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Mean Value Theorem in Surface Integrals

  1. Sep 2, 2009 #1
    I'm reading Div, Grad Curl, and All That, and in coming up with a formula for the divergence, H.M. Schey starts with a small cube centered at (x,y,z), labels the face parallel to the yz-plane as S1 and calculates

    [tex] \int\int_{S_1}\mathbf{F}\cdot\hat{\mathbf{n}}dS=\int\int_{S_1}F_x(x,y,z)dS [/tex]

    He then says that because the cube is small, the integral is equal to Fx evaluated at the center of S1 times the area of S1. The justification apparently comes from some mean value theorem that says "the integral of Fx over S1 is equal to the area of S1 multiplied by the function evaluated somewhere on S1."

    I tried looking up this theorem to no avail. Could someone point me to a proof or maybe outline why it's true?

    Thanks in advance.
     
  2. jcsd
  3. Sep 3, 2009 #2

    Landau

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    Take a look at your message, you'll see that the word 'mean value theorem' has become a link. It's essentially the two-dimensional version of
    [tex]\exists\ c\,\in\,(a,b): \int_a^b f(t)\ =\ f(c)\,(b\ -\ a)[/tex]
     
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