$\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2} dx dydz$

其妙

计算三重积分：$\displaystyle\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\sqrt{x^2+y^2+z^2}{\kern 5pt }{\rm dx}{\rm dy}{\rm dz}$

isee

难道在进修？

其妙

回复 2# isee
没有，
也不想进修

pxchg1200

这不是西西的题么。。。。

pxchg1200

其妙

回复 5# pxchg1200
大牛笔！

nijupeng

-(\[Pi]/24) + 1/4 (Sqrt[3] + Log[7 + 4 Sqrt[3]])

其妙

回复 7# nijupeng

没过程

青青子衿

回复 8# 其妙

回复 7# nijupeng
没过程
其妙 发表于 2015-8-16 16:07

应该是软件算的！
而且更像是Mathematica，因为Mathematica的语言代码的首字母都是大写的，并且Mathematica对\(\ln\)没有定义,只有\(Log\)

其妙

回复 9# 青青子衿

血狼王

这个积分……好像只能一层一层去算了，区域不是柱状体或球体。
要不要我一步一步弄出来？
好像已经有人了

青青子衿

回复 10# 其妙

\[ \int_0^a\frac{1}{\sqrt{x^2+y^2+z^2}}{\rm\,d}x=\ln\left(\dfrac{a}{\sqrt{y^2+z^2}}+\sqrt{\dfrac{a^2}{y^2+z^2}+1}\,\right) \]

\[ \int_0^a\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x=\dfrac{a}{2}\sqrt{a^{\overset{\,}{2}}+y^2+z^2}+\dfrac{1}{2}\left(y^2+z^2\right)\ln\left(\dfrac{a}{\sqrt{y^2+z^2}}+\sqrt{\dfrac{a^2}{y^2+z^2}+1}\,\right) \]

\[ \int_0^a\bigg(x^2+y^2+z^2\bigg)^{\frac{3}{2}}{\rm\,d}x=\dfrac{a}{8}\left(2a^2+5y^2+5z^2\right)\sqrt{a^{\overset{\,}{2}}+y^2+z^2}+\dfrac{3}{8}\left(y^2+z^2\right)^2\ln\left(\dfrac{a}{\sqrt{y^2+z^2}}+\sqrt{\dfrac{a^2}{y^2+z^2}+1}\,\right) \]
\[ \int_0^b\int_0^a\dfrac{1}{\sqrt{x^2+y^2+z^2}}{\rm\,d}x{\rm\,d}y=
\begin{array}{r}
-z\arctan\left(\dfrac{ab}{z\sqrt{a^2+b^2+z^2}}\right)\\
+a\ln\left(\dfrac{b}{\sqrt{a^2+z^2}}+\sqrt{1+\dfrac{b^2}{a^2+z^2}}\,\right)\\
+b\ln\left(\dfrac{a}{\sqrt{b^2+z^2}}+\sqrt{\dfrac{a^2}{b^2+z^2}+1}\,\right)\\
\end{array} \]
\[ \int_0^b\int_0^a\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y=
\begin{array}{r}
\dfrac{ab}{3}\sqrt{a^{\overset{\,}{2}}+b^2+z^2}-\dfrac{1}{3}z^3\arctan\left(\dfrac{ab}{z\sqrt{a^2+b^2+z^2}}\right)\\
+\dfrac{a}{6}\left(a^2+3z^2\right)\ln\left(\dfrac{b}{\sqrt{a^2+z^2}}+\sqrt{1+\dfrac{b^2}{a^2+z^2}}\,\right)\\
+\dfrac{b}{6}\left(b^2+3z^2\right)\ln\left(\dfrac{a}{\sqrt{b^2+z^2}}+\sqrt{\dfrac{a^2}{b^2+z^2}+1}\,\right)\\
\end{array}
\]
\[ \int_0^b\int_0^a\bigg(x^2+y^2+z^2\bigg)^{\frac{3}{2}}{\rm\,d}x{\rm\,d}y=
\begin{array}{r}
\dfrac{ab}{40}\left(7a^2+7b^2+18z^2\right)\sqrt{a^{\overset{\,}{2}}+b^2+z^2}-\dfrac{1}{5}z^5\arctan\left(\dfrac{ab}{z\sqrt{a^2+b^2+z^2}}\right)\\
+\dfrac{a}{40}\left(3a^4+10a^2z^2+15z^4\right)\ln\left(\dfrac{b}{\sqrt{a^2+z^2}}+\sqrt{1+\dfrac{b^2}{a^2+z^2}}\,\right)\\
+\dfrac{b}{40}\left(3b^4+10b^2z^2+15z^4\right)\ln\left(\dfrac{a}{\sqrt{b^2+z^2}}+\sqrt{\dfrac{a^2}{b^2+z^2}+1}\,\right)\\
\end{array} \]
\[ \int_0^c\int_0^b\int_0^a\dfrac{1}{\sqrt{x^2+y^2+z^2}}{\rm\,d}x{\rm\,d}y{\rm\,d}z=
\begin{array}{r}
ab\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+ac\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)
+bc\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^2}{2}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^2}{2}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^2}{2}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array} \]
\[ \int_0^c\int_0^b\int_0^a\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z=
\begin{array}{r}
\frac{abc}{4}\sqrt{a^2+b^2+c^2}+\frac{ab\left(a^2+b^2\right)}{6}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+\frac{ac\left(a^2+c^2\right)}{6}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)
+\frac{bc\left(b^2+c^2\right)}{6}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^4}{12}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^4}{12}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^4}{12}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array}
\]

