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The gradient theorem

  1. Oct 9, 2013 #1


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    i am not sure if this post should be under calculus or not, but i think i'll get a more "complete" answer here. at any rate, i'm wondering if anyone can clarify the intuition behind the gradient theorem: [tex]\iiint\limits_V \nabla \psi dV=\iint\limits_S \psi \vec{dS}[/tex] by intuition, i refer to a physical interpretation. for instance, i understand the divergence theorem states (rate of expansion of a vector field) = (rate the vector field leaks out of the edges). but this extended gradient theorem is more difficult. please help! i doubt i need to clarify this notation, but please let me know if i've been ambiguous.

  2. jcsd
  3. Oct 9, 2013 #2
    This is just the 3D version of [itex]\int_a^b{\frac{df}{dx}dx}=f(b)-f(a)[/itex]
  4. Oct 9, 2013 #3


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    yes, i do realize this, but im a little shakey on the intuition. the slope vectors added up make sense to check the difference of endpoints, but the 2 and 3 dimensions are not as obvious.
  5. Oct 12, 2013 #4
    When you post in the differential geometry part of the forum, you get a differential geometry answer.

    Note first, though, that your "gradient theorem" is actually just a form of the Divergence Theorem. It turns out that all these integral theorems that you learn from multivariate calculus are actually results of a single theorem called the generalized Stokes' Theorem from differential geometry.

    The intuition for your problem can be seen in 2 dimensions. From a topological standpoint, we can integrate over chains, and in the two dimensional case we'll look at a 2-chain and its boundary and see what integration over the two might be like. Look at the pictures below (Ignore the rainbows. Life in mathland is just that happy).

    The first shows the oriented boundary of a region in two dimensional space. The second shows finer and finer oriented tilings of the region. Note that the interior arrows of each tiling go in opposite directions from their neighbors, "cancelling" each other out. The integral over the 2-chain would be equal to, in a sense, the contribution of the boundary.

    Does this help?

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