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Homework Help: Center of Mass and Surface Area (multivariable)

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data

    1. Verify the given moments of inertia and find the center of mass. Assume each lamina has a density of p=1. The problem gives a circle with a radius a.

    2. Find the area of the surface of the portion of the sphere x^2 + y^2 + z^2 = 25 inside the cylinder x^2 + y^2 = 9.

    2. Relevant equations

    3. The attempt at a solution

    1. I already verified the moment, it's the center of mass that's a problem. It gives a picture of a circle with its center at the origin. I would say the center of mass is (0, 0) since its center is at the origin and has a uniform density. Also, I did the math and all and it works out to 0 for x and y as well. The book however says the answer is (a/2, a/2)

    2. I wasn't sure about this one. I tried this double integral.

    [tex]\int^{2\pi}_{0}\int^{3}_{0}\frac{5}{\sqrt{25 - r^{2}}}r*dr*d\theta[/tex]

    (with all the partial derivative and plugging formula and converting to polar steps taken out)
    which comes out to 10pi, but the book says the answer is 20pi.
  2. jcsd
  3. Mar 31, 2010 #2


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    I really can't argue with your logic on the first one. But on the second one, I think you are only getting the area for the positive z part. There's also a negative z part.
  4. Apr 3, 2010 #3
    Does this mean I have to add another double integral with the equation of -5/sqrt 25-r^2 to get the right answer?
  5. Apr 3, 2010 #4


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    Not exactly. You did the z=sqrt(25-r^2) part. If you do the z=(-sqrt(25-r^2)) part (partial derivatives etc.) you'll get the same integral form.
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