Finding volume of the region enclosed by $x^2+y^2-6x=0, z = \sqrt{36-x^2-y^2}$ and $z=0$ I'm trying to calculate the volume of the region enclosed by the cylinder $x^2+y^2-6x=0$, the semicircle $z = \sqrt{36-x^2-y^2}$ and the plane $z=0$. I tried moving the region towards -x so that the cylinder has it's center at zero. I ended up using these equations and proposing this integral using cylindrical coordinates:
New cylinder --> $x^2+y^2=9$
New semicircle --> $z = \sqrt{36-(x+3)^2-y^2}$
$$\int_{0}^{2\pi}\int_{0}^{3}\int_{0}^{\sqrt{36-(rcos(\theta)+3)^2-(rsin(\theta))^2}} r\cdot dzdrd\theta$$
For some reason I can't solve it this way and I don't know what am I doing wrong.

If you take cylindrical coordinates $x = r \cos \theta, y=r \sin \theta, z=z$ with Jacobian $J=r$, then you'll have
$$\int_{0}^{2 \pi}\int_{0}^{6 \cos \theta}\int_{0}^{\sqrt{36-r^2}}rd\theta drdz$$
