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Homework Help: Surface Integral

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    Evaluate the surface integral [tex]\vec{F}\cdot\vec{n}\, dS[/tex]

    where [itex]\vec{F}=<-y,x,0>[/itex] and S is the part of the plane [itex]z=8x-4y-5[/itex] that lies below the triangle with vertices at (0,0,0,), (0,1,0,) and (1,0,0). The orientation of S is given by the upward normal vector. answer: 2


    I am not sure if I am just making a careless mistake or a conceptual one.

    3. The attempt at a solution

    i.) Parametrizing S gives [itex]\vec{r}(x,y)=<x, y, 8x-4y-5>[/itex]

    ii.) Finding [tex]\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y}=<1,0,8>\times<0,1,-4>=-8,4,1[/tex]

    iii.) Thus, [itex] \vec{F}(\vec{r}(x,y))\cdot (\frac{\partial r}{\partial x}\times \frac{\partial r}{\partial y})=<-y,x,0>\cdot<-8,4,1>=<4x+8y>[/itex]

    iv) Therefore [tex]I=\int\int_D (4x+8y)\, dA[/tex]

    [tex]=\int_{x=0}^1 \int_{y=0}^x (4x+8y)\,dy\, dx[/tex]

    I belive that if I made an error, it was made somewhere in here and not in my integration.

    Any major blunders here?

    Thanks,
    Casey
     
  2. jcsd
  3. Nov 30, 2008 #2

    Dick

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    Look at your integral limits. You are integrating over the triangle formed by (0,0,0), (0,1,0) and (1,1,0).
     
  4. Nov 30, 2008 #3
    Not sure I follow. My x bounds are correct right?
     
  5. Nov 30, 2008 #4
    Is my upper y bound supposed to be (1-x)?
     
  6. Nov 30, 2008 #5

    Dick

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    It sure is!!
     
  7. Nov 30, 2008 #6
    :redface: That's what I get for thinking I can do everything in my head...Teehee....
     
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