I have a problem where I have to find volume of figure formed, when $x^2+y^2+z^2=16$ and $ x^2+y^2=6z$ intersects if $z\geq 0$.
Here is a graphic for clarity:
So far I have transformed the problem to cylindrical coordinates (given the fact that we are dealing with sphere):
$x=r\cos{\varphi}; \quad y=r\sin{\varphi}; \quad z=z \quad \Rightarrow \begin{cases} x^2+y^2+z^2=16 \quad \Rightarrow & r^2+z^2=4^2 \\ x^2+y^2=6z \quad \Rightarrow& r^2=6z\end{cases}$
Then I can say that $-z^2+16=6z \quad \Rightarrow -z^2-6z+16=0 \quad \Rightarrow z_1=2$ and $z_2=-8$ (we don't need this one)
Then $r^2=6z \quad \Rightarrow r=\sqrt{12}=2\sqrt{3}$
So the volume would now be: $$\int _0^{2 \pi }\int _0^{2 \sqrt{3}}\int _{\frac{r^2}{6}}^{\sqrt{-r^2-16}}dzdrd\varphi=\int _0^{2 \pi }\int _0^{2 \sqrt{3}}\left(\sqrt{-r^2-16}-\frac{r^2}{6}\right)drd\varphi$$
At this moment I realize that I do not know how to integrate this integral. I was hoping to see Step-by-step solution in WolframAlpha, but no success there (The definite integral of inner integrand is complex)
How would I have to proceed with my problem?
EDIT:
People pointed out that I have forgotten to transform $dxdydz$ to $rd\varphi dr dz$ Also I should have $\sqrt{16-r^2}$ instead of $\sqrt{-16-r^2}$.
So now the volume would be calculated by:
$$\int _0^{2 \pi }\int _0^{2 \sqrt{3}}\int _{\frac{r^2}{6}}^{\sqrt{16-r^2}}rdzdrd\varphi=\int _0^{2 \pi }\int _0^{2 \sqrt{3}}\left (r \cdot \left (\sqrt{16-r^2}-\frac{r^2}{6} \right )\right)drd\varphi$$
Thank to @Lucian I realized, that, the problem could be solved easily, by calculating volume when rotating $x=\sqrt{16-z^2}$ and $x=\sqrt{6z}$ about the $Oz$ axis. Therefore my volume would be: $$\int_{2 \sqrt{3}}^4 \pi \left(16-z^2\right) \, dz+\int_0^{2 \sqrt{3}} \pi 6 z \, dz=\frac{236 \pi }{3}-24 \sqrt{3} \pi$$
(If anyone sees something wrong, please indicate my mistakes)