I want to set the triple integral in cylindrical coordinates which represents the volume of the solid region beteen the hyperboloid $z=\sqrt{12+x^2+y^2}$ and the parabloid $ z=8-x^2-y^2$.
What i have done:
The triple integral which represents the volume in rectangular coordinates is: $$V=\int\int\int{f(x,y,z)dxdydz}$$
But i have trouble in setting the boundaries of each dimension, so i can change to cylindrical:
$x,y$ have to be between a radio $R$ so i can set $-R<=x<=R$ and $y=\sqrt{R^2-x}$. but that $R$ must decreece as $z$ grows.
As your diagram shows the paraboloid is above the hyperboloid so we go from $\sqrt{12+r^2}$ to $8-r^2$ in the $z$ direction and by radial symmetry we go from $0$ to $2\pi$ in the $\theta$ direction.
The $r$ involves a little calculation. Equating $z^2$ gives us $$12+r^2=(8-r^2)^2\implies 12+r^2=r^4-16r^2+64\implies r^4-17r^2+52=0$$
From this we get $$(r^2-4)(r^2-13)=0\implies (r+2)(r-2)(r+\sqrt{13})(r-\sqrt{13})=0$$ and since we are only interested in positive $r$ we are left with $r=2$ and $r=\sqrt{13}$ as candidates. But, as can be seen by plugging $\sqrt{13}$ into the equation of the paraboloid, $z\lt 0$ for this value of $r$ so $r=2$ is the one we want.
Therefore, we go from $0$ to $2$ in the $r$ direction and since $dv=rdrdzd\theta=rdzdrd\theta$ in cylyndrical coordinates, our integral looks as follows.
$$\int_{0}^{2\pi}\int_{0}^{2}\int_{\sqrt{12+r^2}}^{8-r^2}rdzdrd\theta$$ Note that we sum in the $z$ direction first because its boundaries depend upon $r$.