<s>[INST] 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.
 [/INST] An easier way (with the correct result!): by using  cylindrical coordinates we have
$$\begin{align*}\text{Area}&=\int_SdS=\int_E\sqrt{1+f_x^2+f_y^2}\,dxdy\\
&=
\int_E\sqrt{2}\,dxdy=\sqrt{2}|E|=\sqrt{2}\left(\pi\cdot 4\cdot 2\sqrt{2}\right)=\boxed{16\pi}
\end{align*}$$
where $$S=\{(x,y,f(x,y)):(x,y)\in E\}$$ with $f(x,y)=\sqrt{x^2+y^2}$
and $E$ is the $xy$-domain bounded by the ellipse $x^2 + 2y^2 = 16$ which has semiaxes $4$ and $2\sqrt{2}$.
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