# Closed line integral gradient relation proof

Prove the following relation:

$$\oint f \vec{\bigtriangledown}g \cdot d\vec{l} = -\oint g \vec{\bigtriangledown}f \cdot d\vec{l}$$

where f, g are scalar functions.

I've tried a lot of work, but can't seem to figure this relation out. I initially tried using stokes theorem thinking it would give me some kind of result but I ended up just proving the relation:

$$\oint f \vec{\bigtriangledown}g \cdot d\vec{l} = \int_S ((\vec{\bigtriangledown}f)\times(\vec{\bigtriangledown}g))$$

Which is correct, apparently, but I still want to prove the top relation.

I tried taking the gradient of g, and distributing f as a scalar, but I can't see any relation that would re-arrange the del operator.

$$\oint f \vec{\bigtriangledown}g \cdot d\vec{l} = \oint \langle f \frac{\partial g}{\partial x},f\frac{\partial g}{\partial y},f\frac{\partial g}{\partial z}\rangle \cdot\langle dxdydz\rangle$$

Everything I've seen related to closed line integrals states that it is equivalent to 0, but I just don't see how to get this result. I've even thought of using the gradient theorem, by replacing with g, and distributing f (this is the general equation):

$$\int_{a}^{b} (\vec{\bigtriangledown}f)\cdot d\vec{l} = f(\vec{b})-f(\vec{a})$$

Any help would be appreciated.

Notice that you can rewrite the original relation as $$\oint (f\cdot \nabla g + \nabla f\cdot g)dl = 0.$$ Also notice that $$\nabla(f\cdot g) = f\cdot\nabla g + \nabla f\cdot g.$$ Using the fact that gradients are conservative vector fields, and what you know about integrating conservative vector fields about closed paths, can you make the desired conclusion?
Edit: Here's an alternative to the product rule approach. You said you've shown that $$\oint f\cdot\nabla g dl = \int_S((\nabla f)\times (\nabla g)).$$ Presumably this also means that $$\oint \nabla f\cdot g dl = \int_S((\nabla g)\times (\nabla f)).$$ But when we reverse order in a cross product we pick up a factor of $-1$, so $$\oint f\cdot\nabla g dl = \int_S((\nabla f)\times (\nabla g)) = -\int_S((\nabla g)\times (\nabla f)) = -\oint \nabla f\cdot g dl.$$
• The way I read it, we're integrating $f$ times the gradient of $g$ on the left, and $g$ times the gradient of $f$ on the right, so the gradient operator acts on both $f$ and $g$. Notice that the second equation above is a sort of product rule for gradients. Commented Sep 1, 2016 at 16:51
• Okay, I just figured it was like $$f (\vec{\bigtriangledown} g)$$ and not that it was applied to both, I wasn't sure though. Commented Sep 1, 2016 at 17:07
• Gotcha. No, the notation $f\nabla g$ means $f$ (a scalar-valued function) times $\nabla g$ (a vector-valued function). Commented Sep 1, 2016 at 17:09
• Should we please state differently for vector operator from scalar function? For instance in $\nabla(f\cdot g) = f\cdot\nabla g + \nabla f\cdot g$? Commented Aug 1, 2021 at 16:15