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Surface Area of Cylinder inside of sphere

  1. Oct 31, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the area of that part of the cylinder x^2 + y^2 = 2ay that lies inside the sphere x^2 + y^2 + z^2 = 4a^2.

    2. Relevant equations

    If a surface S can be parametrized in terms of two variables u and v, then dS = Norm[dR(u,v)/du x dR(u,v)/dv].

    The surface area is given by ∫∫dS where the integral is evaluated over the given region (here it would be over the region of the cylinder that lies within the sphere).

    3. The attempt at a solution

    I completed the square of the cylinder to get x^2 + (y-a)^2 = a^2

    I parametrized the cylinder as follows (where t stands for theta):

    R(t,z) = (rcos(t),rsin(t),z)
    r = 2asin(t)

    So R(t,z) = (2asintcost,2asintsint,z)

    I computed dS = Norm[dR(t,z)/dt x dR(t,z)/dz] and got 2adtdz, which I know is correct (same answer as textbook).

    For the given cylinder, t goes from 0 to Pi. By equating x^2 + y^2 in the equations given (i.e. sphere and cylinder), we get:

    2ay = 4a^2 - z^2
    z^2 = 4a^2 - 2ay
    z^2 = 4a^2 - 2a(2a(sint)^2)
    z^2 = 4a^2(1-(sint))^2
    z^2 = 4a^2((cost)^2)
    z = 2acost

    So z goes from -2acost to 2acost.

    So the Area = ∫∫1dS
    = ∫∫2adzdt, where t goes from 0 to Pi and z from -2acost to 2acost.

    We get:

    A = ∫2a(2acost + 2acost)dt
    A = ∫2a(4acost)dt
    A = ∫8a^2costdt
    A = 8a^2∫cost

    But the integral of cost from 0 to pi = 0.

    So A = 0.

    The textbook uses symmetry and integrates from 0 to pi/2 (i.e. finds the area of half of the cylinder under the sphere), and then multiplies by 2. Which makes sense. But why doesn't this method work? Furthermore, ∫cost from 0 to pi is NOT EQUAL to 2∫cost from 0 to pi/2, so the two methods are not compatible. So what did I do wrong?

  2. jcsd
  3. Oct 31, 2014 #2


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    Homework Helper

    Your integral equals zero becuase you are taking negative values into account. For the volume, your should have positive values. The volumes on either side of the axis should not cancel eachother out, they should add together. That is why doubling half is better.
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