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Finding surface area with a function. Double Integration.

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate ∫∫f(x,y,z)DS for the given surface and function.

    Part of the plain x+y+z=0, contained in the cylinder x^2+y^2=1 f(x,y,z)=z^2.
    2. Relevant equations

    ∫∫F(x,y,z)Ds=∫∫F(g(u,v)*||n(u,v)||

    N= TuXTv
    Tu= G(u,v)(du); Tv is the same only the derivative is with respect to v

    3. The attempt at a solution
    So, where I confused... is how to find the vector to use for G(u,v) from the cylinder and the plain. Aside from that I think I pretty much understand. Oh, and whould the upper and lower bounds be 1? Or do I use polar coordinates?
     
  2. jcsd
  3. Nov 21, 2012 #2

    LCKurtz

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    You have choices about how to parameterize the surface. You could choose ##x## and ##y## for your parameters$$
    x = x,\, y = y,\, z = 1-x-y\,\quad \vec R(x,y)=\langle x,y,1-x-y\rangle$$or you could choose ##r## and ##\theta## and parameterize it like this$$
    x=r\cos\theta,\,y=r\sin\theta,\,z = 1 - r\cos\theta-r\sin\theta$$and parameterize it accordingly. And, of course, the limits of integration depend on your variables. Even if you start with the ##x,y## parameterization you will likely change the resulting integral to polar coordinates because the domain is a circle.
     
  4. Nov 24, 2012 #3
    on the z=1-rcos(theta)-rsin(theta)... how did you get the 1
    if x+y+z=0
    So are you saying that the g path function would be (x,y,z)=(rcos(theta),rsin(theta),-rcos(theta)-rsin(theta)
     
  5. Nov 24, 2012 #4

    LCKurtz

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    That was a typo. I had read the equation as ##x+y+z=1##. So put 0 instead of 1 in the appropriate places. Sorry.

    That isn't the "g path". What it is is a parameterization of the portion of the plane inside the cylinder in terms of ##r## and ##\theta##.
     
    Last edited: Nov 24, 2012
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