Volume of region inside sphere and cone 
Let $R$ consist of the points lying inside of the sphere $$ x^2 + y^2 + z^2 = 3^2 $$ and inside the cone $$ z = \cot(\alpha)\sqrt{x^2 +y^2} $$ where $\alpha$ is $\arccos(\frac15)$ Find the volume of $R$.

So I used cylindrical coordinates and set the two equations for $z$ equal to each other, and got $z = \pm \sqrt{9-r^2}$ as the bounds for $z$. Then for $r$ I replaced the expression for $z^2$ which I got from the the cone equation with the $z^2$ in the sphere equation. I got $r = \pm \frac{3}{\sqrt{1+\cot^2(\alpha)}}$ as the bounds. I think the bounds for r is where I made some mistake but I don't know what exactly.
I'm aware you can solve this using spherical coordinates as well, but I would like to solve the problem using cylindrical coordinates, and particularly learn where i went wrong in this problem.
The answer should be some rational number times $\pi$.

HINT


*

*The intersection between the cone and the sphere is for $z=\cot \alpha \sqrt{9-z^2}\ge0\\\implies z=\frac{3\cot \alpha}{\sqrt{1+\cot^2 \alpha}}$

*For symmetry we can set up the integral in two parts for $z\ge0$, notably


$$V=\int_0^{2\pi} \int_0^{\frac{3\cot \alpha}{\sqrt{1+\cot^2 \alpha}}}\int_0^{r_1(z)}r\,dr\,dz\,d\theta+\int_0^{2\pi} \int_{\frac{3\cot \alpha}{\sqrt{1+\cot^2 \alpha}}}^3\int_0^{r_2(z)}r\,dr\,dz\,d\theta$$
