# Properties of Line Integral

• I
• WMDhamnekar

#### WMDhamnekar

MHB
TL;DR Summary
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?

Last edited:
Delta2
What will ## \displaystyle\oint_C \vec \nabla (fg) \cdot d \vec r ## become?

WMDhamnekar and Delta2
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?

Delta2 and malawi_glenn
I think ∮C∇(fg)⋅dr=∮Cf∇g⋅dr+∮Cg∇f⋅dr for any closed curve.
This is valid for all lines.

WMDhamnekar and Delta2