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Verification of the Divergence Theorem

  1. Mar 21, 2009 #1
    The question I was given asks to verify the divergence theorem by showing that both sides of the theorem show the same result. With the divergence theorem obviously being [itex] \iint_S\mathbf{F}\cdot\mathbf{n}\,dS = \iiint_V \nabla\cdot\mathbf{F}\,dV [/itex].

    The vector field is [itex] \mathbf{F}=\mathbf{i}+\mathbf{j}+\mathbf{k} [/itex] and the surface through which it is flowing is the cone [itex] z=\sqrt{x^2+y^2} [/itex] below the plane [itex] z = 1 [/itex].

    Firstly, as far as I can tell, [itex] \iiint_V \nabla\cdot\mathbf{F}\,dV = 0 [/itex] since [itex] \nabla\cdot\mathbf{F} = 0 [/itex] which covers the right hand side of the divergence theorem.

    So then I divided the surface into two, that is the top of the cone and the rest of the cone.

    For the top of the cone I got the surface integral as [itex] \frac{\pi}{2} [/itex] since [itex] \mathbf{F}\cdot\mathbf{n} = (\mathbf{i}+\mathbf{j}+\mathbf{k})\cdot(\mathbf{k}) = 1 [/itex] so then you're just left with the surface integral of 1 which by definition is the area of the surface. And since the radius is [itex] \sqrt{\frac{1}{2}} [/itex], the area is [itex] \frac{\pi}{2} [/itex].

    Now for the other surface I found the unit normal to be [itex] \frac{\cos{t}+\sin{t}-1}{\sqrt{2}} [/itex] after parametrizing the surface as [itex] \mathbf{r} = z\cos{t}\mathbf{i} + z\sin{t}\mathbf{j} + z\mathbf{k} [/itex] thus finding [itex] \iint_S\mathbf{F}\cdot\mathbf{n}\,dS = -\sqrt{2}\pi [/itex].

    Obviously I wanted the result to be [itex] -\frac{\pi}{2} [/itex] so that the total surface integral would be equal to zero which would verify the divergence theorem for this problem. But it's not and I can't figure out where I went wrong. Since [itex] -\sqrt{2}\pi [/itex] is fairly close to [itex] -\frac{\pi}{2} [/itex] I figure I'm on the right track at least with my working.

    Hope this makes sense and thank you in advance for any assistance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 21, 2009 #2
    on the upper surface, z=1 so you get [itex]x^2+y^2=1[/itex] is eqn of circle radius 1 and so your upper integral should be [itex]\pi[/itex].

    how did you find the normal - you haven't specified a direction for the unit normal so at the moment it's just a scalar.
     
  4. Mar 21, 2009 #3
    Oh yes of course. Stupid mistake on my behalf. But if the integral of the top surface is [itex]\pi[/itex] I still don't get the correct answer.

    I found the unit normal using [itex] \frac{\partial{\mathbf{r}}}{\partial{t}}\times\frac{\partial{\mathbf{r}}}{\partial{z}} [/itex] then dividing the result by the magnitude. There's supposed to be an i, j and k on the 3 different terms in the unit normal, my bad.
     
    Last edited: Mar 21, 2009
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