Checking the divergence of a function

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SUMMARY

The discussion focuses on verifying the divergence theorem using the vector function V = r² &hat;r + sin(θ) &hat;θ in spherical coordinates, specifically over a unit hemisphere above the xy-plane. The right-hand side (RHS) was calculated as 2π, while the left-hand side (LHS) yielded an unexpected result of 3π after integrating the divergence of V, which was found to be 4r + 2cos(θ)/r. The discrepancy indicates a potential oversight in the boundary surface integration, prompting a request for assistance in identifying the error.

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Homework Statement


Check the divergence theorem using the vector function V = r^2 \hat{r} + sin(θ) \hat{θ} 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)

Homework Equations


∫(Div V) d\tau = \oint V da
(Divergence Theorem)

The Attempt at a Solution


I did the RHS first because it's simpler. given da = R^2 sin(θ) dθ dψ \hat{r} and R = 1
\oint V da → ∫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\tau = 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!
 
Physics news on Phys.org
Are you sure you're integrating over the entire boundary surface? Perhaps you're forgetting something...
 
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Done, thanks. =)
 

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