How to calculate this surface area? (portion of a cylinder inside a sphere ) 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!

$A = \iint dS$
$S: x^2 + y^2 = 8y$
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.
$16\sin^2 2\theta + 16 -32\cos 2\theta + 16\cos^2 2\theta + z^2 = 64\\
z^2 = 32 + 32\cos 2\theta = 64\cos^2\theta$
$2\int_0^{\pi}\int_0^{8\cos\theta} 8 \ dz\ d\theta$
