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Homework Help: Surface Area 1

  1. Apr 21, 2008 #1
    The part of the plane 3x+2y+z=6 that lies in the first octant.

    [tex]A(s)=\int_0^2\int_0^3\sqrt{14}dydy[/tex]

    Are my limits not correct? B/c my answer is just off by a little.

    me: 6\sqr(t14), answer: 3\sqrt(14)
     
  2. jcsd
  3. Apr 21, 2008 #2

    Dick

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    No, not correct. The domain of integration is a triangle in the x-y plane, not a rectangle. So the y limits should depend on x (or vice versa).
     
  4. Apr 21, 2008 #3
    dope! Gotcha, thanks.
     
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