The sphere and the cylinder intersect where $r=2$ and $\rho=5$, so $r=\rho\sin\phi\implies5\sin\phi=2\implies$
$\phi=\sin^{-1}\frac{2}{5}=\cos^{-1}\frac{\sqrt{21}}{5}$.
(Alternatively, $x^2+y^2=4$ and $x^2+y^2+z^2=25\implies$
$z^2=21\implies\rho^{2}\cos^{2}\phi=21\implies25\cos^{2}\phi=21\implies\cos\phi=\frac{\sqrt{21}}{5}\implies\phi=\cos^{-1}\frac{\sqrt{21}}{5}.)$
A line segment starting at the origin and passing through the cylinder intersects the cylinder where
$r=2$, so $\rho\sin\phi=2\implies \rho=\frac{2}{\sin\phi}=2\csc\phi$, so
this gives the lower limit for $\rho$.
Your approach is fine but the integral does not give the volume of the sphere inside the cylinder - i) the way polar angle works, it will leave out parts of the sphere inside the cylinder on the sides unless you specifically address it in your integral. ii) you take $0 \leq \rho \leq 1$ but that will not take out the correct volume for this sphere.
As you are trying to find the volume of the combined region between $0 \leq z \leq 4$, here is what I would suggest.
You have already found that the cylinder and sphere intersect at $z = \sqrt3$. So we can add the volume of the hemisphere ($V_1$) AND the volume of the cylinder between $\sqrt3 \leq z \leq 4 \, (V_2)$. We have to then just subtract from the hemisphere the volume of the spherical cap that is inside the cylinder above $z = \sqrt3 \,, \, (V_3)$.
$V_3$ can be calculated with the below integral,
$V_3 = \displaystyle \int_{0}^{2 \pi} \int_0^{\pi /6} \int_{(\sqrt3 / \cos \phi)}^{2} \, \rho^2 \sin(\phi) \, d \rho \, d \phi \, d \theta$
$V_1 = \frac{16 \pi}{3} \,$ as you mentioned
$V_2 = \pi \times 1^2 \times (4 - \sqrt3) = (4 - \sqrt3) \pi$
Total volume of the combined solid is $V_1 + V_2 - V_3$.
Best Answer
With usual shperical coordinates you would get it with $\sin(\varphi)$, so $$\int_0^{\pi} \int_{0}^{2\pi}\int_{t}^{1}(r^2)^{-\frac{p}{2}} \cdot r^2 \sin(\varphi) \ dr \ d\theta \ d\varphi=$$ $$= 2\pi \biggl[ -\cos(\varphi) \biggr]_0^{\pi}\int_{t}^{1}r^{2-p} \ dr =$$ which converges if and only if $p-2<1$ which means $p<3$, an it is: $$= 4\pi\int_{t}^{1}r^{2-p}= 4\pi \biggl[ \frac{r^{3-p}}{3-p} \biggr]_0^1=\frac{4\pi}{3-p}$$