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Surface Integral

  1. Apr 10, 2005 #1
    Could someone take a look at this please? Thanks

    Q. Evaluate Integral A.n dS for the following case:
    A=(6z, 2x+y, -x) and S is the entire surface of the region bounded by the cylinder x^2 + z^2 = 9, x=0, y=0, z=0 and y=8.

    Using Gauss' (or Divergence) Theorem:

    Integral A.n dS = Integral Div A dV

    Div A = 1, therefore

    Integral 1 dV

    When I last did this question, I said that dV = pi*r^2*h where r^2=9 and h=8. However, I've also got a factor of 1/4 to give me the final answer of 18pi. Where the hell did the quarter come from?! I'm guessing it has something to do with the limits and the fact that it's bounded by x=0, y=0 and z=0, but I don't understand why! Anyone offer some advice?!
  2. jcsd
  3. Apr 10, 2005 #2


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    You think it right. The bases of the cylinder are parallel to the (xy) plane at y =0 and y=8. The (z,y) plane at x=0 cuts the cylinder into half. The other plane, (x,y) at z=0 does the same, so you have a quarter of a cylinder of length h=8.

    Last edited: Jun 29, 2010
  4. Apr 10, 2005 #3
    thanks ever so much! :)
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