How do you evaluate the integral by changing to cylindrical coordinates?
$\int_{-2}^{2}{\int_{-\sqrt{4-{{y}^{2}}}}^{\sqrt{4-{{y}^{2}}}}{\int_{\sqrt{{{x}^{2}}+{{y}^{2}}}}^{2}{\left( xz \right)dzdxdy}}}$

Hint: The integral given in the above question is in the form of the rectangular coordinates $\left( x,y,z \right)$. As stated in the question, we have to change them to the cylindrical coordinates which are $\left( r,\theta ,z \right)$. For this, we have to use the relations $x=r\cos \theta $, $y=r\sin \theta $ and $z=z$. The limits of the integration are also to be changed using these relations.

Complete step by step answer:
Let us write the integral given in the question as
$I=\int_{-2}^{2}{\int_{-\sqrt{4-{{y}^{2}}}}^{\sqrt{4-{{y}^{2}}}}{\int_{\sqrt{{{x}^{2}}+{{y}^{2}}}}^{2}{\left( xz \right)dzdxdy}}}$
The above question is directing us to change the rectangular coordinates $\left( x,y,z \right)$ to the cylindrical coordinates $\left( r,\theta ,z \right)$. The cylindrical coordinates are related to the cylindrical coordinates as
$x=r\cos \theta $, $y=r\sin \theta $ and $z=z$
From the order of the above integral, we can observe that the variable for the innermost integral is $z$, for the middle is $x$, and for the outermost is $y$. So the limits for these are noted from the above integral as
\[\begin{align}
  & z\Rightarrow \left( \sqrt{{{x}^{2}}+{{y}^{2}}},2 \right) \\
 & x\Rightarrow \left( -\sqrt{4-{{y}^{2}}},\sqrt{4-{{y}^{2}}} \right) \\
 & z\Rightarrow \left( -2,2 \right) \\
\end{align}\]
Now, substituting $x=r\cos \theta $, $y=r\sin \theta $ and $z=z$ in the lower limit of $z$ we get
$\begin{align}
  & \Rightarrow z=\sqrt{{{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}} \\
 & \Rightarrow z=\sqrt{{{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta } \\
 & \Rightarrow z=\sqrt{{{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)} \\
\end{align}$
We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$. So we get
$\begin{align}
  & \Rightarrow z=\sqrt{{{r}^{2}}\left( 1 \right)} \\
 & \Rightarrow z=r \\
\end{align}$
So the lower limit of $z$ is equal to \[r\]. The upper limit of $z$ in the cylindrical coordinates will be the same, since the $z$ coordinate is same. So the upper limit of $z$ is equal to $2$.
Now, substituting $x=r\cos \theta $, $y=r\sin \theta $ in the upper limit for $x$ we get
\[\begin{align}
  & \Rightarrow r\cos \theta =\sqrt{4-{{\left( r\sin \theta \right)}^{2}}} \\
 & \Rightarrow r\cos \theta =\sqrt{4-{{r}^{2}}{{\sin }^{2}}\theta } \\
\end{align}\]
Squaring both the sides, we get
\[\begin{align}
  & \Rightarrow {{r}^{2}}{{\cos }^{2}}\theta =4-{{r}^{2}}{{\sin }^{2}}\theta \\
 & \Rightarrow {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta =4 \\
 & \Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=4 \\
 & \Rightarrow {{r}^{2}}=4 \\
\end{align}\]
On solving we get
$r=2$
Thus, the lower limit of $r$ is $0$ and the upper limit of $r$ is $2$.
Lastly, we know that the range for the cylindrical coordinate $\theta $ is $\left( 0,2\pi \right)$.
Hence, the given integral can now be written in the cylindrical coordinates as
\[\begin{align}
  & \Rightarrow I=\int_{0}^{2\pi }{\int_{0}^{2}{\int_{r}^{2}{\left( r\cos \theta z \right)rdzdrd\theta }}} \\
 & \Rightarrow I=\int_{0}^{2\pi }{\int_{0}^{2}{\int_{r}^{2}{zdz\left( {{r}^{2}} \right)dr\left( \cos \theta \right)d\theta }}} \\
\end{align}\]
Firstly, solving the innermost integral, we have
\[\begin{align}
  & \Rightarrow I=\int_{0}^{2\pi }{\int_{0}^{2}{\left[ \dfrac{{{z}^{2}}}{2} \right]_{r}^{2}\left( {{r}^{2}} \right)dr\left( \cos \theta \right)d\theta }} \\
 & \Rightarrow I=\int_{0}^{2\pi }{\int_{0}^{2}{\left[ \dfrac{{{2}^{2}}-{{r}^{2}}}{2} \right]_{r}^{2}\left( {{r}^{2}} \right)dr\left( \cos \theta \right)d\theta }} \\
 & \Rightarrow I=\int_{0}^{2\pi }{\int_{0}^{2}{\left( 2-\dfrac{{{r}^{2}}}{2} \right)\left( {{r}^{2}} \right)dr\left( \cos \theta \right)d\theta }} \\
 & \Rightarrow I=\int_{0}^{2\pi }{\int_{0}^{2}{\left( 2{{r}^{2}}-\dfrac{{{r}^{4}}}{2} \right)dr\left( \cos \theta \right)d\theta }} \\
\end{align}\]
Now, we solve the integration with respect to $r$ as
\[\begin{align}
  & \Rightarrow I=\int_{0}^{2\pi }{\left[ \dfrac{2{{r}^{3}}}{3}-\dfrac{{{r}^{5}}}{10} \right]_{0}^{2}\left( \cos \theta \right)d\theta } \\
 & \Rightarrow I=\int_{0}^{2\pi }{\left( \dfrac{2{{\left( 2 \right)}^{3}}}{3}-\dfrac{{{\left( 2 \right)}^{5}}}{10}-0 \right)\left( \cos \theta \right)d\theta } \\
 & \Rightarrow I=\int_{0}^{2\pi }{\left( \dfrac{16}{3}-\dfrac{16}{5} \right)\left( \cos \theta \right)d\theta } \\
 & \Rightarrow I=\dfrac{32}{15}\int_{0}^{2\pi }{\left( \cos \theta \right)d\theta } \\
\end{align}\]
Now, we know that $\int{\cos \theta d\theta }=\sin \theta $. So we get
\[\begin{align}
  & \Rightarrow I=\dfrac{32}{15}\left[ \sin \theta \right]_{0}^{2\pi } \\
 & \Rightarrow I=\dfrac{32}{15}\left( \sin 2\pi -\sin 0 \right) \\
 & \Rightarrow I=\dfrac{32}{15}\left( 0-0 \right) \\
 & \Rightarrow I=0 \\
\end{align}\]
Hence, the given integral is equal to zero.

Note:
The differentials of the rectangular coordinates $dz dx dy$ in the above question are not simply replaced with the differentials of the cylindrical coordinates $dzdrdd\theta $. The cylindrical coordinate $r$ is also multiplied with them and they are written as $rdzdrdd\theta $ instead of $dzdrdd\theta $. Also, after solving the equation ${{r}^{2}}=4$, we will get the two values for $r$, which will be $2$ and $-2$. But since $r$ denotes the distance, so it cannot be negative and will begin from zero only. It is for this reason that we have taken the limits of $r$ from zero to $2$.