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Checking the divergence of a function

  1. Sep 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Check the divergence theorem using the vector function V = r^2 [itex]\hat{r}[/itex] + sin(θ) [itex]\hat{θ}[/itex] which is expressed in spherical coordinates. For the volume use a hemisphere of unit radius above the xy-plane (see figure below) (picture not shown, but I integrated r: 0 to 1, theta: 0 to pi/2, and phi 0 to 2pi)

    2. Relevant equations
    ∫(Div V) d[itex]\tau[/itex] = [itex]\oint V da[/itex]
    (Divergence Theorem)

    3. The attempt at a solution
    I did the RHS first because it's simpler. given da = R^2 sin(θ) dθ dψ [itex]\hat{r}[/itex] and R = 1
    [itex]\oint V da[/itex] → ∫sinθd from 0 to pi/2 * ∫dψ from 0 to 2pi, which is of course, 2pi.

    I assume that because of it's simplicity, this side must absolutely be correct. Unfortunately, the LHS is much harder.

    First I find Div V = 4r + 2cos(θ)/r, then I use d[itex]\tau[/itex] = r^2 * sinθ dr dθ dψ
    to obtain ∫ ( 4r + 2cos(θ)/r ) * r^2 * sinθ dr dθ dψ.

    Next I integrate with respect to r, 0 to 1, to get sinθ (cos+1) dθ dψ. This is where it gets weird.

    I integrate theta from 0 to pi/2 but I get 3/2. Since phi is from 0 to 2pi, and there are 0 phi terms, the answer must be 3pi. But this does not equal the LHS, which must be correct!

    I have been doing this for a couple of hours and I cannot think of a problem. Please help!
  2. jcsd
  3. Sep 13, 2013 #2
    Are you sure you're integrating over the entire boundary surface? Perhaps you're forgetting something...
  4. Sep 13, 2013 #3
    Done, thanks. =)
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