I am instructed to convert this iterated integral in cartesian coordinates to spherical coordinates. $$ \int^{1}_{0} \int^{\sqrt{1-x^2}}_{-\sqrt{1-x^2}} \int^{\sqrt{4-x^2-y^2}}_{\sqrt{3x^2+3y^2}}\, dzdydx $$
This seems that the bounds of the region are between the equations $x^2+y^2+z^2=4$, $z^2=3x^2+3y^2$, and the y=axis. I think that the region is also bounded by the y-axis because of the lower bound of $x = 0$.
In spherical coordinates, I know that the $x^2+y^2+z^2=4$ is equal to $\rho = 2$. The $z^2=3x^2+3y^2$ will be transformed into $$ \rho^2 \cos^2\phi = 3(\rho^2-\rho^2 \cos^2\phi)\implies \phi = \frac{\pi}{6}. $$ At the latter, I conclude that the iterated integral will be $$ \int^{\frac{\pi}{6}}_0 \int^\pi_0\int^2_0 \rho^2 \sin\phi\, d\rho d\theta d\phi. $$
I got the conclusion about the bounds of $\theta$ because of the sphere. Overall, the resulting iterated integral in cartesian coordinates and the spherical coordinates is equal to $\frac{\pi}{2}$. I want to know now if my understanding about the conversion is correct. Is there a visual representation of this integral to fully understand on how triple integral in spherical coordinates works?