与正三角形均匀带电薄片电场强度有关的曲面积分

葡萄糖

2.28  正三角形带电薄片（带正电荷）位于\(\,\Sigma\colon\,x+y+z=-a\)（其中\(\,a>0\)）的平面上，且带电薄片限定于\(-a\le x\le 0\)与\(-a\le y\le 0\)之间，
        其电荷面密度为\(\sigma\)，试求出原点处的电场强度\(\boldsymbol{E}\)（矢量）
\[ \left|\overrightarrow{\boldsymbol{E}}\,\right|\,=\sqrt{3}\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)}\dfrac{\left|x\right|}{\sqrt{x^2+y^2+z^2}} \]

\[
\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{i}}\rangle=\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|x\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}
\]
\[
\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{j}}\rangle=\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|y\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}
\]
\[
\big|\overrightarrow{\boldsymbol{E}}\big|\,\cos\langle\overrightarrow{\boldsymbol{E}},\overrightarrow{\boldsymbol{k}}\rangle=\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|z\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}
\]

\begin{align*}  
\big|\overrightarrow{\boldsymbol{E}}\big|\,&=\sqrt{3}\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\sigma\,\left|x\right|\,{\rm\,d}S}{4\pi\varepsilon_0\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}\\  
&=\frac{\sqrt{3}\,\sigma}{4\pi\varepsilon_0}
\displaystyle\iint\limits_{\Sigma\colon\,x+y+z=-a}\dfrac{\,\left|x\right|\,}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}}{\rm\,d}S\\
&=\frac{\sqrt{3}\,\sigma}{4\pi\varepsilon_0}
\int_{-a}^0\int_{-a-y}^0 \dfrac{\sqrt{3}\,\left|x\right|}{\left(x^2+y^2+\left(-a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&=\frac{\sqrt{3}\,\sigma}{4\pi\varepsilon_0}
\int_{-a}^0\int_{-a-y}^0 \dfrac{\sqrt{3}\,\left|x\right|}{\left(x^2+y^2+\left(a+x+y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&=\frac{\sqrt{3}\,\sigma}{4\pi\varepsilon_0}
\int_0^a\int_0^{a-y} \dfrac{\sqrt{3}\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&=\frac{3\,\sigma}{4\pi\varepsilon_0}
\int_0^a\int_0^{a-y} \dfrac{\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
\end{align*}
当最后一行那个二重积分中的\(\,a=1\,\)时，用Wolfram Mathematica可以计算出来，利用// FullSimplify化简得到的却是一个含有复数的数值表达式。

1. \!\(
   
2. \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1\)]\(
   
3. \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(1 - y\)]
   
4. \*FractionBox[\(x\),
   
5. SuperscriptBox[\((
   
6. \*SuperscriptBox[\(x\), \(2\)] +
   
7. \*SuperscriptBox[\(y\), \(2\)] +
   
8. \*SuperscriptBox[\((1 - x - y)\), \(2\)])\),
   
9. FractionBox[\(3\), \(2\)]]] \[DifferentialD]x \[DifferentialD]y\)\) \
   
10. // FullSimplify

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zeroieme

因为结果是实数常数，可以让Mathematica先分离实虚部，再逐层简化。于是最后一句代码改成

1. //ComplexExpand//FullSimplify//@#&

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zeroieme

1. AbsoluteTiming[ \!\(
   
2. \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a\)]\(
   
3. \*SubsuperscriptBox[\(\[Integral]\), \(0\), \(a - y\)]
   
4. \*FractionBox[\(x\),
   
5. SuperscriptBox[\((
   
6. \*SuperscriptBox[\(x\), \(2\)] +
   
7. \*SuperscriptBox[\(y\), \(2\)] +
   
8. \*SuperscriptBox[\((a - x - y)\), \(2\)])\),
   
9. FractionBox[\(3\), \(2\)]]] \[DifferentialD]x \[DifferentialD]y\)\) //
   
10.     ComplexExpand // FullSimplify[#, a > 0] & //@ # &]

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直接对\(a\)二重积分而不代入数值 ，算出来的也是常数\(\frac{\pi }{6}\)。

zeroieme

如果

1. //FullSimplify[#, a > 0] & //@ # &

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不够力就上大招

1. //(#//FunctionExpand//PowerExpand[#,Assumptions->True]&//FullSimplify[#,a>0]&)&//@#&

