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Calculate integral using Gauss theorem

  1. Nov 29, 2013 #1
    The problem statement, all variables and given/known data.

    Let
    ##D =\left\{\left(x,y\right)\ \in\ {\mathbb R}^{2}: x^{2} + y^{2} \leq 1\right\}## and let ##{\rm f}:D \to {\mathbb R}## defined as
    ##{\rm f}\left(x,y\right)
    = \left(1 - x^{2} -y^{2}\right)\exp\left(x^{4}y^{10}\right)##.
    Consider the surface ##S## given by the graph of ##{\rm f}## restricted to ##D##, oriented with the exterior normal vector. Let ##G## be the vector field
    ##G:{\mathbb R}^{3} \to {\mathbb R}^{3}## given by

    ##G(x,y,z) = \left(\,-y,\, x,\, x^{2} + y^{2}\,\right)\,,\qquad\mbox{Calculate}\ \int_{S} G\cdot dS##


    The attempt at a solution.

    I am having a hard time trying to visualize the surface ##S##.

    One possibility is to use Gauss theorem, which says that if ##W## is an elementary symmetric region in space where ##\partial W## is a closed oriented surface and ##F## is a function of class ##C^1##, then

    ##\iiint_w (divF).dV=\iint_{\partial W} F.dS##.

    I've calculated the divergence of ##G## and it gives ##0##. If I could find a region ##W## bounded by ##S##, then

    ##0=\iiint_w (divG).dV=\iint_{\partial W} G.dS=\int_S G.dS##

    Another doubt that I have is: how do I know ##S## is a closed surface?
     
  2. jcsd
  3. Nov 29, 2013 #2

    LCKurtz

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    It isn't a closed surface. And as such, the "exterior" normal doesn't make sense. Perhaps you mean the upward directed normal. Your surface is defined on the domain ##D:~x^2+y^2\le 1##. As a hint, what does your ##f(x,y)## look like on the boundary of ##D##?
     
  4. Nov 29, 2013 #3
    Yes, I've meant the upward directed normal, sorry, I am translating from spanish to english and sometimes I pick the wrong words. Hmm, ##f(x,y)=0## on the boundary of ##D##, and this tells me...? (I need a little more push).
     
    Last edited: Nov 29, 2013
  5. Nov 29, 2013 #4

    LCKurtz

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    It tells you that your open surface and your domain ##D## share the same boundary. So if you add the domain ##D## to your picture, you would have a closed surface. And that buys you ...
     
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