How to find $\phi$ for triple integral spherical coordinates Having trouble calculating bounds and how to do it generally for other terms.

I've been asked to find the volume of the solid bounded by the sphere 

$x^2 + y^2 + z^2 = 4$ 

and the cones

$z = \sqrt{x^2 + y^2}$

$z = -\sqrt{3x^2 + 3y^2}$

I assume that $0 &lt;= \rho &lt;= 2$ and $0 &lt;= \theta &lt;= 2\pi$, but I don't know how to find $\phi$ for either cones.
Using cylindrical coordinates, we have the equations $$\begin{align}
z^2&amp;=4-r^2,\\
z&amp;=r\\
z&amp;=-\sqrt 3 r
\end{align}.$$

The first equation gives $4-r^2\ge 0,$ or $0\le r\le 2.$ Then the last two give $-\sqrt 3 r\le z\le r.$ Finally, we have $0\le \phi\le 2π,$ since all the equations are independent of $\phi.$

So you want to calculate the $$\int_0^{2π}\int_0^2\int_{-\sqrt 3 r}^r{r\mathrm dz \mathrm dr\mathrm d\phi.}$$
Using cylindrical coordinates, we have the equations $$\begin{align}
z^2&amp;=4-r^2,\\
z&amp;=r\\
z&amp;=-\sqrt 3 r
\end{align}.$$

The first equation gives $4-r^2\ge 0,$ or $0\le r\le 2.$ Then the last two give $-\sqrt 3 r\le z\le r.$ Finally, we have $0\le \phi\le 2π,$ since all the equations are independent of $\phi.$

So you want to calculate the $$\int_0^{2π}\int_0^2\int_{-\sqrt 3 r}^r{r\mathrm dz \mathrm dr\mathrm d\phi.}$$