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Confirming divergence theorm example

  1. Dec 21, 2011 #1
    Hello, I am having trouble confirming that the flux integral is equal to the divergence over a volume. I am making a silly mistake & its just one of those days that I can't eyeball it. Here is the problem.

    I want to compute the flux integral for

    [itex]
    \vec{ F}=x\hat i+y\hat j-z\hat k
    [/itex]

    out of the closed cone

    [itex]
    x^2+y^2=z^2 \,\,\,\,\,\,\,0\leq z\leq1
    [/itex]

    Let us first do this the easy way using the divergence theorm.

    [itex]
    \int\int \vec F \cdot d \textbf{S}=\int \int \int \nabla \cdot \vec F dV
    [/itex]

    [itex]
    \nabla \cdot \vec F=1
    [/itex]

    [itex]
    \int \int \int \nabla \cdot \vec F dV=\int_0^{2\pi} d\theta \int_0^1 r dr \int_r^1 dz=\frac{\pi}{3}
    [/itex]

    This agrees with the formula for the volume of a cone of radius one and height one. Now, let us apply the flux integral directly.

    First lets look at the cone.

    [itex]
    \int\int \vec F \cdot d \textbf{S}=\int\int \vec F \cdot d \textbf{S}_1+\int\int \vec F \cdot d \textbf{S}_2
    [/itex]

    where S1 is the cone and S2 is the lid of the cone, namely the unit disk z=1.

    [itex]
    f=x^2+y^2-z^2=0
    [/itex]

    [itex]
    \nabla f=(2x,2y,-2z)
    [/itex]

    [itex]
    \hat n=\frac{\nabla f}{|\nabla f|}=\frac{(2x,2y,-2z)}{2 \sqrt{x^2+y^2+z^2}}=\frac{(2x,2y,-2z)}{2 \sqrt{2}z}
    [/itex]

    [itex]
    d\textbf{S}_1=\hat n \sqrt{\frac{\partial f}{\partial x}^2+\frac{\partial f}{\partial y}^2+\frac{\partial f}{\partial z}^2} dx dy=\hat n 2 \sqrt{x^2+y^2+z^2} dx dy= \hat n 2 \sqrt{2}z dx dy =(2x,2y,-2z)dx dy
    [/itex]

    [itex]
    \int\int \vec F \cdot d \textbf{S}_1=\int\int \vec F \cdot (2x,2y,-2z)dx dy=\int\int 2x^2+2y^2+2z^2 dx dy =\int\int 4x^2+4y^2 dx dy=\int_0^{2\pi}d\theta \int_0^1 r dr 4 r^2=2\pi
    [/itex]

    Over the disk [itex]\hat n=\hat k[/itex] and z=1

    [itex]
    \int\int \vec F \cdot d \textbf{S}_2=-\int_0^{2\pi}d\theta\int_0^1 r dr =-\pi
    [/itex]

    Hence,

    [itex]
    2\pi-\pi=\pi
    [/itex]

    So this is the wrong answer. I don't see where I went wrong. Can someone please help? Thank you.
     
  2. jcsd
  3. Dec 24, 2011 #2
    You applied the divergence theorem correctly. The flux over the circle is also correct.
    The mistake you made is when you computed the flux over S1: the cone. To define
    the normal for z=f(x,y) first write it in a format as follows:
    [tex]\sqrt{x^{2}+y^{2}}-z=0[/tex] Then the normal [itex] N[/itex] will be
    [tex]N=<\frac{x}{z},\frac{y}{z},-1>[/tex] From here you can do the work and show
    that the flux you get will be [tex]\frac{4\pi}{3}[/tex] This will confirm the divergence
    theorem. Let me know if this makes sense or if you have any further questions.
     
  4. Dec 24, 2011 #3
    I worked it out and I got the correct answer. Thank you!
     
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