I'm trying to find the surface area of the cone $x^2 + y^2 = z^2$ above the $xy$-plane and under the cylinder $y^2 + z^2 = 16$. I thought of using spherical coordinates $x = \rho \sin \phi \cos \theta,\quad y = \rho \sin \phi \sin \theta,\quad z= \rho \cos \theta$. In this case $\phi=\pi/4$. I know $0\leq \theta < 2\pi$ and $\rho$ ranges from 0 but the upper bound depends on $\theta$. To solve for this upper bound I calculate 

$$x^2+y^2=z^2,\quad y^2+z^2 = 16 \Rightarrow x^2 + 2y^2 = 16$$

$$\frac{1}{2}\rho^2\cos^2\theta + \rho^2\sin^2\theta = 16$$

$$\rho =\frac{4}{\sqrt{\frac{1}{2}\cos^2\theta +\sin^2\theta}}$$

Now I think I integrate

$$\int\_0^{2\pi}\int\_0^{\frac{4}{\sqrt{\frac{1}{2}\cos^2\theta+\sin^2\theta}}}\rho^2\sin\theta\, d\rho d\theta$$

But I'm not sure this is the right integral or even the right approach since that integral looks like hell.