\[ \int_0^c\int_0^b\int_0^a\bigg(x^2+y^2+z^2\bigg)^{\frac{3}{2}}{\rm\,d}x{\rm\,d}y{\rm\,d}z=
\begin{array}{r}
\frac{2abc\left(a^2+b^2+c^2\right)}{15}\sqrt{a^2+b^2+c^2}+\frac{ab\left(9a^4+10a^2b^2+9b^4\right)}{120}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+\frac{ac\left(9a^4+10a^2c^2+9c^4\right)}{120}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)\\
+\frac{bc\left(9b^4+10b^2c^2+9c^4\right)}{120}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^6}{30}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^6}{30}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^6}{30}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array}
\]

青青子衿

回复 5# pxchg1200

pxchg1200 发表于 2014-6-4 09:02

七楼给的答案是正确的，五楼的结果有错误

错误还不只一处！！！
积分计算表达式的“第四行”有错误、“第十二行”到“第十三行”的过程也有错误：

2. \frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\left(\cos\varphi\right)^4}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\frac{1}{\left(\sin\theta\right)^3}d\theta d\varphi
3. \frac{3}{8}\int_0^{\frac{\pi}{4}}\left(\frac{\sqrt{1+\left(\cos\varphi\right)^2}}{\left(\cos\varphi\right)^3}+\frac{\ln\left(\left(\cos\varphi\right)+\sqrt{1+\left(\cos\varphi\right)^2}\right)}{\left(\cos\varphi\right)^4}\right)d\varphi

复制代码

\[
\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\left(\cos\varphi\right)^4}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\frac{1}{\sin\theta}d\theta d\varphi\\
\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\left(\cos\varphi\right)^4}\int_0^{\frac{\pi}{2}-\arctan\left(\sec\varphi\right)}\frac{1}{\cos\theta}d\theta d\varphi
\]

青青子衿

回复 13# 青青子衿
\begin{align*}
\int_0^1\int_0^1\int_0^1\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z
&=\,\iiint\limits_{\substack{0\,\leqslant\,x\,\leqslant\,1 \\ 0\,\leqslant\,y\,\leqslant\,1\\ 0\,\leqslant\,z\,\leqslant\,1}}\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\
&=\,{\color{red}{3}}\iiint\limits_{\substack{0\,\leqslant\,y\,\leqslant\,x\,\leqslant\,1 \\ 0\,\leqslant\,z\,\leqslant\,x\,\leqslant\,1}}\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\
\,\\
\overset{\begin{cases}
x=r\sin\theta\cos\varphi\\
y=r\sin\theta\sin\varphi\\
z=r\cos\theta\\
\end{cases}}{\overline{\overline{\hspace{4cm}}}}\quad&\,3\iiint\limits_{V_1}r^3\sin\theta{\rm\,d}r{\rm\,d}\theta{\rm\,d}\varphi\\
&=3\int_0^{\frac{\pi}{4}}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sin\theta\cos\varphi}}r^3\sin\theta{\rm\,d}r{\rm\,d}\theta{\rm\,d}\varphi\\
&=3\int_0^{\frac{\pi}{4}}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\frac{1}{4\sin^4\theta\,\cos^4\varphi}\sin\theta{\rm\,d}\theta{\rm\,d}\varphi\\
&=\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\cos^4\varphi}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\frac{1}{\sin^3\theta}{\rm\,d}\theta{\rm\,d}\varphi\\
&=\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\cos^4\varphi}\int_{0}^{\frac{\pi}{2}-\arctan\left(\sec\varphi\right)}\frac{1}{\cos^3\theta}{\rm\,d}\theta{\rm\,d}\varphi\\
&=\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\cos^4\varphi}\int_{0}^{\frac{\pi}{2}-\arctan\left(\sec\varphi\right)}\frac{1}{\cos\theta}{\rm\,d}(\tan\theta){\rm\,d}\varphi\\
&=\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\cos^4\varphi}\int_{0}^{\frac{\pi}{2}-\arctan\left(\sec\varphi\right)}\sqrt{1+\tan^{\overset{\,}{2}}\theta}{\rm\,d}(\tan\theta){\rm\,d}\varphi\\
&=\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\cos^4\varphi}\int_{0}^{\frac{1}{\sec\varphi}}\sqrt{1+u^{\overset{\,}{2}}}{\rm\,d}u{\rm\,d}\varphi\\
&=\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\cos^4\varphi}\int_{0}^{\cos\varphi}\sqrt{1+u^{\overset{\,}{2}}}{\rm\,d}u{\rm\,d}\varphi\\
&=\frac{3}{4}\int_0^{\frac{\pi}{4}}\frac{1}{\cos^4\varphi}\left(\frac{\cos\varphi\,\sqrt{1+\cos^{\overset{\,}{2}}\varphi}}{2}+\frac{\ln\left(\cos\varphi+\sqrt{1+\cos^{\overset{\,}{2}}\varphi}\,\right)}{2}\right){\rm\,d}\varphi\\
\end{align*}

