(2012年中科院考研题)设$ ho (x,y,z)$是原点$O$到椭球面$frac{x^2}2+frac{y^2}2+z^2=1$的上半部分(即满足$zgeq 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{ ho (x,y,z)}dS.]
所求积分为
[I=iint_{Sigma}{frac{z}{ ho left( x,y,z ight)}dS}=frac{1}{2}iint_{Sigma}{zsqrt{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
egin{align*}dS&=sqrt{1+left(frac{partial varphi}{partial x} ight)^2+left(frac{partial varphi}{partial x} ight)^2}dxdy=sqrt{1+left(frac{-x}{2z} ight)^2+left(frac{-y}{2z} ight)^2}dxdy\&=frac1{2z}sqrt{x^2+y^2+4z^2}dxdy.end{align*}
因此
egin{align*}I&=iint_{Sigma}{frac{z}{ ho left( x,y,z ight)}dS}=frac{1}{2}iint_{Sigma}{zsqrt{x^2+y^2+z^2}dS}\&=frac{1}{4}iint_Dsqrt{x^2+y^2+z^2}sqrt{x^2+y^2+4z^2}dxdy=frac{1}{4}iint_Dsqrt{1+frac{x^2}2+frac{y^2}2}sqrt{4-x^2-y^2}dxdy\&=frac14int_0^{2pi}d hetaint_0^{sqrt{2}}rsqrt{1+frac{r^2}2}sqrt{4-r^2}dr=fracpi4int_0^{sqrt{2}}sqrt{1+frac{u}2}sqrt{4-u}du\&=fracpi{4sqrt{2}}int_0^{sqrt{2}}sqrt{8+2u-u^2}du=frac{sqrt{2}pi}{16}left(sqrt{10-6sqrt{2}}+9arcsin frac{sqrt 2-1}3+2sqrt{2}+9arcsin frac13 ight).end{align*}
这是因为
egin{align*}&int{sqrt{8+2u-u^2}du}=usqrt{8+2u-u^2}-int{frac{u-u^2}{sqrt{8+2u-u^2}}du}\&=usqrt{8+2u-u^2}-int{frac{left( 8+2u-u^2
ight) -8-u}{sqrt{8+2u-u^2}}du}\
&=usqrt{8+2u-u^2}-int{sqrt{8+2u-u^2}du}+int{frac{8+u}{sqrt{8+2u-u^2}}du}\
&=left( u-1
ight) sqrt{8+2u-u^2}-int{sqrt{8+2u-u^2}du}+int{frac{9}{sqrt{9-left( u-1
ight) ^2}}du}\
&=left( u-1
ight) sqrt{8+2u-u^2}-int{sqrt{8+2u-u^2}du}+9arcsin frac{u-1}{3}+C,
end{align*}
即
$$
int{sqrt{8+2u-u^2}du}=frac{u-1}{2}sqrt{8+2u-u^2}+frac{9}{2}arcsin frac{u-1}{3}+C.
$$