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Surface Integral

  1. Dec 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Let S be the part of the paraboloid [itex]z=1+x^2+y^2[/itex] lying above the rectangle
    x between 0 and 1; y between -1 and 0 and oriented by the upward normal. Compute

    [itex]\int\int_SF\cdot n\,dS[/itex] where F=<xz, xy, yz>


    So I have Parametrized the surface S as r(x,y,z)=<x, y, 1+x2+y2>

    Then I have found dr\dx cross dr/dy =f

    then I found F(r(x,y)) dot f

    Now I need to integrate this over the domain of E but I am having trouble finding my bounds for x and y?

    I need to project the paraboloid downward onto x-y plane right? This gives a curve, oh wait, the curve is just the equation of the paraboloid with z=0 right?


    So the curve is 1+x2+y2=0

    why does that not sit well with me?
     
  2. jcsd
  3. Dec 17, 2008 #2
    Something is not right here. If my curve is 1+x2+y2=0

    That means that x=sqrt[-1-x2] which is just stupid.... wtf am I doing? :mad: 10 hours until the final and I'm effing up surface integrals:cry:

    And did I compute the cross product wrong? Should it be dr/dy "cross" dr/dx instead?
     
  4. Dec 17, 2008 #3
    I think I got it. If I project the paraboloid downwards, I get a circle of radius 1. So x=sqrt[1-y6] which is my UPPEr bound for x and y=sqrt[1-x6] which is my LOWER bound for y.

    Yes???:biggrin:

    Either way, the integral is retarded; F dot (dr/dy x dr/dx) is a huge mess. Converting to polar coordinates will help a little, but am I right to say that this is a long-a$$ integral?
     
    Last edited: Dec 17, 2008
  5. Dec 18, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The problem says " lying above the rectangle
    x between 0 and 1; y between -1 and 0" Those are your limits of integration.

    You should have
    [tex]\int_{x=0}^1\int_{y= -1}^0 (2x^2- 2x^4- 3x^2y^2+ 2xy^2+ y^2- y^4)dydx[/tex]
    not all that hard, surely.
     
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