I have to find the volume between the plane $z=3-2y$ and the paraboloid $z=x^2+y^2$. I understand that the domain of this region is the disk centred at (0,-1) with radius 2 if I set the equations equal to each other.
However I'm having a lot of trouble finding the correct limits for $r$ in cylindrical coordinates. I think that I have two options: the first is to take the coordinates from the centre of the domain as $x=rcos\theta$, $y=rsin\theta -1$. Then my integration bounds for $r$ become $0 \leq r \leq 2$. Then substituting the polar forms into the equations for the plane and paraboloid, I get that z should be integrated from $z = r^2 - 2rsin\theta + 1$ to $z=1-2rsin\theta$. Finally $0 \leq \theta \leq 2\pi$.
This seemed fine until I evaluated the triple integral and got a negative answer. So my next method was to take the polar coordinates to be from the origin rather than the centre of the domain, so I get that $z = r^2$ to $z=3 - 2rsin\theta$. The point that I get confused is where I try to find the bounds for $r$: since the domain is a disk not centred at the origin, the upper bound of $r$ in the integral should be a function in terms of $\theta$ but I can't find $r(\theta)$ explicitly.
I know that the maximum value of $r$ is given by the equation of the domain, which in polar form is the disk $r^2 = 3-2rsin\theta$. However it's the constant $3$ that is giving me trouble; I can't simply divide both sides by $r$ to find $r(\theta)$, so what can I do?
