By the structure of the conditions, I assume cylindrical coordinates is a suitable option. However I'm quite newby in that.
I want to use the change of coordinates:
$(\rho,\theta, z) \mapsto x=\rho \cos\theta , y=\rho \sin\theta, z=z$, with Jacobian $J=\rho$.
So that the volume is $\iiint\rho \:d\rho d \theta d z$.
From my the definition of $K$, I can guess that $\rho$ goes from $0$ to $\sqrt{1-\frac{z}{2}}$.
Also I managed to show that $4x+4y\leq z\leq 2-2(x^2+y^2)$.
I'm confused about what is the range for $\theta$, with that I will be able to construct the integral and find the volume of $K$.
Can someone explain a general approach to this kind of problems?
Note that\begin{align}4x+4y\leqslant2-2(x^2+y^2)&\iff x^2+y^2+2x+2y\leqslant1\\&\iff(x+1)^2+(y+1)^2\leqslant3.\end{align}So, use this change of coordinates: $x=-1+\rho\cos\theta$ and $y=-1+\rho\sin\theta$. Then $\theta$ can take any value in $[0,2\pi]$ and $\rho$ can take any value in $\left[0,\sqrt3\right]$. Finally\begin{multline}4x+4y\leqslant z\leqslant2-2(x^2+y^2)\iff\\\iff4\rho(\cos\theta+\sin\theta)-8\leqslant z\leqslant4\rho(\cos\theta+\sin\theta)-2\rho^2-2.\end{multline}So, the volume of $K$ is\begin{align}\int_0^{2\pi}\int_0^{\sqrt3}\int_{4\rho(\cos\theta+\sin\theta)-8}^{4\rho(\cos\theta+\sin\theta)-2\rho^2-2}\rho\,\mathrm dz\,\mathrm d\rho\,\mathrm d\theta&=\int_0^{2\pi}\int_0^{\sqrt3}6\rho-2\rho^3\,\mathrm d\rho\,\mathrm d\theta\\&=9\pi.\end{align}