问题及解答

计算 $\iiint_{\Omega}z\sqrt{x^2+y^2+z^2}\mathrm{d}V$, 其中 $\Omega=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2+z^2\leqslant 1, z\geqslant\sqrt{3(x^2+y^2)}\}$.

令 

\[
\begin{cases}
x=r\sin\varphi\cos\theta,\\
y=r\sin\varphi\sin\theta,\\
z=r\cos\varphi.
\end{cases}
\]

则 $\theta\in[0,2\pi]$, $\varphi\in[0,\frac{\pi}{6}]$. 于是

\[
\begin{split}
\iiint_{\Omega}z\sqrt{x^2+y^2+z^2}\mathrm{d}V&=\int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{\frac{\pi}{6}}\mathrm{d}\varphi\int_{0}^{1}r\cos\varphi\cdot r\cdot r^2\sin\varphi\mathrm{d}r\\
&=2\pi\int_{0}^{\frac{\pi}{6}}\sin\varphi\cdot\cos\varphi\mathrm{d}\varphi\int_0^1 r^4\mathrm{d}r\\
&=\pi\int_{0}^{\frac{\pi}{6}}\sin(2\varphi)\mathrm{d}\varphi\cdot\frac{1}{5}r^5\biggr|_{0}^{1}\\
&=\frac{\pi}{5}\cdot\frac{-1}{2}\cos(2\varphi)\biggr|_{0}^{\frac{\pi}{6}}\\
&=\frac{\pi}{5}\cdot\frac{-1}{2}\cdot(\cos\frac{\pi}{3}-\cos 0)\\
&=\frac{\pi}{20}.
\end{split}
\]