\begin{align*}  
\int_0^1\int_0^1\int_0^1\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\kern 2pt}{\rm\,d}x{\rm\,d}y{\rm\,d}z   
&=\frac{3}{8}\int_0^{\frac{\pi}{4}}\left(\frac{\sqrt{1+\cos^{\overset{\,}{2}}\varphi}}{\cos^3\varphi}+\frac{\ln\left(\cos\varphi+\sqrt{1+\cos^{\overset{\,}{2}}\varphi}\,\right)}{\cos^4\varphi}\right){\rm\,d}\varphi\\   
&=\frac{3}{8}\int_0^{\frac{\pi}{4}}\frac{\sqrt{1+\cos^{\overset{\,}{2}}\varphi}}{\cos^3\varphi}{\rm\,d}\varphi+\frac{3}{8}\int_0^{\frac{\pi}{4}}\frac{\ln\left(\cos\varphi+\sqrt{1+\cos^{\overset{\,}{2}}\varphi}\,\right)}{\cos^4\varphi}{\rm\,d}\varphi\\  
&=\frac{3}{8}\left(\,\underline{\frac{\sqrt{3}}{2}+\ln\left(1+\sqrt{3}\,\right)-\frac{1}{2}\ln2}
+\underline{\frac{\sqrt{3\,}}{6}-\frac{\pi}{9}+\frac{5}{3}\ln\left(1+\sqrt{3}\,\right)-\frac{5}{6}\ln2}\,\right)\\  
&=\frac{3}{8}\left(\frac{2\sqrt{3}}{3}-\frac{\pi}{9}+\frac{8}{3}\ln\left(1+\sqrt{3}\,\right)-\frac{4}{3}\ln2\right)\\  
&=\frac{\sqrt{3}}{4}-\frac{\pi}{24}+\ln\left(1+\sqrt{3}\,\right)-\frac{1}{2}\ln2\\
&=\frac{\sqrt{3}}{4}-\frac{\pi}{24}+\ln\left(\frac{1}{\sqrt{2}}+\sqrt{1+\frac{1}{2}}\,\right)=\color{red}{\frac{\sqrt{3}}{4}-\frac{\pi}{24}+\operatorname{arsinh}\left(\frac{1}{\sqrt{2}}\right)}\\
&=\color{red}{\frac{\sqrt{3}}{4}-\frac{\pi}{24}+\ln\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)}\\  
&=\color{red}{\frac{\sqrt{3}}{4}-\frac{\pi}{24}+\frac{1}{4}\ln\left(7+4\sqrt{3}\,\right)}\\
\end{align*}
.

1. \int_0^1\int_0^1\int_0^1\sqrt{x^2+y^2+z^2}dxdydz
2. \frac{\sqrt{3}}{4}-\frac{\pi}{24}+\frac{1}{4}\ln\left(7+4\sqrt{3}\right)

复制代码

\begin{align*}
\int_0^{\frac{\pi}{4}}\frac{\sqrt{1+\cos^{\overset{\,}{2}}\varphi}}{\cos^3\varphi}{\rm\,d}\varphi
&=\int_0^{\frac{\pi}{4}}\frac{\sqrt{1+\cos^{\overset{\,}{2}}\varphi}}{\cos\varphi}{\rm\,d}(\tan\varphi)\\
&=\int_0^{\frac{\pi}{4}}\sqrt{1+\frac{1}{\cos^2\varphi}}{\rm\,d}(\tan\varphi)\\
&=\int_0^{\frac{\pi}{4}}\sqrt{1+1+\tan^\overset{\,}{2}\varphi}{\rm\,d}(\tan\varphi)\\
&=\int_0^1\sqrt{2+v^\overset{\,}{2}}{\rm\,d}v=\frac{\sqrt{3}}{2}+\ln\left(\frac{1}{\sqrt2}+\sqrt{1+\frac{1}{2}}\,\right)\\
&=\frac{\sqrt{3}}{2}+\ln\left(1+\sqrt{3}\,\right)-\frac{1}{2}\ln2\\
\end{align*}
...

1. \int_0^{\frac{\pi}{4}}\frac{\sqrt{1+\left(\cos\varphi\right)^2}}{\left(\cos\varphi\right)^3}d\varphi
2. \frac{\sqrt{3}}{2}+\ln\left(1+\sqrt{3}\right)-\frac{1}{2}\ln2

复制代码

\begin{align*}   
\int_0^{\frac{\pi}{4}}\frac{\ln\left(\cos\varphi+\sqrt{1+\cos^{\overset{\,}{2}}\varphi}\,\right)}{\cos^4\varphi}{\rm\,d}\varphi   
&=\int_0^{\frac{\pi}{4}}\frac{\ln\left(\cos\varphi+\sqrt{1+\cos^{\overset{\,}{2}}\varphi}\,\right)}{\cos^2\varphi}{\rm\,d}(\tan\varphi)\\   
&=\int_0^{\frac{\pi}{4}}\left(1+\tan^2\varphi\right)\ln\left(\frac{1}{\sqrt{1+\tan^2\varphi}}+\sqrt{1+\frac{1}{1+\tan^2\varphi}}\,\right){\rm\,d}(\tan\varphi)\\   
&=\int_0^1\left(1+v^2\right)\ln\left(\frac{1}{\sqrt{1+v^2}}+\sqrt{1+\frac{1}{1+v^2}}\,\right){\rm\,d}v\\   
&=\int_0^1\left(1+v^2\right)\ln\left(1+\sqrt{1+v^{\overset{\,}{2}}+1}\,\right){\rm\,d}v-\frac{1}{2}\int_0^1\left(1+v^2\right)\ln\left(1+v^2\right){\rm\,d}v\\  
&=\int_0^1\left(1+v^2\right)\ln\left(1+\sqrt{2+v^{\overset{\,}{2}}}\,\right){\rm\,d}v-\frac{1}{2}\int_0^1\left(1+v^2\right)\ln\left(1+v^2\right){\rm\,d}v\\
&=\boxed{-\frac{7}{9}+\frac{\sqrt{3\,}}{6}+\frac{\pi}{18}+\frac{5}{3}\ln\left(1+\sqrt{3}\,\right)-\frac{1}{6}\ln2\,}-\frac{1}{2}\boxed{\left(\frac{4}{3}\ln2-\frac{14}{9}+\frac{\pi}{3}\right)}\\
&=\frac{\sqrt{3\,}}{6}-\frac{\pi}{9}+\frac{5}{3}\ln\left(1+\sqrt{3}\,\right)-\frac{5}{6}\ln2
\end{align*}
...

