Surface integral Watch

Dan R
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#1
Report Thread starter 13 years ago
#1
Let S be the surface, x² + y² <= 1, z = 0 (ie a disk in the x-y plane).
and let F be the vector field, F = (x,y,z).

Take the normal to be directed in the negative z-direction.

Then what is the surface integral, ∫∫ (over S) F.dS ?

I get 0. Does this sound right or not?
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Neapolitan
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It has to be 0 by inspection as r.dS will contribute negatively what (-r).dS does to the integral
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Dan R
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#3
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cheers.
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