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Verifying Divergence Theorem

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Let E be the solid region defined by [tex]0 \leq z \leq 9+x^2+y^2[/tex] and [tex] x^2+y^2 \leq 16[/tex].
    Let S be the boundary surface of E, with positive (outward) orientation.
    Also, consider the vector field [tex]F(x,y,z)=<x,y,x^4+y^4+z>[/tex]

    There are five parts to the problem
    A) Compute the [tex]\int\int\int (div F)dV[/tex]

    B) Compute the flux of [tex]\int\int F \bullet dS[/tex]of F, across the (oriented) upper boundary S1 of E.

    C) Compute the flux of [tex]\int\int F \bullet dS[/tex]of F, across the (oriented) side boundary S2 of E.

    D) Compute the flux of [tex]\int\int F \bullet dS[/tex]of F, across the (oriented) bottom boundary S3 of E.

    E) Use A,B,C,D to verify the Divergence Theorem for F on E.

    2. Relevant equations



    3. The attempt at a solution
    For A)
    The Div F = 3. Also, I converted to cylindrical coords, so the limits of integration would be
    [tex]\\0 \leq z \leq 25;
    \\0 \leq r \leq 4;
    \\0 \leq \Theta \leq 2\Pi\\[/tex]

    [tex]\int_{0}^{2\Pi}\int_{0}^{4}\int_{0}^{25}(3r) dz dr d\Theta[/tex]
    This integral equals 1200Pi.

    For B)
    [tex]\iint_{S1} F \bullet dS[/tex]
    I parametrized the boundary of the curve as [itex]h(r, \Theta)=<r\cos\Theta, r\sin\Theta,25>[/itex]
    Then parametrized the vector field with those parameters.
    The partials with respect to each variable of h...
    [tex]h^{}_{r}=<cos\Theta, sin\Theta, 0>[/tex]
    and [tex]h^{}_{\Theta}=<-r\sin\Theta, r\cos\Theta,0>[/tex]
    Also, I determined the cross product of [itex]h^{}_{r}X h^{}_{\Theta} = <0,0,r>[/itex]

    So...(this is after parametrizing and dotting with the normal vector)
    Question Here: There are factors in this integral that I noticed are odd functions over a symmetric surface. That means that when I integrate them, won't they just go to Zero, and I can save myself time and not do them now correct?
    Assuming that is correct, this is the integral I came up with.
    [tex]\int_{0}^{2\Pi}\int_{0}^{4}(25r)*rdrd\Theta[/tex]

    I obtained an answer of [tex]\frac{1600\Pi}{3}[/tex]

    Now, I'm not exactly sure that is correct. I know I need to follow the same basic steps in order to determine the side flux and bottom flux integrals. I am having trouble figuring out the flux integral for the side boundary.
    Any assistance would be great!
     
  2. jcsd
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