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*}
\begin{align*}
\color{red}{\boxed{\quad\color{black}{\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\,=\,\frac{\pi}{6}}\quad}}
\end{align*}
如图,分别以$\triangle ABC$的边$AB,AC$为一边在三角形外作正方形$ABEF$和$ACGH$, $M$为$FH$上的中点,求证: $MA\perp BC$.
如图,分别以$\triangle ABC$的边$AB,AC$为一边在三角形外作正方形$ABEF$和$ACGH$, $MA\perp BC$,求证: $M$为$FH$上的中点.