- #1

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[tex]\int\int\int_D \bigtriangledown \times F dV = 0[/tex]

in the sense that each component of [tex]\bigtriangledown \times F[/tex] has integral 0 over D."

I'm really stumped on this one.

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- Thread starter Treadstone 71
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- #1

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[tex]\int\int\int_D \bigtriangledown \times F dV = 0[/tex]

in the sense that each component of [tex]\bigtriangledown \times F[/tex] has integral 0 over D."

I'm really stumped on this one.

- #2

Physics Monkey

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[tex] (\vec{\nabla}\times\vec{F})_x = \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} [/tex]. Since integration is linear you can consider each piece in turn. Can you find a way to represent

[tex]

\frac{\partial F_z}{\partial y}

[/tex]

as a divergence. Hint: to make progress, apply Gauss' Theorem to the result.

- #3

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[tex]\int\int\int_D\bigtriangledown(0,F_3,0)dV=\int\int_{\partial D}(0,F_3,0)d\vec{S}[/tex].

If the field were tangent to the surface, the dot product between the normal vector and it would be zero, but since it's already perpenticular...

- #4

Physics Monkey

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[tex]\int\int _{\partial D}(0,F_3,0)d\vec{S}[/tex]

equals zero. However, when you combine this term with the other term in the curl, then maybe you can say something ...

- #5

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It boils down to

[tex]\int\int \bigtriangledown(0,F_3,0)-\bigtriangledown(0,0,F_2)\vec{N}dS[/tex]

The integrand becomes

[tex]\bigtriangledown(0,F_3,F_2)\vec{N}[/tex]

Since the normal is parallel to the field, it becomes (F3F2-F2F3)|N|, and this is 0.

Did I get this right? If so thanks for the help!

[tex]\int\int \bigtriangledown(0,F_3,0)-\bigtriangledown(0,0,F_2)\vec{N}dS[/tex]

The integrand becomes

[tex]\bigtriangledown(0,F_3,F_2)\vec{N}[/tex]

Since the normal is parallel to the field, it becomes (F3F2-F2F3)|N|, and this is 0.

Did I get this right? If so thanks for the help!

Last edited:

- #6

Physics Monkey

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I don't know why you have the gradient operator still there, but yes, you've pretty much got it.

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