<s>[INST] Evaluate an integral by changing to cylindrical coordinates I have to evaluate the following integral by changing to cylindrical coordinates.
Should I just substitute all the $x,y,z$ to cylindrical coordinates?
$\int_{-2}^2\int_{-\sqrt{4-y^2}}^\sqrt{4-y^2}\int_\sqrt{x^2+y^2}^2 xzdzdxdy$
 [/INST] The region  $(x,y,z) :-2 \leq y \leq 2, -\sqrt{4-y^2} \leq x \leq \sqrt{4-y^2}, \sqrt{x^2 + y^2} \leq z \leq 2$ is the region above the cone $ z = \sqrt{x^2 + y^2}$ and below the plane $z = 2$. In polar coordinates, this region is $(r, \theta, z): 0 \leq r \leq 2, 0 \leq \theta \leq 2\pi, r \leq z \leq 2$. If $I$ is the value of the integral,
$$\begin{align}
I &= \int_0^{2\pi} \int_0^2 \int_r^2 z \cdot \underbrace{r \cos \theta}_{x} \,dz \,r \, dr \,d\theta \\
&= \left[ \int_0^{2\pi} \cos \theta \,d\theta \right] \cdot \left[ \int_0^2 \int_r^2 r^2 z \,dz  \,dr \right] \\
&= \left[ \sin \theta \right]_0^{2\pi} \cdot \left[ \int_0^2 \int_r^2 r^2 z \,dz  \,dr \right] \\
&= 0
\end{align}$$
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