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用FunctionExpand和PowerExpand解开ArcTan与Log

.·.·.

zeroieme 发表于 2018-12-6 12:03
如果不够力就上大招

用FunctionExpand和PowerExpand解开ArcTan与Log

所以\(\pi +\sqrt{2} \log \left(17-12 \sqrt{2}\right)-\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)-2 \tan ^{-1}\left(2 \sqrt{2}\right)+4 \sqrt{2} \sinh ^{-1}(1)=0\)么……
或者说这里给出了一个pi的表达式
本来觉得很神奇
忽然想起高中有个东西叫倍角公式……
瞬间觉得自己被教做人了……

zeroieme

.·.·. 发表于 2018-12-7 18:47
所以\(\pi +\sqrt{2} \log \left(17-12 \sqrt{2}\right)-\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)-2 ...

\(\sqrt{2} \log \left(17-12 \sqrt{2}\right)+4 \sqrt{2} \sinh ^{-1}(1)=0\)
反三角函数反双曲三角函数转对数后可以合并

.·.·.

zeroieme 发表于 2018-12-7 20:52
\(\sqrt{2} \log \left(17-12 \sqrt{2}\right)+4 \sqrt{2} \sinh ^{-1}(1)=0\)
反三角函数反双曲三角函 ...

才发现Mathematica好坑

1. FullSimplify[4 Sqrt[2] ArcSinh[1]+Sqrt[2] Log[17-12 Sqrt[2]]]
   
2. FullSimplify[\[Pi]-ArcTan[(4 Sqrt[2] )/7]-2ArcTan[2Sqrt[2]]]
   
3. FullSimplify[(\[Pi]-ArcTan[(4 Sqrt[2])/7]-2ArcTan[2Sqrt[2]])+(Sqrt[2] Log[17-12 Sqrt[2]]+4 Sqrt[2] ArcSinh[1])]

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这玩意的输出是
\(0\)
\(0\)
\(\pi +\sqrt{2} \log \left(17-12 \sqrt{2}\right)-\tan ^{-1}\left(\frac{4 \sqrt{2}}{7}\right)-2 \tan ^{-1}\left(2 \sqrt{2}\right)+4 \sqrt{2} \sinh ^{-1}(1)\)

