Stokes and Divergence theorem questions

In summary, the conversation discusses the evaluation of a surface integral using different methods, including the divergence theorem, Stoke's theorem, and direct computation. It is noted that the surface must be closed in order to use the divergence theorem, and the conversation also touches on the relationship between the two theorems. Ultimately, the conversation concludes with a better understanding of how to use the divergence theorem to compute surface integrals.
  • #1
EngageEngage
208
0

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

Homework Equations



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

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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  • #2
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?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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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  • #6
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?
 
  • #7
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.
 
  • #8
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.
 
  • #9
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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  • #10
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?
 
  • #11
Thank you for all the help!
 
  • #12
EngageEngage said:
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.
EngageEngage said:
Thank you for all the help!
It's a pleasure :smile:
 
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What is the Stokes Theorem?

The Stokes Theorem is a mathematical theorem that relates the integral of a vector field over a surface to the line integral of the same vector field over the boundary of that surface.

What is the Divergence Theorem?

The Divergence Theorem is a mathematical theorem that relates the volume integral of a vector field to the surface integral of the same vector field over the boundary of that volume.

What is the significance of the Stokes and Divergence Theorems?

The Stokes and Divergence Theorems are important tools in vector calculus, allowing for the conversion of complicated volume or surface integrals into simpler line integrals and vice versa. They are also used in many applications in physics and engineering, such as calculating fluid flow and electromagnetic fields.

What are the conditions for applying the Stokes and Divergence Theorems?

For the Stokes Theorem, the surface must be smooth and oriented, and the vector field must be continuous and differentiable. For the Divergence Theorem, the volume must be bounded by a closed surface and the vector field must be continuous and differentiable within the volume.

Can the Stokes and Divergence Theorems be used in any dimension?

Yes, the Stokes Theorem can be applied in any dimension, while the Divergence Theorem is specific to three-dimensional space.

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