1. \int_0^{\frac{\pi}{4}}\frac{\ln\left(\cos\varphi+\sqrt{1+\left(\cos\varphi\right)^2}\right)}{\left(\cos\varphi\right)^4}d\varphi
2. \frac{\sqrt{3}}{6}-\frac{\pi}{9}+\frac{5}{3}\ln\left(1+\sqrt{3}\right)-\frac{5}{6}\ln2

复制代码

\begin{align*}      
\int_0^1\left(1+v^2\right)\ln\left(1+\sqrt{2+v^{\overset{\,}{2}}}\,\right){\rm\,d}v      
&=\int_0^1\ln\left(1+\sqrt{2+v^{\overset{\,}{2}}}\,\right){\rm\,d}\left(v+\frac{v^3}{3}\right)\\      
&=\left.\left(v+\frac{v^3}{3}\right)\ln\left(1+\sqrt{2+v^{\overset{\,}{2}}}\,\right)\right|_0^1-\int_0^1\left(v+\frac{v^3}{3}\right){\rm\,d}\left[\ln\left(1+\sqrt{2+v^{\overset{\,}{2}}}\,\right)\right]\\      
&=\frac{4}{3}\ln\left(1+\sqrt{3}\,\right)-\int_0^1\left(v+\frac{v^3}{3}\right)\frac{v}{\sqrt{2+v^2}\left(\sqrt{2+v^2}+1\right)}{\rm\,d}v\\     
&=\frac{4}{3}\ln\left(1+\sqrt{3}\,\right)-\int_0^{\operatorname{arsinh}\frac{1}{\sqrt{2\,}}}\frac{\sqrt{2\,}\sinh t\left(\sqrt{2\,}\sinh t+\frac{2\sqrt{2\,}\sinh^3t}{3}\right)}{\sqrt{2+2\sinh^2t}\left(\sqrt{2+2\sinh^2t}+1\right)}{\rm\,d}\left(\sqrt{2\,}\sinh t\right)\\      
&=\frac{4}{3}\ln\left(1+\sqrt{3}\,\right)-\int_0^{\ln\left(1+\sqrt{3}\right)-\frac{1}{2}\ln2}\frac{2\sinh^2t+\frac{4\sinh^4t}{3}}{\sqrt{2\,}\cosh t\left(\sqrt{2\,}\cosh t+1\right)}\sqrt{2\,}\cosh t{\rm\,d}t\\   
&=\frac{4}{3}\ln\left(1+\sqrt{3}\,\right)-\int_0^{\ln\left(\frac{1}{\sqrt{2\,}}+\sqrt{\frac{3}{2}}\,\right)}\frac{2\sinh^2t+\frac{4\sinh^4t}{3}}{\sqrt{2\,}\cosh t+1}{\rm\,d}t\\   
&=\frac{4}{3}\ln\left(1+\sqrt{3}\,\right)-\int_0^{\ln\left(\frac{1}{\sqrt{2\,}}+\sqrt{\frac{3}{2}}\,\right)}\frac{2\left(\frac{e^t-e^{-t}}{2}\right)^2+\frac{4}{3}\left(\frac{e^t-e^{-t}}{2}\right)^4}{\sqrt{2\,}\left(\frac{e^t+e^{-t}}{2}\right)+1}{\rm\,d}t\\
&=\frac{4}{3}\ln\left(1+\sqrt{3}\,\right)-\int_0^{\ln\left(\frac{1+\sqrt{3}}{\sqrt{2\,}}\,\right)}\frac{\left(e^{2x}-1\right)^2\left(e^{4t}+4e^{2t}+1\right)}{6e^{4t}\left(\sqrt{2\,}e^{2t}+2e^t+\sqrt{2\,}\,\right)}{\rm\,d}\left(e^t\right)\\
&=\frac{4}{3}\ln\left(1+\sqrt{3}\,\right)-\frac{1}{6}\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\frac{\left(w^2-1\right)^2\left(w^4+4w^2+1\right)}{w^4\left(\sqrt{2\,}w^2+2w+\sqrt{2\,}\,\right)}{\rm\,d}w\\
&=\frac{4}{3}\ln\left(1+\sqrt{3}\,\right)-\frac{1}{6}\left(\frac{14}{3}-\sqrt{3\,}-2\ln\left(1+\sqrt{3}\,\right)+\ln2-\frac{\pi}{3}\right)\\
&=\boxed{-\frac{7}{9}+\frac{\sqrt{3\,}}{6}+\frac{\pi}{18}+\frac{5}{3}\ln\left(1+\sqrt{3}\,\right)-\frac{1}{6}\ln2\,}\\
\end{align*}
.

