<s>[INST] 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 $\pi$
 [/INST] $\textbf{Second method}$ (I presume that this is the method you want): 
Show that the volume enclosed by the paraboloid $z=x^2+y^2$, the $xy$-plane $z=0$, and the cylinder $x^2+(y-1)^2=1$ is $\frac{3\pi}{2}$. 
This time, use the change of coordinates: 
$$
\begin{align*}
x &= r\cos \theta, \\ 
y &=r\sin\theta, \\ 
z &=z, \\ 
\end{align*}
$$
where 
$$
\begin{align*}
0 \leq &z\leq r^2, \\ 
0\leq &r \leq 2\sin\theta, \mbox{ and }\\ 
0 \leq &\theta \leq \pi. \\ 
\end{align*}
$$
Let $E$ be the solid defined in this prompt. 
Then the volume of the enclosed object is 
$$
\begin{align*}
V&=\iiint_E 1 \: dV \\ 
&= \int_{0}^{\pi}\int_{0}^{2\sin\theta} \int_{0}^{r^2} r\: dzdrd\theta \\
&= \int_{0}^{\pi}\int_{0}^{2\sin\theta} r^3 \: dr d\theta \\ 
&= \int_{0}^{\pi} \frac{r^4}{4}\Bigg|_0^{2\sin\theta} \: d\theta \\ 
&= \int_{0}^{\pi} 4\sin^4\theta\: d\theta \\ 
&= \int_{0}^{\pi} 4\left( \frac{1-\cos(2\theta) }{2}\right)^2 \: d\theta \\ 
&= \int_{0}^{\pi} \left( 1-\cos(2\theta) \right)^2 \: d\theta \\ 
&= \int_{0}^{\pi} \left( 1-2\cos(2\theta)+\cos^2(2\theta) \right) \: d\theta \\ 
&= \int_{0}^{\pi} \left( 1-2\cos(2\theta)+\frac{1}{2}+\frac{1}{2}\cos(4\theta) \right) \: d\theta \\ 
&=  \theta +\sin(2\theta)+\frac{1}{2}\theta +\frac{1}{8}\sin(4\theta) \Bigg|_0^{\pi} \\ 
&=\frac{3}{2}\pi + 0 + 0 \\ 
&=\frac{3}{2}\pi. 
\end{align*}
$$
</s>