Surface integral and divergence theorem over a hemisphere

  • Thread starter marineric
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  • #1
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Homework Statement



Please evaluate the integral [itex]\oint[/itex] d[itex]\vec{A}[/itex][itex]\cdot[/itex][itex]\vec{v}[/itex], where [itex]\vec{v}[/itex] = 3[itex]\vec{r}[/itex] and S is a hemisphere defined by |[itex]\vec{r}[/itex]| [itex]\leq[/itex]a and z ≥ 0,

a) directly by surface integration.

b) using the divergence theorem.

Homework Equations



-Divergence theorem in spherical coordinates

The Attempt at a Solution



Another one where the [itex]\vec{r}[/itex] messes me up. Simple enough if it was in regular xyz. Plus the [itex]\vec{v}[/itex]... and I don't really know where to start.
 

Answers and Replies

  • #2
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Well, [itex] \mathbf{r} = x \mathbf{e}_x + y \mathbf{e}_y + z \mathbf{e}_z [/itex] shouldn't be too confusing...

In this case, the shape of the integration area suggests that you might want to use spherical coordinates instead of x, y and z. Do you know what the area element is in spherical coordinates? (or even better, can you calculate it?) What about volume element?
 
  • #3
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Ok so attempt at a solution:

∫∫ 3[itex]\vec{r}[/itex] r[itex]^{2}[/itex]sinθdθd[itex]\phi[/itex]

limits are 0 to 2∏ for θ, and 0 to ∏/2 for [itex]\phi[/itex], or I could just do 3r^3 time the surface area of a hemisphere, which is 2*∏*r^2, so, 6*∏*a^5?

for divergence... do I just take the divergence in spherical coordinates and multiply by the volume of a hemisphere, which is 2/3*∏*r^3?
 
  • #4
938
9
Okay, couple of things

1) The surface area element is a vector.
2) Be careful with the angles. Right now your definitions do not work.
3) As you noticed, the integrand does not depend on the angles. Be careful about jumping over the integral though; now you're getting too many factors of r.



for divergence... do I just take the divergence in spherical coordinates and multiply by the volume of a hemisphere, which is 2/3*∏*r^3?
Is the divergence constant? If it is, then it should of course work.
 

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