# 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:

$y^2 + z^2 = 16\\ \rho^2\sin^2\theta + \rho^2 = 16\\ \rho = \sqrt {\frac{16}{1+\sin^2\theta}}$

$\int_{0}^{2\pi}\int_{0}^{\sqrt \frac{16}{1+\sin^2\theta}} 2\rho \;d\rho\;d\theta$

$\int_{0}^{2\pi}\rho^2|_0^{\sqrt \frac{16}{1+\sin^2\theta}} \;d\theta\\ \int_{0}^{2\pi}\frac{16}{1+\sin^2\theta} \;d\theta\\ \int_{0}^{2\pi}\frac{16\sec^2\theta}{\sec^2\theta+\tan^2\theta} \;d\theta\\ \int_{0}^{2\pi}\frac{16\sec^2\theta}{1+2\tan^2\theta} \;d\theta\\ 8{\sqrt2}\tan^{-1}(\sqrt 2 \tan\theta)|_0^{2\pi}\\ 16\sqrt 2$