<s>[INST] Volume above a cone and within a sphere, using triple integrals and cylindrical polar coordinates Consider the region above the cone $z = \sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2 = 16$. 
Use cylindrical polar coordinates to show that the volume of region $R$ is $\frac{64\pi}{3}(2-\sqrt{2})$.
To solve this, I took the limits of $\theta$ as $0$ to $2π$.
I then took the limits of $z$ by substituting $z^2 = x^2+y^2$ (cone) into the equation of the sphere, giving me $z = 2\sqrt{2}$ as my upper bound (and $0$ as the lower bound).
I finally then took the limits of $r$ by changing the equation of the sphere into $r = \sqrt{16-z^2}$ (using $r^2 = x^2+y^2$), and $r = z$ (using $r^2 = x^2+y^2 = z^2$, the equation of the cone).
This gave me
$$\int_{0}^{2\pi}\int_{0}^{2\sqrt{2}}\int_{z}^{\sqrt{16-z^{2}}}rdrdzd\theta$$

"∫2π 0 ∫2√2 0 ∫(√16-Z^2) Z  rdrdZdθ"

However, the answer I got from this integral was $2π(16√2 - 16/3(√2))$, which was incorrect.
 [/INST] The volume is give by
$$V=\iiint_{E}\, {\rm d}V$$
where $E$ is the solid built by above of the cone $z=\sqrt{x^{2}+y^{2}}$ and inside of the sphere $x^{2}+y^{2}+z^{2}=4^{2}$.
Using cylindrical coordinates $(r,\theta,z)$:
$$x=r\cos\theta,\quad y=r\sin\theta,\quad z=z$$
The intersection is a projection over $(X,Y)$ is given by
$$\{x^{2}+y^{2}+z^{2}=4^{2}\}\cap\{z=+\sqrt{x^{2}+y^{2}}\}\implies x^{2}+y^{2}=(\sqrt{8})^{2}$$
Then,
\begin{align*}
V&=\int_{0}^{2\pi}\int_{0}^{\sqrt{8}}\left(\sqrt{16-r^{2}}-\sqrt{r^{2}}\right)\color{red}{r}\, {\rm d}r{\rm d}\theta,\\
&=\int_{0}^{2\pi}\frac{32}{3}(2-\sqrt{2})\, {\rm d}\theta,\\
&=\boxed{\frac{64\pi}{3}(2-\sqrt{2})}
\end{align*}
Using spherical coordinates $(r,\theta,\phi)$:
$$x=\rho\cos\theta\sin\phi,\quad y=\rho\sin\theta\sin\phi,\quad z=\rho\cos\phi$$
Then,
\begin{align*}
V&=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{4}\color{red}{\rho^{2}\sin\phi}\,{\rm d}\rho\, {\rm d}\phi\, {\rm d}\theta,\\
&=\int_{0}^{2\pi}\frac{4^{3}}{3}\left(-\cos\frac{\pi}{4}+\cos 0\right)\, {\rm d}\theta,\\
&=\boxed{\frac{64\pi}{3}(2-\sqrt{2})}
\end{align*}
as desired.
</s>