theta}\rho^2\cos^2\theta\rho\rho \rmd\rho\rmd\theta + \int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}}\int_0^{b\csc\theta}\rho^2\cos^2\theta\rho\rho \rmd\rho\rmd\theta\\
=&\int_0^{\arctan\frac{b}{a}}\cos^2\theta\int_0^{a\sec\theta}\rho^4\rmd\rho\rmd\theta + \int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}}\cos^2\theta\int_0^{b\csc\theta}\rho^4\rmd\rho\rmd\theta\\
=&\frac{a^5}{5}\int_0^{\arctan\frac{b}{a}}\sec^3\theta\rmd\theta+\frac{b^5}{5}\underbrace{\int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}}\cot^2\theta\csc^3\theta\rmd\theta}_{I_2} \\
=&\frac{a^5}{5}\times\frac{1}{2}\left(\sec\theta\tan\theta+\ln\left|\sec\theta+\tan\theta\right|\right)\biggr\rvert_0^{\arctan\frac{b}{a}} + \frac{b^5}{5}I_2 \\
=&\frac{a^5}{10}\left(\frac{b\sqrt{a^2+b^2}}{a^2}+\ln\left(\sqrt{1+\frac{b^2}{a^2}}+\frac{b}{a}\right)\right)+\frac{b^5}{5}I_2 \\
I_2=& \int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}}\cot^2\theta\csc^3\theta\rmd\theta \\
=&-\int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}}\cot\theta\csc^2\theta\rmd\csc\theta \\
=&-\cot\theta\csc^3\theta\biggr\rvert_{\arctan\frac{b}{a}}^{\frac{\pi}{2}} + \int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}}\csc\theta\left(-\csc^4\theta-2\cot\theta\csc\theta\csc\theta\cot\theta\right)\rmd\theta \\
=&\frac{a}{b}\frac{\left(a^2+b^2\right)\sqrt{a^2+b^2}}{b^3} -\int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}} \left(1+\cot^2\theta\right)\csc^3\theta\rmd\theta-2\int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}}\cot^2\theta\csc^3\theta\rmd\theta \\
=&\frac{a^3\sqrt{a^2+b^2}}{b^4} + \frac{a\sqrt{a^2+b^2}}{b^2} -\int_{\arctan\frac{b}{a}}^{\frac{\pi}{2}} \csc^3\theta\rmd\theta-3I_2 \\
I_2=&\frac{a^3\sqrt{a^2+b^2}}{4b^4} + \frac{a\sqrt{a^2+b^2}}{4b^2} - \frac{1}{4}\times\frac{1}{2}\left(-\csc\theta\cot\theta\biggr\rvert_{\arctan\frac{b}{a}}^\frac{\pi}{2}+\ln\left|\csc\theta -\cot\theta\right|\biggr\rvert_{\arctan\frac{b}{a}}^\frac{\pi} {2}\right)\\
=&\frac{a^3\sqrt{a^2+b^2}}{4b^4} + \frac{a\sqrt{a^2+b^2}}{4b^2} - \frac{1}{8}\left(\frac{a\sqrt{a^2+b^2}}{b^2}-\ln\left(\sqrt{1+\frac{a^2}{b^2}}-\frac{a}{b}\right)\right) \\
I =&\frac{3a^3b\sqrt{a^2+b^2}}{20}+\frac{ab^3\sqrt{a^2+b^2}}{40}+\frac{a^5}{10}\ln\left(\sqrt{1+\frac{b^2}{a^2}}+\frac{b}{a}\right)+\frac{b^5}{40}\ln\left(\sqrt{1+\frac{a^2}{b^2}}-\frac{a}{b}\right) \\
＝&\frac{3a^3b\sqrt{a^2+b^2}}{20}+\frac{ab^3\sqrt{a^2+b^2}}{40}+\frac{a^5}{10}\mathrm{arsh}\left(\frac{b}{a}\right)+\frac{b^5}{40}\mathrm{arsh}\left(-\frac{a}{b}\right) \\
＝&\frac{3a^3b\sqrt{a^2+b^2}}{20}+\frac{ab^3\sqrt{a^2+b^2}}{40}+\frac{a^5}{10}\mathrm{arsh}\left(\frac{b}{a}\right)-\frac{b^5}{40}\mathrm{arsh}\left(\frac{a}{b}\right)
\end{align*}
\begin{align}
\therefore I＝&\frac{3a^3b\sqrt{a^2+b^2}}{20}+\frac{ab^3\sqrt{a^2+b^2}}{40}+\frac{a^5}{10}\mathrm{arsh}\left(\frac{b}{a}\right)-\frac{b^5}{40}\mathrm{arsh}\left(\frac{a}{b}\right)\quad\left(a,b\gt0\right) \label{eq1}
\end{align}
若a, b至少有一个为$0$显然$I=0$.
将表达式 \ref{eq1} 中的$a,b$分别替换成$\abs{a},\abs{b}$, 记为$I_0$.
根据被积函数的对称性,若$ab\gt0$ 则$\int_0^{a}\int_0^bx^2\sqrt{x^2+y^2}\rmd y\rmd x=I_0$,若$ab\lt0$则$\int_0^{a}\int_0^bx^2\sqrt{x^2+y^2}\rmd y\rmd x=-I_0$ 所以最终结果可以写成:
\begin{align}
I\left(a,b\right) = \led
& \frac{3a^3b\sqrt{a^2+b^2}}{20}+\frac{ab^3\sqrt{a^2+b^2}}{40}+ \frac{a^5}{10}\mathrm{arsh}\left(\frac{b}{\abs{a}}\right)-\frac{b^5}{40}\mathrm{arsh}\left(\frac{a}{\abs{b}}\right), && ab\ne0 \\
& 0, && ab=0
\endled
\end{align}