- #1

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##\bf a = ∫_S d \bf a##, where S is the surface and ##\bf a ##is the vector area of it.

Please proof that ##\bf a = \frac{1}2\oint \! \bf r \times d\bf l##, where integration is around the boundary line.

Any help would be very appreciated!

- Thread starter rbwang1225
- Start date

- #1

- 118

- 0

##\bf a = ∫_S d \bf a##, where S is the surface and ##\bf a ##is the vector area of it.

Please proof that ##\bf a = \frac{1}2\oint \! \bf r \times d\bf l##, where integration is around the boundary line.

Any help would be very appreciated!

- #2

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Strokes theorem?

hmmm

Well say you perform a surface integral, if the vector field in question is the normal vector of the surface, then the only thing left in the integrand is dA (scalar).

So I guess using strokes theorem, you have to find a vector field who's curl is the normal vector of the surface.

hmmm

Well say you perform a surface integral, if the vector field in question is the normal vector of the surface, then the only thing left in the integrand is dA (scalar).

So I guess using strokes theorem, you have to find a vector field who's curl is the normal vector of the surface.

Last edited:

- #3

- 606

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Strokes theorem?

hmmm

Well say you perform a surface integral, if the vector field in question is the normal vector of the surface, then the only thing left in the integrand is dA (scalar).

So I guess using strokes theorem, you have to find a vector field who's curl is the normal vector of the surface.

Strokes Theorem is what we get sometimes from our loved ones.

DonAntonio

- #4

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Oops

- #5

chiro

Science Advisor

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What's this new theorem and who proved it lover boy?Strokes Theorem is what we get sometimes from our loved ones.Theorem is, perhaps, what you mean.Stokes

DonAntonio

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