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Surface Integral With Divergence Thm

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##\mathit{F}(x,y,z) = (e^y\cos z, \sqrt{x^3 + 1}\sin z, x^2 + y^2 + 3)## and let ##S## be the graph of ##z = (1-x^2-y^2)e^{(1-x^2-3y^2)}## for ##z \ge 0##, oriented by the upward unit normal. Evaluate ##\int_{S} \mathit{F} \ dS##. (Hint: Close up this surface and use the Divergence Theorem)


    2. Relevant equations



    3. The attempt at a solution
    It's clear that ##div \ F = 0##, so if I was working with a closed surface, ##\int_{S} \mathit{F} \ dS## would equal 0. For the graph of ##S##, I am unsure as to what "closing up" a surface means. How would one close up a surface?
     
  2. jcsd
  3. Mar 26, 2013 #2
    Add the surface z=0 to creates closed surface. Then, the divergence theorem gives the surface integral of the sum of the surfaces; subtract the integral over what you added to get the surface integral of the desired surface.
     
  4. Mar 26, 2013 #3
    Ok, so what I think you're saying is I have to do:

    ##0 - \int_{plane \ z = 0} F \ dS##

    which will get me only the surface integral of the graph of z?
     
  5. Mar 26, 2013 #4

    Dick

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    Yes. And you'll need to pick the correct direction for the normal vector for the surface z=0.
     
  6. Mar 26, 2013 #5
    For it to be outward pointing, would the z-component have to be negative, ie, (0,0,-1)?
     
  7. Mar 26, 2013 #6

    Dick

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    Exactly.
     
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