Surface Integrals

  • #1

Homework Statement


Consider the surface [tex] S_1 [/tex] described by the equations

[tex] x = (1-w)^3cos(u), y = (1-w)^3sin(u), z = w, 0 <= u < 2\pi, 0 <= w < 1 [/tex]

The first few parts of the question were quite simple. Firstly we had to calculate dS and then compute the surface integral for the vector field [tex] \mathbf{a}= (y-z)\mathbf{i} + (z-x)\mathbf{j} + 2z \mathbf{k} [/tex]

The following part was a surface [tex] S_2 [/tex] which is a circle of radius 1 centered at the origin at the z = 0 plane

The Attempt at a Solution



Now, I'm unsure if I've just missed something blatantly obvious, but I was attempting to to find [tex] d\mathbf{S} [/tex] which I get to be [tex] d\mathbf{S} = -\mathbf {k}dS [/tex]. But when I dot this with [tex] {\boldmath a} [/tex], since z has reduced to zero in the vector field I get 0

Can anyone give me a push in the right direction? Thanks
 
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Answers and Replies

  • #2
vela
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I think the only thing obvious you missed is that the integral over S2 is equal to 0. As you noted, in the z-plane, the vector field a lies in the plane, so there's no flux crossing that area.
 
  • #3
Ah great thanks. After using the divergence theorem to compute the volume of the body, (which I did by adding the integral of s1 to the integral of s2) the next part is to verify the result by explicitly calculating the volume in polar coordinates.
To do so I calculated [tex] \int_v (\nabla \cdot \mathbf{a})dV [/tex] which gave me the same volume as I calculated above but negative

Is that the correct approach?
 
  • #4
vela
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Yes, that's the right approach. If you're off by a sign, it may mean you got the direction of the normal backwards when calculating the surface integral. Make sure it points outward from the volume.
 
  • #5
I was comparing homework questions with a friend earlier and they've confused me now. They were assuring me that for the part where we have the two surfaces and have to use the divergence theorem, that we add the two surfaces together on the RHS (as I did) but on the LHS calculate [tex] (\nabla \cdot \mathbf{a}) [/tex] so that we are left with [tex] 2\int_vdV [/tex] which we integrate to get V and divide the RHS by 2 to get the volume. Whereas after calculating the sum of the two surface integrals I left my answer for volume at that. The following part (using their method) would then involve simply evaluating [tex] \int_vdV [/tex] in cylindrical coordinates. But I excplicitly calculated [tex] \int_v (\nabla \cdot \mathbf{a})dV [/tex] at this point.

I've since worked through both methods and the approach they adopted gives a volume of [tex] \pi/7 [/tex] whereas my method gave a volume [tex] 2\pi/7 [/tex]

We haven't done any similar examples and rushed through the topic quite quickly that I'm in quite a muddle over which one is correct.

Thanks for all of your help so far
 
  • #6
vela
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You're not calculating the volume. The theorem says

[tex]\oint_S \mathbf{a}\cdot d\mathbf{S} = \int_V (\nabla\cdot\mathbf{a})\,dV[/tex]

The righthand side isn't generally equal to the volume. For example, in physics, we have ∇⋅E=ρ, where E is the electric field and ρ is the electric charge density. The righthand side is then the integral of the charge density over a volume, which yields the total charge contained in the volume. The only time the RHS will give you the volume is if the divergence of the vector field is equal to 1, which it isn't in general.
 
  • #7
Okay, thanks. That makes sense

One final point I wanted to ask about (sorry to have dragged this out so much!) in calculating the volume using spherical polar coordinates, i'm getting a negative answer still...

I think it could be perhaps a problem with my limits, which I've used (in cylindrical coordinates)

[tex] {\rho = 1 - 0} [/tex],
[tex] {\phi = 0 - 2\pi} [/tex]
[tex] {z = 0 - \rho^{1/3}} [/tex]

In particular its that last limit that I'm unsure about, we are told at the start that z = w, so I deduced that [tex] {\rho = (1-w)^3} [/tex] and substituted z in and rearranged it
 
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  • #8
vela
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I get z=1-ρ1/3 as the upper limit, not simply ρ1/3, but I think the reason for the negative sign of your answer is because you have the limits on ρ backwards. It should be from 0 to 1, not 1 to 0.
 
  • #9
Oh sorry that was supposed to be

[tex] z = 0 - (1 - \rho^{1/3}) [/tex]

I had considered the limit of [tex] \rho [/tex] to be the problem, but isn't the lower limit of [tex] \rho = 1 [/tex] when w's lower limit is 0?
 
  • #10
vela
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Yes, but w is just a parameter. You'd get the same surface if you said w runs from 1 to 0. Once you have the surfaces that bound the volume, then you just want to write the limits of the integrals so that they'll give you a positive answer when calculating the volume.
 
  • #11
Excellent. I'm happy with the question now! Thanks a lot for your very helpful input =)
 

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