1. \int_0^1\left(1+v^2\right)\ln\left(1+\sqrt{2+v^2}\right)dv
2. -\frac{7}{9}+\frac{\sqrt{3}}{6}+\frac{\pi}{18}+\frac{5}{3}\ln\left(1+\sqrt{3}\right)-\frac{1}{6}\ln2

复制代码

\begin{align*}   
\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\frac{\left(w^2-1\right)^2\left(w^4+4w^2+1\right)}{w^4\left(\sqrt{2\,}w^2+2w+\sqrt{2\,}\,\right)}{\rm\,d}w   
&=   
\begin{split}   
+\frac{1}{\sqrt{2\,}}\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\frac{1}{w^4}{\rm\,d}w-\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\frac{1}{w^3}{\rm\,d}w+\frac{3}{\sqrt{2\,}}\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\frac{1}{w^2}{\rm\,d}w\\   
-2\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\frac{1}{w}{\rm\,d}w-\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}w{\rm\,d}w+\frac{1}{\sqrt{2\,}}\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}w^2{\rm\,d}w\\   
+\frac{3}{\sqrt{2\,}}\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}{\rm\,d}w-8\int_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\frac{{\rm\,d}\left(\sqrt{2}w+1\right)}{\left(\sqrt{2}w+1\right)^2+1}  
\end{split}\\   
&=   
\begin{split}   
+\frac{1}{\sqrt{2\,}}\left.\left(\frac{-1}{3w^3}\right)\right|_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}
-\left.\left(\frac{-1}{2w^2}\right)\right|_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}
+\frac{3}{\sqrt{2\,}}\left.\left(\frac{-1}{w}\right)\right|_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\\   
-2\left.\bigg(\ln w\bigg)\right|_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}
-\left.\left(\frac{w^2}{2}\right)\right|_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}
+\frac{1}{\sqrt{2\,}}\left.\left(\frac{w^3}{3}\right)\right|_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}\\   
+\frac{3}{\sqrt{2\,}}\left.\bigg(w\bigg)\right|_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}-8\left.\bigg(\arctan\left(\sqrt{2}w+1\right)\bigg)\right|_1^{\frac{1+\sqrt{3}}{\sqrt{2\,}}}
\end{split}\\
&=   
\begin{split}   
+\frac{5+\sqrt{2\,}-3\sqrt{3}}{6}
-\frac{-1+\sqrt{3}}{2}
+\frac{3\left(1+\sqrt{2}-\sqrt{3}\,\right)}{2}\\   
-2\ln\bigg(1+\sqrt{3}\bigg)+\ln2
-\frac{1+\sqrt{3\,}}{2}  
+\frac{5-\sqrt{2\,}+3\sqrt{3}}{6}\\   
+\frac{3\left(1-\sqrt{2}+\sqrt{3}\,\right)}{2}
-\frac{\pi}{3}
\end{split}\\
&=\frac{14}{3}-\sqrt{3\,}-2\ln\left(1+\sqrt{3}\,\right)+\ln2-\frac{\pi}{3}
\end{align*}
...

1. \int_1^{\frac{1+\sqrt{3}}{\sqrt{2}}}\frac{\left(w^2-1\right)^2\left(w^4+4w^2+1\right)}{w^4\left(\sqrt{2}w^2+2w+\sqrt{2}\right)}dw
2. \frac{14}{3}-\sqrt{3}-2\ln\left(1+\sqrt{3}\right)+\ln2-\frac{\pi}{3}

复制代码

...
\begin{align*}
\int_0^1\left(1+v^2\right)\ln\left(1+v^2\right){\rm\,d}v
&=\int_0^1\ln\left(1+v^2\,\right){\rm\,d}\left(v+\frac{v^3}{3}\right)\\
&=\left.\left(v+\frac{v^3}{3}\right)\ln\bigg(1+v^2\bigg)\right|_0^1-\int_0^1\left(v+\frac{v^3}{3}\right){\rm\,d}\left[\ln\left(1+v^2\right)\right]\\        
&=\frac{4}{3}\ln2-\int_0^1\left(v+\frac{v^3}{3}\right)\frac{2v}{1+v^2}{\rm\,d}v\,=\frac{4}{3}\ln2-2\int_0^1\frac{3v^2+v^4}{3(1+v^2)}{\rm\,d}v\\
&=\frac{4}{3}\ln2-2\int_0^1\frac{2+3v^2+v^4-2}{3(1+v^2)}{\rm\,d}v=\frac{4}{3}\ln2-2\int_0^1\frac{(1+v^2)(2+v^2)-2}{3(1+v^2)}{\rm\,d}v\\
&=\frac{4}{3}\ln2-\frac{2}{3}\int_0^1(2+v^2){\rm\,d}v+\frac{4}{3}\int_0^1\frac{1}{1+v^2}{\rm\,d}v\\
&=\boxed{\frac{4}{3}\ln2-\frac{14}{9}+\frac{\pi}{3}}\\
\end{align*}
.

