Đáp án:
$I =- \dfrac{162}{5}$
Giải thích các bước giải:
\(\begin{array}{l}
\quad I = \displaystyle\iiint\limits_{\Omega}xzdxdydz\\
\text{Đặt}\ \begin{cases}x = r.\sin\theta.\cos\varphi\\
y=r.\sin\theta.\sin\varphi\\
z= r.\cos\theta \end{cases}\\
\text{Miền $\Omega$ được biểu diễn:}\\
\Omega: \begin{cases}0\leqslant r \leqslant 3
\\ \dfrac{\pi}{2}\leqslant \theta \leqslant \pi\\ 0 \leqslant \varphi \leqslant 2\pi \end{cases}\\
\text{Ta được:}\\
\quad I = \displaystyle\int\limits_0^{2\pi}d\varphi\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\pi}d\theta\displaystyle\int\limits_0^3 (r.\sin\theta.\cos\varphi). (r.\cos\theta).r^2\sin\theta dr\\
\to I = \displaystyle\int\limits_0^{2\pi}d\varphi\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\pi}d\theta\displaystyle\int\limits_0^3 r^4\sin^2\theta.\cos\theta.\cos\varphi dr\\
\to I = \displaystyle\int\limits_0^{2\pi}d\varphi\displaystyle\int\limits_{\tfrac{\pi}{2}}^{\pi}\dfrac{243}{5}\sin^2\theta.\cos\theta d\theta\\
\to I = \displaystyle\int\limits_0^{2\pi}\left(-\dfrac{81}{5}\right)d\varphi\\
\to I = - \dfrac{162}{5}
\end{array}\)
