Using spherical coordinates to evaluate $\iiint_{E}z dV$ where $E$ lies above paraboloid $z = x^2 + y^2$ and below the plane $z=2y$ I was initially stuck trying to solve this question using spherical coordinates. I eventually got it by using cylindrical coordinates: $$\int_0^\pi \int_0^{2\sin\theta}\int_{r^2}^{2r\sin\theta}rz dz dr d\theta = \frac{5\pi}6$$
Here, we know that the paraboloid $z=x^2+y^2$ described in cylindrical coordinates is $z=r^2$ and the plane $z=2y$ is $z=2r\sin\theta$, since $r=\sqrt{x^2+y^2}$ and $y=r \sin\theta$. The parabloid and the plane intersect within the circle $r=2 \sin\theta.$
On my first attempt though I tried solving it using spherical coordinates explained as following.
Referring to the image ( https://i.stack.imgur.com/O5a94.png, where the curves are plotted on the $yz$ plane where $x=0$), $\phi_{min}$ can be determined by $\phi_{min} =\frac{\pi}{2}-\arcsin(\frac{4}{2{\sqrt{5}}})=\arcsin(\frac{\sqrt{5}}{5})$. $\rho_{max}$ is limited by the paraboloid $z=x^2+y^2$, so converting it to spherical coordinates by substituting  $x=\rho\sin\phi\cos\theta$, $y=\rho\sin\phi\sin\theta$, and $z=\rho\cos\phi$ becomes $$\rho\cos\phi=\rho^2\sin^2\phi\cos^2\theta+\rho^2\sin^2\phi\sin^2\theta=\rho^2\sin^2\phi(\cos^2\theta+\sin^2\theta)=\rho^2\sin^2\phi$$ Cancellation and isolation of $\rho$ gives $$\rho_{max}=\frac{\cos\phi}{\sin^2\phi}$$ Therefore, we get that the $E$ is defined as $$\{0\le\theta\le\pi, \arcsin\frac{\sqrt{5}}{5}\le\phi\le\frac{\pi}{2}, 0\le\rho\le\frac{\cos\phi}{\sin^2\phi}\}$$ Now we get that the original integral described in spherical coordinates is as following, where $z=\rho \cos\phi$ and the jacobian is $\rho^2\sin\phi$. $$\int_0^{\pi}\int_{\arcsin\frac{\sqrt{5}}{5}}^{\frac{\pi}{2}}\int_0^{\frac{\cos\phi}{\sin^2\phi}}\rho^3\sin\phi\cos\phi 
 d\rho d\phi d\theta \approx 8.37758040957$$ which is not correct. Either I have set up the integral wrong or this question is not possible to solve using spherical coordinates. If it's the former, can somebody please explain where I've messed up? Likewise, if it's the latter, can somebody please explain why it's not possible to solve this question using spherical coordinates?
For reference, here is the convention that I use for working with spherical coordinates (Stewart_Calculus_Early_Transcendentals_9e_p1102_fig5)
 A: Your mistake is that you have bounds of $\phi$ and $\theta$ independent of each other. In the order you have written, the integral should be,
$$ \int_0^{\pi} \int_{\arctan ((\csc \theta)/2)}^{\pi/2}\int_0^{\cot\phi \csc\phi}  \rho^3\sin\phi\cos\phi ~d\rho ~d\phi ~d\theta $$
In spherical coordinates,
$x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
Equation of the plane is, $z = 2y~$ or $~\rho \cos\phi = 2 \rho \sin\theta \sin\phi$
$$\implies \sin\theta = \frac 12 \cot\phi \tag1$$
You can use the above to find lower bound of $\phi$ in terms of $\theta$. That is the integral I have written above. Alternatively, using the symmetry about yz-plane,
$$\arcsin \left(\frac 12 \cot\phi\right) \leq \theta \leq \pi - \arcsin\left(\frac 12 \cot\phi\right)$$
From $(1)$, we can see that $$~\arctan \left(\frac 12\right) \leq \phi \le \frac {\pi} 2$$
Equation of the paraboloid is, $z = x^2 + y^2~$ or $~\rho \cos\phi = \rho^2 \sin^2\phi$ and that leads to the bounds of $\rho$ as,
$$ 0 \le \rho \le \cot\phi \csc\phi$$
Using symmetry about yz-plane, the integral becomes
$ \displaystyle  \int_{\arctan(1/2)}^{\pi/2}\int_{\arcsin \left(\frac 12 \cot\phi\right)}^{\pi/2} \int_0^{\cot\phi \csc\phi} 2 \rho^3 \sin \phi \cos\phi ~ d\rho ~ d\theta ~ d\phi$
