Triple integral bounded above by $z=6-x^2-y^2$ and below by $z=\sqrt{x^2+y^2}$ in addition: $x+y\geq0$
As I understand it I should find the volume of a kind of ice cream cone(see image)
My setup in cylindrial coordinates:
$\int\int\int\,r\,dz\,dr\,d\theta$
$0 \leq r \leq 2$ (as the intersection between the two plots are the unit cirle with radius $2$
the integral of $dz$ with limits from $z= \sqrt{x^2+y^2}$ to $z=6-x^2-y^2$
Integrate the angle $\displaystyle{-\frac{\pi}{4}\leq\theta\leq\frac{3\pi}{4}}$.
I can integrate this, but can't get the correct answer I am told in the problem to integrate the region
Triple integral $xdV$.
$dV$ must be $r\,dz\,dr\,d\theta$ but why the $x$ there?