# Find the volume of the solid bounded below by the circular cone and above the sphere

I'm asked to find the volume of the solid bounded below by the circular cone $$z= 1.5\sqrt(x^2+y^2)$$and above the sphere $$x^2+y^2+z^2=7.5z.$$

I tried to solve this using spherical coordinates, which gave me the bounds $$0 <= \rho <= 7.5 cos(\phi)$$ $$0 <= \phi <= arccot(1.5)$$ $$0<= \theta <= 2\pi$$ Are my bounds correct?

The spherical coordinates: $$x= \rho \cos(\theta)\sin(\phi)$$ $$y= \rho \sin(\theta)\sin(\phi)$$ $$z= \rho \cos(\phi)$$ We have $$x^2+y^2+z^2\leq7.5z \Rightarrow 0 \leq \rho \leq 7.5 \cos(\phi)$$ $$1.5\sqrt{x^2+y^2}=z \Rightarrow 1.5\sqrt{\rho^2 \sin^2(\phi)}=\rho \cos(\phi)\Leftrightarrow \operatorname{ctg}(\phi)=1.5$$ We have to find volume of the solid BELOW $z=1.5\sqrt{x^2+y^2}$ and ABOVE $x^2+y^2+z^2\leq7.5z$, so for azimuthal angle $\phi$ we have $$\operatorname{arcctg}(1.5) \leq \phi \leq \frac{\pi}{2},$$ and $$0 \leq \theta \leq 2 \pi.$$
P.S In case we have to find volume of the solid ABOVE $z=1.5\sqrt{x^2+y^2}$ and BELOW $x^2+y^2+z^2\leq7.5z$ , then $0 \leq \phi \leq \operatorname{arcctg}(1.5)$.