Evaluate this integral using cylindrical coordinates Find the volume of the solid bounded above by the paraboloid of revolution 
$z^{2}=x^{2}+y^{2}$
And below by the $xy$ plane, and on the sides by the cylinder $x^{2}+y^{2}=2ax$ 
We take $a>0$.
I'm struggling to understand what this would look like graphically, I understand how to find limits of integration for $x$ and $y$ but struggling to find them for $z$. So far i have equated the two terms, but i have got no where with that. 
Thanks 

This is what I have right now, so correct me if I'm wrong: 
I think you meant $z=x^2+y^2$ if you are trying to indicate a paraboloid of revolution. The graph of $z^2=x^2+y^2$ would be two cones. 
The cylinder $x^2 + y^2 = 2ax$ can be represented as follows: 
$\begin{align} x^2 + y^2 &= 2ax \\ 
x^2 - 2ax + y^2 &= 0 \\
\left(x^2 - 2ax + a^2\right) + y^2 &= a^2 \\
\left(x - a\right)^2 + y^2 &= a^2 \end{align}$
Since integrating over this circular region would be very complicated, let's translate both objects to the left by $a$ units and move the centre of the circle to the origin. The cylinder would become $x^2 + y^2 = a^2$ and your paraboloid of revolution would be $z = \left(x + a\right)^2 + y^2$. 
Now we evaluate the integral: 
$\begin{align} \iint\limits_{R}{f\left(x, y\right)\,dA} &= \iint\limits_{R}{r\,f\left(r\cos\theta, r\sin\theta\right)\,dr\,d\theta} \\ 
&= \int^{2\pi}_{0}{\int^{a}_{0}{r^3 - 2ar^2\cos\theta + a^2r\,dr\,d\theta}} \\ 
\, &= \int^{2\pi}_{0}{a^{4}\left(\frac{3}{4} - \frac{2\cos{\theta}}{3}\right)d\theta} \\ 
&= a^{4}\int^{2\pi}_{0}{\frac{9 - 8\cos{\theta}}{12}\,d\theta} \\
&= \boxed{\boxed{\frac{3\pi a^4}{2}.}} \end{align}$