wayne

与$a$无关是因为  这个积分是 无量纲的。

葡萄糖

与正三角形均匀带电薄片电场强度有关的曲面积分
\begin{align*}
&&&\int_0^a\int_0^{a-y} \dfrac{\,x}{\left(x^2+y^2+\left(a-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
&&\overset{\begin{cases}  
x\,=\,a\,\cdot\,u\\  
y\,=\,a\,\cdot\,v\\   
\end{cases}}{\overline{\overline{\hspace{3cm}}}}&\int_0^a\int_0^{a-av} \dfrac{\,a\cdot u}{\left(a^2u^2+a^2v^2+\left(a-au-av\right)^2\right)^{\frac{3}{2}}}{\rm\,d}\left(au\right){\rm\,d}\left(av\right)\\
&&=&\int_0^1\int_0^{1-v} \dfrac{u}{\left(u^2+v^2+\left(1-u-v\right)^2\right)^{\frac{3}{2}}}{\rm\,d}u{\rm\,d}v\\
&&=&\int_0^1\int_0^{1-y} \dfrac{\,x}{\left(x^2+y^2+\left(1-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x{\rm\,d}y\\
\end{align*}
\begin{align*}
&&&\int_0^{1-y} \dfrac{x}{\left(x^2+y^2+\left(1-x-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x\\  
&&\overset{
b=\frac{1-y}{2}
}{\overline{\overline{\hspace{2cm}}}}&\int_0^{2b} \dfrac{x}{\left(x^2+y^2+\left(-x+1-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_0^{2b} \dfrac{x}{\left(x^2+y^2+x^2-2\left(1-y\right)x+\left(1-y\right)^2\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_0^{2b} \dfrac{x}{\left(2x^2-2\left(1-y\right)x+2\left(\frac{1-y}{2}\right)^2+y^2+\frac{\left(1-y\right)^2}{2}\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_0^{2b} \dfrac{x}{\left(2\left(x-\frac{1-y}{2}\right)^2+y^2+\frac{\left(1-y\right)^2}{2}\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_0^{2b} \dfrac{x}{\left(2\left(x-b\right)^2+y^2+2b^2\right)^{\frac{3}{2}}}{\rm\,d}x\\
&&=&\int_{-b}^b \dfrac{t+b}{\left(2t^2+y^2+2b^2\right)^{\frac{3}{2}}}{\rm\,d}t=\int_0^b \dfrac{2b}{\left(2t^2+y^2+2b^2\right)^{\frac{3}{2}}}{\rm\,d}t\\
&&=&\left.\dfrac{2bt}{\left(y^2+2b^2\right)\sqrt{2t^2+y^2+2b^2}}\right|_0^b=\dfrac{2b^2}{\left(y^2+2b^2\right)\sqrt{2b^2+y^2+2b^2}}\\
&&=&\dfrac{2b^2}{\left(y^2+2b^2\right)\sqrt{y^2+4b^2}}=\dfrac{2\left(\frac{1-y}{2}\right)^2}{\left(y^2+2\left(\frac{1-y}{2}\right)^2\right)\sqrt{y^2+4\left(\frac{1-y}{2}\right)^2}}\\
&&=&\dfrac{\left(1-y\right)^2}{\left(2y^2+\left(1-y\right)^2\right)\sqrt{y^2+\left(1-y\right)^2}}\\
\end{align*}
\begin{align*}
&&&\int_0^1\dfrac{\left(1-y\right)^2}{\left(2y^2+\left(1-y\right)^2\right)\sqrt{y^2+\left(1-y\right)^2}}{\rm\,d}y\\  
&&=&\int_0^1\dfrac{y^2}{\left(2\left(1-y\right)^2+y^2\right)\sqrt{\left(1-y\right)^2+y^2}}{\rm\,d}y\\
&&=&\int_0^1\dfrac{y^2}{\left(2\left(y^2-2y+1\right)+y^2\right)\sqrt{2y^2-2y+1}}{\rm\,d}y\\
&&=&\int_0^1\dfrac{y^2}{\left(3y^2-4y+2\right)\sqrt{2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}}}{\rm\,d}y\\
&&\overset{
y-\frac{1}{2}=\frac{1}{2}\sinh t
}{\overline{\overline{\hspace{3cm}}}}&\int_{\ln\left(\sqrt{2}-1\right)}^{\ln\left(\sqrt{2}+1\right)}\dfrac{\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)^2}{\left(3\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)^2-4\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)+2\right)\sqrt{\frac{1}{2}\sinh^2t
+\frac{1}{2}}}{\rm\,d}\left(\frac{1}{2}\sinh t
\right)\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\ln\left(\sqrt{2}-1\right)}^{\ln\left(\sqrt{2}+1\right)}\dfrac{\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)^2}{3\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)^2-4\left(\frac{1}{2}\sinh t
+\frac{1}{2}\right)+2}{\rm\,d}t\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\ln\left(\sqrt{2}-1\right)}^{\ln\left(\sqrt{2}+1\right)}\dfrac{\left(e^{2t}+2e^t-1\right)^2}{3e^{4t}-4e^{3t}+6e^{2t}+4e^t+3}{\rm\,d}t\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\ln\left(\sqrt{2}-1\right)}^{\ln\left(\sqrt{2}+1\right)}\dfrac{\left(e^{2t}+2e^t-1\right)^2}{\left(3e^{4t}-4e^{3t}+6e^{2t}+4e^t+3\right)e^t}{\rm\,d}\left(e^t\right)\\
&&\overset{e^t=w}{\overline{\overline{\hspace{3cm}}}}&\,\,\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{\left(w^2+2w-1\right)^2}{\left(w^2-2w+3\right)\left(3w^2+2w+1\right)w}{\rm\,d}w\\
\end{align*}

\begin{align*}
&&&\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{\left(w^2+2w-1\right)^2}{\left(w^2-2w+3\right)\left(3w^2+2w+1\right)w}{\rm\,d}w\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{\left(w^2+2w-1\right)^2}{\left(w^2-2w+3\right)\left(3w^2+2w+1\right)w}{\rm\,d}w\\
&&=&\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{1}{3w}{\rm\,d}w+\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{2\left(1+w\right)}{3\left(w^2-2w+3\right)}{\rm\,d}w-\frac{1}{\sqrt{2\,}\,}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{2\left(1+w\right)}{3w^2+2w+1}{\rm\,d}w\\
&&=&\,\boxed{\frac{\pi}{6}}
\end{align*}