Volume integral help I have a volume integral to compute with the following bounded volume $V\in \mathbb{R}^3$ $$ \frac{x^2+y^2}{4}+z^2\leq 1 \;\;,\;\; \frac{1}{2} \sqrt{x^2+y^2}\leq z\;,\;\; z\geq 0$$
I hadn't a clue how to do it until my lecturer said to use spherical coordinates $$\widetilde{r}(r,\theta,\phi)=\begin{pmatrix} 2rsin{\phi}cos{\theta} \\ 2rsin{\phi}sin{\theta} \\ rcos{\phi} \end{pmatrix}\;\;\;\;\;\; \phi:\left[0,\frac{\pi}{4}\right], \theta:\left[0,2\pi\right], r:[0,1]$$ As the cone cuts out the ellipsoid at $\frac{\pi}{4}$. However I'm having a hard time trying to show that the cone cuts out the ellipsoid at $\frac{\pi}{4}$ mathematically though. Also I don't see why $r$ is only from $[0,1]$ rather than $[0,2]$ as the largest value of $r$ is $2$ and occurs when the ellipsoid and cone intersect. Any help or hints? I've shown that it cuts the ellipsoid at $z= \frac{1}{\sqrt{2}}$ and therefore $x^2+y^2=2$ on that plane, however I can't seem to find the angle.
EDIT: Managed to find a way to compute the volume, answer is below. 
 A: 
Just a picture , a nice problem . I would find the centroid of the 2-d shape  that results from intersection and rotate it around the z-axis and then use the theorem of Pappus to find the volume, but that is just another way of doing the problem, not what you are being asked to do. Hope this helps. 
A: ANSWERING MY OWN QUESTION
Finally worked out a way to calculate it that I understand. :D.
Using cylindrical coordinates, the parametrisation becomes:
$$\widetilde{r}(r,\theta,z)=\begin{pmatrix} r\cos{\theta} \\ r\sin{\theta} \\ z \end{pmatrix}$$ Then rearranging the bounds for $z$ in terms of $r$ I achieve: $$\frac{1}{2}r\leq z \leq \frac{1}{2} \sqrt{4-r^2}$$ Also to find the bounds for $r$ I need all positive values of $r$ that still satisfy $$\frac{1}{2} r\leq \frac{1}{2}\sqrt{4-r^2}$$ Since this is the volume defined above. This gives $r:[0,\sqrt{2}]$. And then $\theta:[0,2\pi]$ as well. The Jacobian for cylindrical coordinates is $r$. Now I can do the integral: $$\iiint_{V} r \;dr d\theta dz$$ With the bounds from above this becomes $$\int_{0}^{2\pi} d\theta \int_{0}^{\sqrt{2}} dr\int_{\frac{1}{2}r}^{\frac{1}{2}\sqrt{4-r^2}} r \;dz$$ $$=2\pi\int_{0}^{\sqrt{2}}\frac{r}{2}\sqrt{4-r^2}-\frac{r^2}{2}\; dr$$ Using substitution $r=2\sin(t)$ we end up with $$2\pi\int_{0}^{\frac{\pi}{4}} 4{\cos^2{t}}\sin{t}-4{\sin^2{t}\cos{t}} \;dt$$ Which can be calculated to be $$\frac{4}{3}\pi(2-\sqrt{2})$$ 
