1. Denote by [tex]\mathcal D[/tex] the region bounded by the cylinder [tex]x^2+y^2=9[/tex] and the planes [tex]z=0[/tex] and [tex]z=6[/tex]. Converting to cylindrical coordinates involves setting
[tex]\begin{cases}x=r\cos u\\y=r\sin u\\z=v\end{cases}[/tex]
and [tex]\mathcal D[/tex] is obtained by varying the parameters over [tex]0\le r\le3[/tex], [tex]0\le u\le2\pi[/tex], and [tex]0\le v\le6[/tex]. The volume element is
[tex]\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm du\,\mathrm dv[/tex]
so the integral becomes
[tex]\displaystyle\iiint_{\mathcal D}xz\,\mathrm dV=\int_{v=0}^{v=6}\int_{u=0}^{u=2\pi}\int_{r=0}^{r=3}r^2v\cos u\,\mathrm dr\,\mathrm du\,\mathrm dv=0[/tex]
2. Now let [tex]\mathcal D[/tex] denote the region bounded by the sphere [tex]x^2+y^2+z^2=4[/tex] and the coordinate planes in the octant in which each coordinate of the point [tex](x,y,z)[/tex] is negative.
Converting to spherical coordinates, we set
[tex]\begin{cases}x=\rho\cos u\sin v\\y=\rho\sin u\sin v\\z=\rho\cos v\end{cases}[/tex]
where we obtain [tex]\mathcal D[/tex] by varying over [tex]0\le\rho\le2[/tex], [tex]\pi\le u\le\dfrac{3\pi}2[/tex], and [tex]\dfrac\pi2\le v\le\pi[/tex]. The volume element is now
[tex]\mathrm dV=\rho^2\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv[/tex]
so the integral becomes
[tex]\displaystyle\iiint_{\mathcal D}z\,\mathrm dV=\int_{v=\pi/2}^{v=\pi}\int_{u=\pi}^{u=3\pi/2}\int_{\rho=0}^{\rho=2}\rho^3\cos v\sin v\,\mathrm d\rho\,\mathrm du\,\mathrm dv=-\pi[/tex]
