Volume of $K = \left\{ (x,y,z)\in\mathbb{R}^3 \bigm| x^2+y^2+z^2\leq 8\,\wedge\, 4z \geq x^2+y^2+4 \right\}$ One needs to find the volume of the following body:
$$K = \left\{ (x,y,z)\in\mathbb{R}^3 \bigm| x^2+y^2+z^2\leq 8\,\wedge\, 4z \geq x^2+y^2+4 \right\}$$
My approach
I might be able to use cylinder coordinates to transform it to $$K = \left\{ (r,z)\in\mathbb{R}^3 \bigm| r^2+z^2\leq 8\,\wedge\, 4z \geq r^2+4 \right\}$$
But I am not really sure how to proceed from here. At the end I should end up with an integral of the form of $$ \int _{0} ^{2\pi} d\theta\int_{z_1} ^{z_2} dz \int_{x_1} ^{x_2} r dr  ~~~\lor ~~~  \int _{0} ^{2\pi} d\theta\ \int_{x_1} ^{x_2} r dr  \int_{z_1} ^{z_2} dz$$
Therefore I must find a parametrization of the area above which allows me to integrate. I am however not sure how I should proceed here or how one even starts such a problem.
 A: I think it's easier to have the integral with respect to $z$ on the inside. This is because it seems easier to me to describe which $z$ lie between the two surfaces $r^2+z^2 = 8$ and $4z = r^2+4$ for a given $r$ than which $r$ values lie between the surfaces for a given $z$ (the latter would require splitting into cases and solving one more quadratic equation).
So I want the outermost integral to be with respect to $r$, which means that I need to find for which $r$ the two surfaces intersect. We solve
$$
4z = r^2+4\\
r^2 = 4z-4
$$
Inserting this into the other equation gives
$$ 4z-4 + z^2 = 8\\
z^2 +4z -12 = 0\\
z^2 + 4z + 4 - 16 = 0\\
(z+2)^2 - 4^2 = 0\\
(z+2-4)(z+2+4) = 0\\
z = 2 \quad\lor\quad z = -6
$$
Since $z$ is clearly positive when the two surfaces intersect, we get $z = 2$, and thus $r = 2$.
So we have
$$
\int_0^{2\pi}d\theta\int_0^2r\,dr\int_{r^2/4 + 1}^{\sqrt{8-r^2}}dz\\
2\pi\int_0^2r\left(\sqrt{8-r^2} - \frac{r^2}4 - 1\right)dr\\
= 2\pi\left(\int_0^2r\sqrt{8-r^2}\,dr - \int_0^2\left(\frac{r^3}4 + r\right)dr\right)
$$
The second integral is immediate, and the first requires a substitution $(8-r^2)\mapsto u$.
