Finding Volume using triple integrals Find the volume common to $y^2+z^2=4ax$ and $x^2+y^2=2ax$
I have problem establishing the common region. I know the region but unable to visualize it to get to the limits of the integral.
 A: As @abiessu hinted and you've got it, we have the below region (for example for $a=2$) shaped on $z=0$. Note that we can use the symmetric here to avoid additional calculations.

So, using the Cylindrical coordinates we have:
$$\theta|_0^{\pi/2},~~r|_0^{2a\cos\theta}$$
Of course: $$z|_0^{\sqrt{4ar\cos\theta-r^2\sin^2\theta}}$$ so $$V=4\int_{\theta}\int_{r}\int_z~rdzdrd\theta$$

A: $x^2 + y^2 = 2 ax \implies (x-a)^2 + y^2 = a^2$ is a cylinder centered at $(a,0)$ and parallel to $z-$ axis.
$y^2+z^2=4ax \implies \frac 1 {4a} (y^2 + z^2) = x $  is a parabloid opening towards $+$ ve $x$-axis. Putting $z=0$, gives two curves are $(x-a)^2 + y^2 = a^2$ and $y^2 = 4ax$,  and the circle completely lines inside the concave part of parabola.
That should give the integral (in Cartesian coordinate) as 
\begin{align*}
 & \; \; \; \int_{0}^{2a}\int_{-\sqrt{2ax -x^2}}^{\sqrt{2ax -x^2}} \int_{-\sqrt{4ax - y^2}}^{\sqrt{4ax - y^2}}  dz dy dx \\ 
 &= 4\int_0^{2a} \int_0^{\sqrt{2ax - x^2}} \int_0^{\sqrt{4 ax -y^2}}  dz dy dx \\ 
 &=  2 \left(\frac{8 a^3}{3}+\pi  a^3\right)
\end{align*}
evaluated by Mathematica.  

Integrate[1, {x, 0, 2 a}, {y, 0, Sqrt[2 a x - x^2]}, {z, 0, 
    Sqrt[4 a x - y^2]}]

On polar coordinates, 
\begin{align*}
 & \; \; \; \int_{-\frac \pi 2 }^{\frac \pi 2 } \int_{0}^{2a \cos(\theta) }\int_{-\sqrt{4a r \cos (\theta) - r^2 \sin^2(\theta)}}^{\sqrt{4a r \cos (\theta) - r^2 \sin^2(\theta)}} r \; dz dr  d\theta \\ 
 &=  4 \int_0^{\frac \pi 2} \int_0^{2a \cos \theta} \int_0^{\sqrt{4ar \cos\theta - r^2 \sin^2 \theta}} r \; dz dr d\theta 
\end{align*}
It seems that both Matlab and Mathematica does not return the value of integral for variable $a$.
For $a=1,2, ... ,5$

Table[Integrate[
    r, {\[Theta], -Pi/2, Pi/2}, {r, 0, 2 a Cos[\[Theta]]}, {z, 0, 
     Sqrt[4 a r Cos[\[Theta]] - r^2 Sin[\[Theta]]^2]} ], {a, 1, 5}]

yields, 
$$\left\{\frac{8}{3}+\pi ,\frac{64}{3}+8 \pi ,9 (8+3 \pi ),\frac{64}{3} (8+3 \pi ),\frac{125}{3} (8+3 \pi )\right\}$$
which is half the value of our integral and seems consistent with the one evaluated from Cartesian coordinate.