1. \int_0^1\left(1+v^2\right)\ln\left(1+v^2\right)dv
2. \frac{4}{3}\ln2-\frac{14}{9}+\frac{\pi}{3}

复制代码

青青子衿

\begin{align*}            
\int_0^{\cos\varphi}\left(1+v^2\right)\ln\left(\cos\varphi+\sqrt{1+\cos^2\varphi+v^2}\,\right){\rm\,d}v           
&=\int_0^{\cos\varphi}\ln\left(\cos\varphi+\sqrt{1+\cos\varphi+v^{\overset{\,}{2}}}\,\right){\rm\,d}\left(v+\frac{v^3}{3}\right)\\            
&=\quad   
\begin{split}        
\left.\left(v+\frac{v^3}{3}\right)\ln\left(\cos\varphi+\sqrt{1+\cos^2\varphi+v^{\overset{\,}{2}}}\,\right)\right|_0^{\cos\varphi}\\   
-\int_0^{\cos\varphi}\left(v+\frac{v^3}{3}\right){\rm\,d}\left[\ln\left(\cos\varphi+\sqrt{1+\cos^2\varphi+v^{\overset{\,}{2}}}\,\right)\right]   
\end{split}\\      
&=\quad   
\begin{split}        
\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\   
-\int_0^{\cos\varphi}\left(v+\frac{v^3}{3}\right)\frac{v}{\sqrt{1+\cos^2\varphi+v^2}\left(\cos\varphi+\sqrt{1+\cos^2\varphi+v^2}\,\right)}{\rm\,d}v   
\end{split}\\        
&=\quad   
\begin{split}        
\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\   
-\int_0^{\operatorname{arsinh}\frac{\sqrt{\eta^2-1}}{\eta}}\frac{\eta\sinh t\left(\eta\sinh t+\frac{\eta^3\sinh^3t}{3}\right)}{\sqrt{\eta^2+\eta^2\sinh^2t}\left(\sqrt{\eta^2-1}+\sqrt{\eta^2+\eta^2\sinh^2t}\,\right)}{\rm\,d}\left(\eta\sinh t\right)   
\end{split}\\   
&=\quad   
\begin{split}        
\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\   
-\int_0^{\operatorname{arsinh}\frac{\sqrt{\eta^2-1}}{\eta}}\frac{\eta^2\sinh^2t+\frac{\eta^4\sinh^4t}{3}}{\eta\cosh t\left(\sqrt{\eta^2-1}+\eta\cosh t\,\right)}\left(\eta\cosh t\right){\rm\,d}t   
\end{split}\\   
&=\quad   
\begin{split}        
\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\   
-\int_0^{\operatorname{arsinh}\frac{\sqrt{\eta^2-1}}{\eta}}\frac{\eta^2\sinh^2t+\frac{\eta^4\sinh^4t}{3}}{\sqrt{\eta^2-1}+\eta\cosh t}{\rm\,d}t   
\end{split}\\  
&=\quad   
\begin{split}        
\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\   
-\int_0^{\ln\left(\frac{\sqrt{\eta^2-1}}{\eta}+\sqrt{1+\frac{\eta^2-1}{\eta^2}}\right)}\frac{\eta^2\left(\frac{e^t-e^{-t}}{2}\right)^2+\frac{\eta^4}{3}\left(\frac{e^t-e^{-t}}{2}\right)^4}{\sqrt{\eta^2-1}+\eta\left(\frac{e^t+e^{-t}}{2}\right)}{\rm\,d}t   
\end{split}\\  
&=\quad   
\begin{split}        
\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\   
-\int_0^{\ln\left(\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}\right)}\frac{\eta^2\left(e^{2t}-1\right)^2\left(\eta^2e^{4t}+2\left(6-\eta^2\right)e^{2t}+\eta^2\right)}{24e^{4t}\left(\eta e^{2t}+2\sqrt{\eta^2-1}\,e^t+\eta\right)}{\rm\,d}\left(e^t\right)   
\end{split}\\  
&=\quad   
\begin{split}        
\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\   
-\frac{\eta^2}{24}\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\frac{\left(w^2-1\right)^2\left(\eta^2w^4+2\left(6-\eta^2\right)w^2+\eta^2\right)}{w^4\left(\eta w^2+2\sqrt{\eta^2-1}\,w+\eta\right)}{\rm\,d}w   
\end{split}\\      
&=\quad      
\begin{split}         
\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\  
-\frac{\left(6+\cos^2\varphi\right)\cos\varphi}{9}   
+\frac{\sqrt{1+2\cos^2\varphi}\,\cos^2\varphi}{6}\\   
+\frac{\left(3-\cos^2\varphi\right)\cos\varphi}{6}\ln\left(\frac{\cos\varphi+\sqrt{1+2\cos^2\varphi}}{\sqrt{1+\cos^2\varphi}}\right)\\        
+\frac{4}{3}\arctan\left(\frac{\cos\varphi+\sqrt{1+2\cos^2\varphi}-\sqrt{1+\cos^2\varphi}}{1+\left(2\cos\varphi+\sqrt{1+2\cos^2\varphi}\,\right)\left(\cos\varphi+\sqrt{1+\cos^2\varphi}\,\right)}\,\right)         
\end{split}\\
\int_0^{\cos\varphi}\left(1+v^2\right)\ln\left(\cos\varphi+\sqrt{1+\cos^2\varphi+v^2}\,\right){\rm\,d}v
&=\quad      
\begin{split}
-\frac{\left(6+\cos^2\varphi\right)\cos\varphi}{9}     
+\frac{\sqrt{1+2\cos^2\varphi}\,\cos^2\varphi}{6}\\     
+\left(\frac{3\cos\varphi}{2}+\frac{\cos^3\varphi}{6}\right)\ln\left(\cos\varphi+\sqrt{1+2\cos^{\overset{\,}{2}}\varphi}\,\right)\\
-\left(\frac{\cos\varphi}{4}-\frac{\cos^3\varphi}{12}\right)\ln\left(1+\cos^{\overset{\,}{2}}\varphi\,\right)\\
+\frac{4}{3}\arctan\left(\frac{\cos\varphi+\sqrt{1+2\cos^2\varphi}-\sqrt{1+\cos^2\varphi}}{1+\left(2\cos\varphi+\sqrt{1+2\cos^2\varphi}\,\right)\left(\cos\varphi+\sqrt{1+\cos^2\varphi}\,\right)}\,\right)         
\end{split}\\
\int_0^1\left(1+v^2\right)\ln\left(1+\sqrt{2+v^{\overset{\,}{2}}}\,\right){\rm\,d}v
&=\boxed{-\frac{7}{9}+\frac{\sqrt{3\,}}{6}+\frac{\pi}{18}+\frac{5}{3}\ln\left(1+\sqrt{3}\,\right)-\frac{1}{6}\ln2\,}\\     
\end{align*}
...

