I have to find the volume of the solid which base is bounded by $$x^{2}+y^{2}+2y=0$$ and it's bounded, above, by the surface $$z=4-x^{2}-y^{2}$$
I tried to use cylindrical coordinates, where $$x=r\cos\theta$$ $$y=r\sin\theta$$ and $$0\leq r\leq-2\sin\theta$$ $$\frac{\pi}{2}\leq\theta\leq\frac{3\pi}{2}$$
My points are: can $r$ be negative? Is the volume given by the double integral $$\int_{\pi/2}^{3\pi/2}\int_{0}^{-2\sin\theta}(4-r^{2})rdrd\theta?$$
Hint: Take shifted cylindrical coordinates where $$\begin{cases}x=r\cos t\\ y=-1+r\sin t\\ z=z\end{cases}$$ with $t \in [0,2\pi].$