[Math] How to calculate this surface area? (portion of a cylinder inside a sphere )

Question

The surface area of ​​the portion of the cylinder $x^2+y^2=8y$ located inside of the sphere $x^2+y^2+z^2=64$

I'm stuck, so any tip will be helpful

Thanks in advance!

Answer 1

$A = \iint dS$

$S: x^2 + y^2 = 8y$

Convert to cylindrical.

$x = r\cos\theta\\ y = r\sin\theta\\ z = z$

Plug these into the equation of the cylinder. $r = 8\sin\theta$

And substitute back for parameterization of the surface

$x = 8\sin\theta\cos\theta = 4\sin 2\theta\\ y = 8\sin^2\theta = 4 - 4\cos 2\theta\\ z = z$

$dS = $$\|(\frac {\partial x}{\partial \theta}, \frac {\partial y}{\partial \theta},\frac {\partial z}{\partial \theta})\times (\frac {\partial x}{\partial z}, \frac {\partial y}{\partial z},\frac {\partial z}{\partial z})\|\\ \|(8\cos2\theta, 8\sin 2\theta, 0) \times (0,0,1)\| = \|(8\sin2\theta, -8\cos 2\theta, 0)\| = 8\ dz\ d\theta$

$\iint 8\ dz\ d\theta$

The sphere will establish the limits for z.

$16\sin^2 2\theta + 16 -32\cos 2\theta + 16\cos^2 2\theta + z^2 = 64\\ z^2 = 32 + 32\cos 2\theta = 64\cos^2\theta$

$2\int_0^{\pi}\int_0^{8\cos\theta} 8 \ dz\ d\theta$