(2012年中科院考研题)设$\rho (x,y,z)$是原点$O$到椭球面$\frac{x^2}2+\frac{y^2}2+z^2=1$的上半部分(即满足$z\geq 0$的部分) $\Sigma$的任一点$(x,y,z)$处的切面的距离,求积分\[\iint_\Sigma \frac z{\rho (x,y,z)}dS.\]
所求积分为
\[I=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}.\]
记$z=\varphi (x,y), (x,y)\in D$,其中$D$为$x^2+y^2=2$.首先有
\begin{align*}dS&=\sqrt{1+\left(\frac{\partial \varphi}{\partial x}\right)^2+\left(\frac{\partial \varphi}{\partial x}\right)^2}dxdy=\sqrt{1+\left(\frac{-x}{2z}\right)^2+\left(\frac{-y}{2z}\right)^2}dxdy\\&=\frac1{2z}\sqrt{x^2+y^2+4z^2}dxdy.\end{align*}
因此
\begin{align*}I&=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}\\&=\frac{1}{4}\iint_D\sqrt{x^2+y^2+z^2}\sqrt{x^2+y^2+4z^2}dxdy=\frac{1}{4}\iint_D\sqrt{1+\frac{x^2}2+\frac{y^2}2}\sqrt{4-x^2-y^2}dxdy\\&=\frac14\int_0^{2\pi}d\theta\int_0^{\sqrt{2}}r\sqrt{1+\frac{r^2}2}\sqrt{4-r^2}dr=\frac\pi4\int_0^{\sqrt{2}}\sqrt{1+\frac{u}2}\sqrt{4-u}du\\&=\frac\pi{4\sqrt{2}}\int_0^{\sqrt{2}}\sqrt{8+2u-u^2}du=\frac{\sqrt{2}\pi}{16}\left(\sqrt{10-6\sqrt{2}}+9\arcsin \frac{\sqrt 2-1}3+2\sqrt{2}+9\arcsin \frac13\right).\end{align*}
这是因为
\begin{align*}&\int{\sqrt{8+2u-u^2}du}=u\sqrt{8+2u-u^2}-\int{\frac{u-u^2}{\sqrt{8+2u-u^2}}du}\\&=u\sqrt{8+2u-u^2}-\int{\frac{\left( 8+2u-u^2 \right) -8-u}{\sqrt{8+2u-u^2}}du}\\
&=u\sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+\int{\frac{8+u}{\sqrt{8+2u-u^2}}du}\\
&=\left( u-1 \right) \sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+\int{\frac{9}{\sqrt{9-\left( u-1 \right) ^2}}du}\\
&=\left( u-1 \right) \sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+9\arcsin \frac{u-1}{3}+C,
\end{align*}
即
$$
\int{\sqrt{8+2u-u^2}du}=\frac{u-1}{2}\sqrt{8+2u-u^2}+\frac{9}{2}\arcsin \frac{u-1}{3}+C.
$$