Find the volume of the solid that lies above the cone : $z^2 = x^2 + y^2$ and inside the sphere $x^2 + y^2 + (z-2)^2 = 4$ Using spherical coordinates find the volume of the solid that lies above the cone: $z^2 = x^2 + y^2$ and inside the sphere $x^2 + y^2 + (z-2)^2 = 4$ 
I'm aware that there's a similar question here using a different method. 
Assuming it's the same problem I should get the same sol. but my attempts with spherical coor. yield a different answer so I need to know where the mistake is. 
My attempt : 
The solid is bounded above by the sphere centered at $(0,0,2)$ and below by the circle at which the sphere and the cone intersect: $x^2+y^2=2z$ (at $z=2$) whose projection on the $xy$-plane is $x^2+y^2=4$  
So the limits are given by : $$ Q= \{ 0 \leq \rho \leq 2 \;,\; 0 \leq \phi \leq \frac{\pi}{2} \;,\; 0 \leq \theta \leq 2\pi \}$$
$$
\begin{align}
V &= \underset{Q}{\iiint} \rho^2 \sin(\phi) \;d\rho \;d\phi \;d\theta \\
&= \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^{2} \rho^2 \sin(\phi) \;d\rho \;d\phi \;d\theta \\
&= \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \sin(\phi) \left[ \frac{1}{3} \rho^3  \right]_0^2 \;d\phi \;d\theta 
= \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \frac{8}{3} \sin(\phi) \;d\phi \;d\theta \\
&= \int_0^{2\pi} \frac{8}{3} \bigg[ \sin(\phi)  \bigg]_0^{\frac{\pi}{2}} \;d\theta \\
&= \frac{8}{3} \int_0^{2\pi} d\theta \\
&= \frac{16}{3}\pi
\end{align}
$$


Edit: Thank you for correcting me on the limits for $\rho$. But I imagined the problem to be computing the volume of the hemisphere $z = \sqrt{4-x^2-y^2}+2$. 
Mainly because it's supported by the answered questions in my textbook. The book claims the answer to be $\dfrac{16}{3}\pi$ but without showing the steps. Therefore I interpreted "the solid that lies above the cone" in the question as " (strictly) above but not inside". 
 Hence,
$$
\begin{align}
V &= \underset{Q}{\iiint} \rho^2 \sin(\phi) \;d\rho \;d\phi \;d\theta \\
&= \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_2^{4\cos(\phi)} \rho^2 \sin(\phi) \;d\rho \;d\phi \;d\theta \\
&(\cdots) \Rightarrow \; V = \frac{16}{3}\pi
\end{align}
$$

I attached an image visualizing my understanding of the problem.

As a check, the volume is the sum of a half-circle and a cone with a circular base, i.e.
$$ V= \frac{2\pi r^3}3 + \frac13 (\pi r^2)r=\pi r^3  \overset{r=2} = 8\pi$$
