Undergrad What is the equation for line integrals involving the gradient of a function?

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The discussion centers on the application of the product rule for gradients in the context of line integrals. The equation ∮C∇(fg)·dr is analyzed, with participants suggesting it can be expressed as ∮Cf∇g·dr + ∮Cg∇f·dr for any closed curve. It is noted that for closed curves in R, the integral ∮C∇(fg)·dr equals zero, leading to the conclusion that ∮Cf∇g·dr = -∮Cg∇f·dr. This reasoning is affirmed as valid for all lines. Overall, the conversation emphasizes the relationship between gradients and line integrals in vector calculus.
WMDhamnekar
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Let f (x, y) and g(x, y) be continuously differentiable real-valued functions in a region R.
Show that ##\displaystyle\oint_C f\nabla g \cdot dr = -\displaystyle\oint_C g\nabla f \cdot dr## for any closed curve in R
I don't have any idea to answer this question. So, any math help will be accepted.
I know ##\nabla fg = f\nabla g + g\nabla f \rightarrow (1) ## But I don't understand to how to use (1) here?
 
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What will ## \displaystyle\oint_C \vec \nabla (fg) \cdot d \vec r ## become?
 
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drmalawi said:
What will ## \displaystyle\oint_C \vec \nabla (fg) \cdot d \vec r ## become?
I think ## \displaystyle\oint_C \nabla (fg) \cdot dr = \displaystyle\oint_C f\nabla g \cdot dr + \displaystyle\oint_C g\nabla f \cdot dr ## for any closed curve.

But, for any closed curve in R , ##\displaystyle\oint_C \nabla (fg) \cdot dr = 0,## Hence, we can deduce that ##\displaystyle\oint_C f\nabla g \cdot dr = -\displaystyle\oint_C g\nabla f \cdot dr ##
Is that logic correct?
 
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WMDhamnekar said:
I think ∮C∇(fg)⋅dr=∮Cf∇g⋅dr+∮Cg∇f⋅dr for any closed curve.
This is valid for all lines.
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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