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Evauate Surface integral

  1. May 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Evauate Surface integral
    [itex]\int\int_{\sigma}(x^2 + y^2)dS[/itex]
    where [itex]\sigma [/itex] is the portion of the sphere [itex]x^2 + y^2 + z^2 = 4 [/itex]
    above the plane z = 1.

    2. Relevant equations
    [itex]
    \int\int_{\sigma} f(x,y) \sqrt{\frac{\partial z}{\partial x}^2+\frac{\partial z}{\partial y}^2+1}
    [/itex]
    3. The attempt at a solution

    The plane and the sphere intercect at z=1 and x2+y2=3.
    so
    [itex]z=\sqrt{4-x^2-y^2}[/itex]

    we have

    [itex]
    \frac{\partial z}{\partial x}= \frac{-x}{\sqrt{4-x^2-y^2}} [/itex]
    and
    [itex]\frac{\partial z}{\partial y}= \frac{-y}{\sqrt{4-x^2-y^2}} [/itex]

    Z gives a projection of a disk onto the xy plane of x2+y2[itex]\leq 3[/itex]

    Which is
    [itex] R: 0 \leq \theta \leq 2\pi [/itex] [itex] 0\leq r \leq \sqrt{3} [/itex] in polar co-ordinates

    Therefore
    [itex]\int\int_{\sigma}(x^2 + y^2)dS[/itex]= [itex]\int\int_{R}(x^2 + y^2)\sqrt{(\frac{\partial z}{\partial x})^2+(\frac{\partial z}{\partial y})^2+1} [/itex]

    which equals


    [itex]\int\int_{R}(x^2 + y^2)dS[/itex]= [itex]\int\int_{R}(x^2 + y^2)\sqrt{(\frac{-x}{\sqrt{4-x^2-y^2}})^2+(\frac{-y}{\sqrt{4-x^2-y^2}})^2+1} [/itex]

    In polar co-ordinates
    We have I think.....

    [itex]\int_{0}^{2\pi}\int_{0}^{\sqrt{3}}\left[(r^2)2\sqrt{\frac{-1}{r^2-4}} (r) \right]dr d\theta[/itex]

    [itex]\int_{0}^{2\pi}\frac{-10}{3}d\theta[/itex]
    [itex] = \frac{-20\pi}{3}[/itex]

    Does this look alright ?
    I'm not sure about if I've changed the [itex]\sqrt{(\frac{-x}{\sqrt{4-x^2-y^2}})^2+(\frac{-y}{\sqrt{4-x^2-y^2}})^2+1}[/itex] to polar co-ordinates correctly
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 29, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That actually looks pretty good. Except where did you get that wacky (-1) factor? You are integrating a nonnegative function, x^2+y^2. The answer had better be nonnegative.
     
  4. May 29, 2009 #3
    Thanks alot!
    I don't think I was concentrating enough.
    It is very easy to get lost doing some of these integrals.
     
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