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Evaluation of Surface Integral in Gauss's Law

  1. Jul 7, 2016 #1
    I am a tenth grader, and a newbie to Advanced Calculus. While working out problems sets for Gauss's Law, I encountered the following Surface Integral:

    upload_2016-7-7_10-54-41.png


    I couldn't attempt anything, having no knowledge over surface integration. So please help.
     
  2. jcsd
  3. Jul 7, 2016 #2

    stevendaryl

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    Well, for the specific problem you are interested in, a surface integral can be thought of as just nested ordinary integrals: If [itex]f(x,y)[/itex] is some function that depends on two variables, [itex]x[/itex] and [itex]y[/itex], then [itex]\int f(x,y) dA = \int dx (\int dy f(x,y))[/itex]. So you first do the integral over [itex]y[/itex], treating [itex]x[/itex] as a constant. That gives you an expression that may involve the variable [itex]x[/itex]. Then you do the integral over [itex]x[/itex]. Or you can do it in the opposite order: first integrate over [itex]x[/itex], then integrate over [itex]y[/itex].
     
  4. Jul 7, 2016 #3
    Since the surface integral is an integration over a square area in the x-y plane, a single variable integration would also work. I'm not sure how far you've gone in your study of calculus, but if you're new to the subject it's possible you've seen the problem of finding area in the x-y plane via integration with respect to x or y. You could use the same basic approach for this problem (although I would personally lean towards stevendaryl 's approach in general):

    [itex]\int x dA = \int x y dx = \int x dx [/itex]
    y = 1 corresponds to the distance between the top of the square and the x-axis
     
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