1. \int_0^{\cos\varphi}\left(1+v^2\right)\ln\left(\cos\varphi+\sqrt{1+\left(\cos\varphi\right)^2+v^2}\right)dv
2. -\frac{\left(6+\left(\cos\varphi\right)^2\right)\cos\varphi}{9}+\frac{\left(\cos\varphi\right)^2\sqrt{1+2\left(\cos\varphi\right)^2}}{6}+\left(\frac{3\left(\cos\varphi\right)}{2}+\frac{\left(\cos\varphi\right)^3}{6}\right)\ln\left(\left(\cos\varphi\right)+\sqrt{1+2\left(\cos\varphi\right)^2}\right)-\left(\frac{\cos\varphi}{4}-\frac{\left(\cos\varphi\right)^3}{12}\right)\ln\left(1+\left(\cos\varphi\right)^2\right)+\frac{4}{3}\arctan\left(\frac{\left(\cos\varphi\right)+\sqrt{1+2\left(\cos\varphi\right)^2}-\sqrt{1+\left(\cos\varphi\right)^2}}{1+\left(2\cos\varphi+\sqrt{1+2\left(\cos\varphi\right)^2}\right)\left(\cos\varphi+\sqrt{1+\left(\cos\varphi\right)^2}\right)}\right)

复制代码

...
\begin{align*}      
\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\frac{\left(w^2-1\right)^2\left(\eta^2w^4+2\left(6-\eta^2\right)w^2+\eta^2\right)}{w^4\left(\eta w^2+2\sqrt{\eta^2-1}\,w+\eta\right)}{\rm\,d}w     
&=      
\begin{split}      
+\frac{8-\eta^2}{\eta}\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}{\rm\,d}w+\frac{8-\eta^2}{\eta}\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\frac{1}{w^2}{\rm\,d}w\\
-2\sqrt{\eta^2-1}\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}w{\rm\,d}w-2\sqrt{\eta^2-1}\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\frac{1}{w^3}{\rm\,d}w\\
+\eta\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}w^2{\rm\,d}w+\eta\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\frac{1}{w^4}{\rm\,d}w\\
-\frac{4\left(4-\eta^2\right)\sqrt{\eta^2-1}}{\eta^2}\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\frac{1}{w}{\rm\,d}w\\
-\frac{32}{\eta^2}\int_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\frac{{\rm\,d}\left(\eta w+\sqrt{\eta^2-1}\,\right)}{\left(\eta w+\sqrt{\eta^2-1}\,\right)^2+1}
\end{split}\\      
&=      
\begin{split}
+\frac{8-\eta^2}{\eta}\left.\bigg(w\bigg)\right|_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}+\frac{8-\eta^2}{\eta}\left.\left(\frac{-1}{w}\right)\right|_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\\
-2\sqrt{\eta^2-1}\left.\left(\frac{w^2}{2}\right)\right|_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}
-2\sqrt{\eta^2-1}\left.\left(\frac{-1}{2w^2}\right)\right|_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\\
+\eta\cdot\left.\left(\frac{w^3}{3}\right)\right|_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}+\eta\cdot\left.\left(\frac{-1}{3w^3}\right)\right|_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\\      
-\frac{4\left(4-\eta^2\right)\sqrt{\eta^2-1}}{\eta^2}\left.\bigg(\ln w\bigg)\right|_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}\\  
-\frac{32}{\eta^2}\left.\bigg(\arctan\left(\eta w+\sqrt{\eta^2-1}\,\right)\bigg)\right|_1^{\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}}   
\end{split}\\
&=         
\begin{split}      
+\frac{8-\eta^2}{\eta}\left(\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}-\frac{\eta}{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}\right)\\   
-\sqrt{\eta^2-1}\left(\frac{\left(\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}\,\right)^2}{\eta^2}-\frac{\eta^2}{\left(\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}\,\right)^2}\right)\\   
+\frac{\eta}{3}\left(\frac{\left(\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}\,\right)^3}{\eta^3}-\frac{\eta^3}{\left(\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}\,\right)^3}\right)\\  
-\frac{4\left(4-\eta^2\right)\sqrt{\eta^2-1}}{\eta^2}\ln\left(\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}\right)\\      
-\frac{32}{\eta^2}\bigg(\arctan\left(2\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}\,\right)-\arctan\left(\eta+\sqrt{\eta^2-1}\,\right)\bigg)     
\end{split}\\   
&=         
\begin{split}      
+\frac{8-\eta^2}{\eta}\cdot\frac{2\sqrt{\eta^2-1}}{\eta}\\   
-\sqrt{\eta^2-1}\cdot\frac{4\sqrt{\eta^2-1}\sqrt{2\eta^2-1}}{\eta^2}\\   
+\frac{\eta}{3}\cdot\frac{2\left(7\eta^2-4\right)\sqrt{\eta^2-1}}{\eta^3}\\  
-\frac{4\left(4-\eta^2\right)\sqrt{\eta^2-1}}{\eta^2}\ln\left(\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}\right)\\      
-\frac{32}{\eta^2}\bigg(\arctan\left(2\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}\,\right)-\arctan\left(\eta+\sqrt{\eta^2-1}\,\right)\bigg)     
\end{split}\\   
&=     
\begin{split}   
+\frac{8\left(5+\eta^2\right)\sqrt{\eta^2-1}}{3\eta^2}   
-\frac{4\left(\eta^2-1\right)\sqrt{2\eta^2-1}}{\eta^2}\\   
-\frac{4\left(4-\eta^2\right)\sqrt{\eta^2-1}}{\eta^2}\ln\left(\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}}{\eta}\right)\\      
-\frac{32}{\eta^2}\arctan\left(\frac{\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}-\eta}{1+\left(2\sqrt{\eta^2-1}+\sqrt{2\eta^2-1}\,\right)\left(\sqrt{\eta^2-1}+\eta\,\right)}\,\right)        
\end{split}\\
\end{align*}

