Given the cone $z^2=x^2+y^2$, $z\geq 0$ and the cylinder $z^2+y^2=64$
I am looking for the surface area of the section of the cone inside of the cylinder.
I can parametrize the cone as:
$r(t,\theta)=\langle tcos(\theta), tsin(\theta), t \rangle$
The cone gives me the inequality:
$z^2 \leq 64-y^2 \iff z \leq \sqrt{64-y^2}$ (since $z \geq 0$)
Plugin in the values of the parametrization:
$t^2 \leq 64-t^2\sin^2(\theta) \iff t \leq \frac{8}{\sqrt{sin^2(\theta)+1}}$
I know the Jacobian of the partial derivatives is:
$|r_\theta\times r_t| = \sqrt2t$
Thus the surface area should be:
$\int^{2\pi}_0\int^{\sqrt{\frac{8}{sin^2(\theta)+1}}}_0\sqrt2t$ $dt d\theta$
However with this method I end up trying to integrate $\frac{1}{sin^2(\theta)+1}$ with respect to $\theta$ which not only do I not know how to do, but it makes me think my method is wrong.
EDIT: attempting to compute that integral with the help of wolfram tells me the surface area is 0, which is trivially false. So I have done something wrong.
Best Answer
If you insist on solving the problem using parametrization, then try to make sense of the following hint.
Hint: A parametrization of your surface can be given by \begin{align} \phi(u, v) = (8u\cos v, 4\sqrt{2} u \sin v, u \sqrt{32+32\cos^2 v}) \end{align} where $0\leq u \leq 1$ and $0\leq v\leq 2\pi$. Remember to check that $\|\phi_u\times \phi_v\| = 64 u$.
Additional Hint: Parametrize the boundary of your generalized cone, then "connect it to the origin" via "scaling".