1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface Integral using Divergence Theorem

  1. Dec 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Evaluate the surface Integral [tex]I=\int\int_S\vec{F}\cdot\vec{n}\,dS[/tex]

    where [tex]\vec{F}=<z^2+xy^2,x^100e^x, y+x^2z>[/tex]

    and S is the surface bounded by the paraboloid [itex]z=x^2+y^2[/itex]

    and the plane z=1; oriented by the outward normal.





    3. The attempt at a solution

    [tex]I=int\int_S\vec{F}\cdot\vec{n}\,dS=\int\int\int_E(div\vec{F})dV[/tex]

    [tex](div\vec{F})=y^2+x^2[/tex]

    [tex]\Rightarrow I=\int\int_D(\int_{z=x^2+y^2}^1(x^2+y^2)\,dz)\,dA[/tex]

    [tex]\Rightarrow I=\int\int_D(1-(x^2+y^2)\,dA[/tex]

    Is it just Polar Coordinates all the way home now?

    Thanks,
    Casey
     
  2. jcsd
  3. Dec 7, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No, I don't think that's right. The integral of f(x,y)dz is z*f(x,y), isn't it? Try evaluating that between the limits again.
     
  4. Dec 7, 2008 #3
    I am sorry, where are you referring to?
    The divergence of F= x^2+y^2 That is f(x,y), now if I integrate that wrt z
    ohh! I forgot to include the factor of f(x,y) so it should be

    [tex] I=\int\int_D(x^2+y^2)*(1-(x^2+y^2)\,dA[/tex]

    yes? Now Polar?
     
  5. Dec 7, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Now polar.
     
  6. Dec 7, 2008 #5
    Polar bears. Polar beers. Sweet numchuck skills Dick. Thanks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?