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Stokes and Divergence theorem questions

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1. Homework Statement
Let
[tex]\vec{F}=xyz\vec{i}+(y^{2}+1)\vec{j}+z^{3}\vec{k}[/tex]
And let S be the surface of the unit cube in the first octant. Evaluate the surface integral:


[tex]
\int\int_{S} \nabla\times \vec{F} \cdot \vec{n} dS
[/tex]
using:
a) The divergence theorem
b) Stoke's theorem
c) Direct computation
2. Homework Equations

Divergence Theorem:
[tex]\int\int\int\nabla \bullet \vec{F} dV = \int\int \vec{F} \bullet \vec{n} dS[/tex]

3. The Attempt at a Solution
I think I have b and c worked out but part a I am not sure of. What I don't understand is how I can use the divergence theorem here. If I substitute [tex] F = curl(F)[/tex] into the divergence theorem I will get zero by identity. So, how is that possible to evaluate? If someone could pelase tell me what I'm doing wrong I would greatly appreciate it.
 
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Answers and Replies

TMM
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You can't substitute some curl(A) for some B unless B's divergence is zero. When you take the divergence of F, is it zero?
 
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The divergence of F is yz + 2y + 3z^2.

I'm still not too sure how I could apply the divergence theorem here to help me evaluate that integral though.
 
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
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I can't see why one would need to take the curl of F? Surely to use the divergence theorem one merely needs to take the divergence of F x n?
 
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I realized I wrote down the probelm wrong here. The problem is actually Evaluate the surface integral:

[tex]
\int\int_{S} \nabla\times \vec{F} \cdot \vec{n} dS
[/tex]
That is why i've been trying to substitute the curl of F into the divergence formula. I've been saying that
[tex] \vec{B} = \nabla\times \vec{F} [/tex]
and then just using that in the divergence theorem, which is a flawed technique because it would always return zero since I would be taking the divergence of solenoidal fields all the time:

[tex]\int\int\int\nabla \cdot (\nabla \times \vec{B}) dV = 0[/tex]
 
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HallsofIvy
Science Advisor
Homework Helper
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It's not clear to me why you call that a "flawed" technique. Yes, the divergence of a solenoidal field is 0 and the integral over the unit cube is obviously 0. Are you saying that you DIDN'T get 0 using the divergence theorem and direct integration?
 
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I would think that this would be flawed because I could argue that any surface integral of a curl would be zero (by using the divergence theorem like i did above) which conflict with Stokes theorem because this is clearly not always the case. I must not be understanding the relationships between the theorems.
 
HallsofIvy
Science Advisor
Homework Helper
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Why would Stoke's theorem conflict? The divergence theorem connects the integral over a closed surface to the integral over the region the surface encloses. Stoke's theorem connect the integral over surface to an integral over the boundary of the surface. If the surface is close, it has no boundary and Stoke's theorem necessarily gives 0 as result.
 
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contnuing my last reply, consider I have:

[tex]\vec{F}= z \vec{i} +x \vec{j} + y \vec{k}[/tex]
[tex]\vec{G} = \nabla\times\vec{F} = <1, 1, 1> [/tex]
So,
[tex]\nabla\cdot\vec{G} = 0[/tex]
So, I would get the divergence theorem telling me that:

[tex]\int\int\nabla\times\vec{F}\cdot \vec{n} dS = 0[/tex]
But, at the same time,
[tex] \int\int\nabla\times\vec{F}\cdot \vec{n} dS = \int \vec{F}\cdot d\vec{R} \neq 0[/tex]
edit: consider my Surface the part of the sphere in the first octant with radius 1 for simplicity
 
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Ahh, disregard my last post. I think I now understand why i was having difficulty with this. So then is the only way I can use the divergence method to compute such an integral is if the surface is closed? And in this case, will it always be zero?
 
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Thank you for all the help!
 
Hootenanny
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Science Advisor
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Ahh, disregard my last post. I think I now understand why i was having difficulty with this. So then is the only way I can use the divergence method to compute such an integral is if the surface is closed?
Yes as Halls has already said, your version of the divergence theorem states that the integral of a continuous vector field, which has continuous first order partial derivatives, over a closed surface is equal to the integral of the divergence of the same vector field over the volume enclosed by the surface. Hence, the surface must be closed. It would perhaps better to write the RHS explicitly, so as to remind yourself that the surface must be closed,

[tex]\iiint_V \text{div}\left(\underline{F}\right) dV = {\bigcirc \hspace{-0.85cm} \int \hspace{-0.55cm} \int}_S\underline{F}\cdot\underline{n}dS[/tex]

It should also be noted that the surface S is oriented by the outward normals n.
Thank you for all the help!
It's a pleasure :smile:
 
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