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!
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By area do you mean surface area? – theyaoster May 22 '17 at 19:30
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yes, I do! surface area – user418360 May 22 '17 at 19:31
1 Answers
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$A = \iint dS$
$S: x^2 + y^2 = 8y$
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.
$16\sin^2 2\theta + 16 -32\cos 2\theta + 16\cos^2 2\theta + z^2 = 64\\ z^2 = 32 + 32\cos 2\theta = 64\cos^2\theta$
$2\int_0^{\pi}\int_0^{8\cos\theta} 8 \ dz\ d\theta$
Doug M
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But I need the cylinder surface area, this would be the surface sphere area – user418360 May 22 '17 at 19:41
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@DougM I think the bounds in the double integral in the last line are wrong, because in the way you set them up the integral evaluates to $0$; using the symmetry of the problem you could write $2\cdot 2\int_{0}^{\pi/2}\left(\int_{-8\cos(\theta)}^{-8\cos(\theta)}8dz\right)d\theta=64.$ – lorenzo Jun 21 '22 at 20:19