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Area of a cylinder inside a sphere (surface integral)

  1. Jun 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the area of the cylinder [itex]x^2 + y^2 -y = 0[/itex] inside the sphere [itex]x^2 + y^2 +z^2 =1[/itex]

    2. Relevant equations

    [itex] dA = sec \gamma dydz [/itex] where [tex]sec \gamma = \frac{|\nabla \phi|}{|\partial \phi/ \partial x|}[/tex]

    3. The attempt at a solution

    The method shown in this section is to calculate the angle of the area element to the plane ([itex]\gamma[/itex]) and integrate over the area of the projection onto the plane, in this case the y-z plane. The final answer will need to be multiplied by 2 to get the total area for both sides of the plane.

    In the above equation for [itex]sec \gamma[/itex], phi is the given expression for the cylinder. Evaluating gives [tex]sec \gamma = \frac {\sqrt{(2x)^2 + (2y-1)^2}}{2x}[/tex] [tex]= \frac{\sqrt{4x^2 + 4y^2 -4y +1}}{2x} = \frac{\sqrt{4(x^2 + y^2 - y) +1}}{2x} = \frac{1}{2x}[/tex][tex]= \frac{1}{2\sqrt{y-y^2}}[/tex]
    Completing the square
    [itex] = \frac{1}{2\sqrt{(y-\frac{1}{2})^2 -\frac{1}{4}}}[/itex]

    So the integral becomes [itex]\int^{1}_{z=-1} \int^{\sqrt{1-z^2}}_{y=0}\frac{1}{2\sqrt{(y-\frac{1}{2})^2 -\frac{1}{4}}}[/itex]

    Now to integrate this expression I used a trig substitution [itex] (y-\frac{1}{2}) = \frac{1}{2}sec \theta[/itex] but this leads to a big dirty expression inside a natural logarithm that I have no idea how to integrate for the second part. I figure I'm missing something fundamental since the answer is simply 4.

    Any suggestions?

    Thanks in advance.

    P.S. I noticed a thread on a topic very similar to this that was resolved using parameterization of the cylinder, but since I haven't covered this in the textbook, and the given method makes no mention of it, I'd like to solve it without using parameterizations if possible.
     
  2. jcsd
  3. Jun 23, 2014 #2
    It's just occurred to me that I completed the square incorrectly. So please ignore that part :D.

    Also, I feel like cylindrical coordinates might be helpful here, but when I change the integral, it seems to get even worse.

    Any assistance would be appreciated!
     
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