Find volume bounded by 3 equations using integration The prompt is to find the volume of the solid bounded by the equations $x^2 + y^2 -2y = 0$, $z = x^2 + y^2$ and $z \ge 0$
Plotting the equations, I get something like this
 
We get a paraboloid and a cylinder.
How to know if I have to use double or triple integration to solve this problem? and if I should use spherical or double polar or cylindrical coordinate system?
Also how to find the limits of integration in this particular problem as an example?
This is what the cross section looks like 
It is clearly visible that $\theta$ goes from $0$ to $\pi$

First, complete the square to see that 
$$
x^2+y^2-2y=x^2 +(y-1)^2-1=0. 
$$
So $x^2+(y-1)^2=1$. The cylindrical change-of-coordinate we will be using is: 
$$
x=r\cos \theta, \hspace{4mm} y=1+r\sin \theta,\hspace{4mm}   z=z, \hspace{4mm} \mbox{ where } 0\leq r\leq 1,\hspace{4mm}  0\leq \theta\leq 2\pi. 
$$
Note that the absolute value of the Jacobian is:
$$
\left|\frac{\partial(x,y)}{\partial(r,\theta)}\right| 
= \left| \pmatrix{
x_r& x_\theta \\
y_r & y_\theta \\ 
}
\right| 
 = r. 
$$
Now, we calculate the volume of the solid: 
$$
\begin{align*}
\iint_{D} \int_0^{x^2+y^2} dz\: dA 
&= \iint_{D} x^2+y^2 \: dA \\ 
&= \int_{0}^{2\pi}\int_{0}^{1} \left( (r\cos\theta)^2 + (1+r\sin \theta)^2  \right) 
r\: dr d\theta \\ 
&= \int_{0}^{2\pi}\int_{0}^{1} \left( r^2\cos^2\theta+ r^2\sin^2 \theta + 1+2 r\sin \theta  \right) 
r\: dr d\theta \\ 
&= \int_{0}^{2\pi}\int_{0}^{1} \left( r^2 + 1+2 r\sin \theta  \right) 
r\: dr d\theta \\ 
&= \int_{0}^{2\pi}\int_{0}^{1} \left( r^3 + r + 2 r^2 \sin \theta  \right) 
  dr d\theta \\ 
&= \int_{0}^{2\pi} \left( \frac{r^4}{4} + \frac{r^2}{2} + 2 \frac{r^3}{3} \sin \theta  \right) \Bigg|_{0}^1 
 d\theta \\ 
&= \int_{0}^{2\pi} \left( \frac{1}{4} + \frac{1}{2} + \frac{2}{3} \sin \theta  \right)  
 d\theta \\ 
&= \frac{\pi}{2}+\pi -\frac{2}{3}\cos\theta\Bigg|_{0}^{2\pi} \\
&= \frac{3\pi}{2}. 
\end{align*}
$$