\begin{align*}
\int_0^{\cos\varphi}\left(1+v^2\right)\ln\left(1+v^2\right){\rm\,d}v
&=\int_0^{\cos\varphi}\ln\left(1+v^2\,\right){\rm\,d}\left(v+\frac{v^3}{3}\right)\\
&=\left.\left(v+\frac{v^3}{3}\right)\ln\bigg(1+v^2\bigg)\right|_0^{\cos\varphi}-\int_0^{\cos\varphi}\left(v+\frac{v^3}{3}\right){\rm\,d}\left[\ln\left(1+v^2\right)\right]\\
&=\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(1+\cos^2\varphi\,\right)-\frac{2}{3}\int_0^{\cos\varphi}(2+v^2){\rm\,d}v+\frac{4}{3}\int_0^{\cos\varphi}\frac{1}{1+v^2}{\rm\,d}v\\
&=\left(\cos\varphi+\frac{\cos^3\varphi}{3}\right)\ln\left(1+\cos^2\varphi\,\right)-\frac{2}{3}\left(2\cos\varphi+\frac{\cos^3\varphi}{3}\right)+\frac{4}{3}\arctan\left(\cos\varphi\right)\\
&=\boxed{\frac{4}{3}\ln2-\frac{14}{9}+\frac{\pi}{3}}\\
\end{align*}

青青子衿

回复 15# 青青子衿
参见：Bestimmte Integrale_维基教程
de.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Mehrfachintegrale#3

hbghlyj

${\displaystyle {\begin{array}{l}x=r\sin \theta \cos \varphi \\y=r\sin \theta \sin \varphi \\z=r\cos \theta \end{array}}\quad \qquad dx\,dy\,dz=r^{2}\sin \theta \,dr\,d\theta \,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}+z^{2}}}}$
变换后的积分域$V_1$
$\displaystyle x\geq y\iff r\sin \theta \cos \varphi \geq r\sin \theta \sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}$
$\displaystyle x\geq z\iff r\sin \theta \cos \varphi \geq r\cos \theta \iff \tan \theta \geq \sec \varphi \iff \arctan(\sec \varphi )\leq \theta \leq {\frac {\pi }{2}}$
$\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\sin \theta \,\cos \varphi }}$

画图(HTML format)
import graph3;settings.render=2;
unitsize(8cm,8cm,1cm);

currentprojection = perspective(2,4,6);

triple f(pair p) {
real theta=pi/2-atan(cos(p.x))*p.y;
return (p.x,theta,1/(sin(theta)*cos(p.x)));
}
draw(surface(f,(0,0), (pi/4,1)),yellow,nolight);
real thetamax=atan(sqrt(2));
draw(surface(graph(new triple(real t){return (pi/4,t,sqrt(2)/sin(t));},thetamax,pi/2)--(pi/4,pi/2,0)--(pi/4,thetamax,0)--cycle),lightblue,nolight);
draw(surface(graph(new triple(real t){return (t,pi/2,1/cos(t));},0,pi/4)--(pi/4,pi/2,0)--(0,pi/2,0)--cycle),lightmagenta,nolight);

xaxis3(Label("$\varphi$"),Bounds,InTicks);
yaxis3(Label("$\theta$"),Bounds,InTicks);
zaxis3(Label("$r$"),Bounds,InTicks);