Surface area of a cone intersecting a horizontal cylinder 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.

$x = \rho \cos\theta \sin\phi\\
y = \rho \sin\theta \sin\phi\\
z = \rho \cos\phi$
On the surface of the cone.
$\phi = \frac {\pi}{4}$
$x = \frac {\sqrt 2}{2}\rho \cos\theta\\
y = \frac {\sqrt 2}{2}\rho \sin\theta \\
z = \frac {\sqrt 2}2\rho$
We can drop the $\frac {\sqrt 2}2$ across terms and still have a good perametarizatinon for the surface.
Next what is dS?
$dS = \|(\frac{dx}{d\rho},\frac{dy}{d\rho},\frac{dz}{d\rho})\times(\frac{dx}{d\theta},\frac{dy}{d\theta},\frac{dz}{d\theta})\|$
$\begin{bmatrix}\cos\theta&\sin\theta&1\\
-\rho \sin\theta&\rho\cos\theta&0\end{bmatrix}$
$dS = \|(\rho\cos\theta, \rho \sin\theta, \rho)\| = 2\rho$
$\iint 2\rho \;d\rho\;d\theta$
Limits:
$y^2 + z^2 = 16\\
\rho^2\sin^2\theta + \rho^2 = 16\\
\rho = \sqrt {\frac{16}{1+\sin^2\theta}}$
$\int_{0}^{2\pi}\int_{0}^{\sqrt \frac{16}{1+\sin^2\theta}} 2\rho \;d\rho\;d\theta$
$\int_{0}^{2\pi}\rho^2|_0^{\sqrt \frac{16}{1+\sin^2\theta}} \;d\theta\\
\int_{0}^{2\pi}\frac{16}{1+\sin^2\theta} \;d\theta\\
\int_{0}^{2\pi}\frac{16\sec^2\theta}{\sec^2\theta+\tan^2\theta} \;d\theta\\
\int_{0}^{2\pi}\frac{16\sec^2\theta}{1+2\tan^2\theta} \;d\theta\\
8{\sqrt2}\tan^{-1}(\sqrt 2 \tan\theta)|_0^{2\pi}\\
16\sqrt 2$
