Proving ∫f ∇g · dr = −∫g ∇f · dr for a closed curve

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charmmy
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Homework Statement


Let f(x,y) and g(x,y) be continuously differentiable real-valued functions in a region R. Show that ∫f ∇g · dr ]= − ∫g ∇f · dr for any closed curve C in R.


Homework Equations





The Attempt at a Solution



I don't really know where to start, so I tried to evaluate the LHS of the equation but how do I do this symbolically? and where do I lead on from ther?
 
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charmmy said:

Homework Statement


Let f(x,y) and g(x,y) be continuously differentiable real-valued functions in a region R. Show that ∫f ∇g · dr ]= − ∫g ∇f · dr for any closed curve C in R.


Homework Equations





The Attempt at a Solution



I don't really know where to start, so I tried to evaluate the LHS of the equation but how do I do this symbolically? and where do I lead on from ther?

Hint: What is [itex]\mathbf{\nabla}\left[f(x,y)g(x,y)\right][/itex]? What is the line integral of the gradient of a function over a closed curve?
 
so is LaTeX Code: \\mathbf{\\nabla}\\left[f(x,y)g(x,y)\\right] =
take h= f(x,y) g(x,y)
then ∇h=dh/dx+dh/dy...

What is the line integral of the gradient of a function over a closed curve? : is this just equal to zero?

i'm quite confused
 
charmmy said:
so is LaTeX Code: \\mathbf{\\nabla}\\left[f(x,y)g(x,y)\\right] =
take h= f(x,y) g(x,y)
then ∇h=dh/dx+dh/dy...

Well, yes, that's the basically definition of gradient. However, it isn't all that useful to you here...there is a product rule that should be in your textbook/notes that tells you how to take the gradient of a product of two scalar functions...Use that.

What is the line integral of the gradient of a function over a closed curve? : is this just equal to zero?

Yes, this is a direct consequence of the fundamental theorem of gradients.

[tex]\oint \mathbf{\nabla}(fg)\cdot d\textbf{r}=0[/tex]

You can also express the integrand in terms of [itex]f[/itex], [itex]g[/itex], [itex]\mathbf{\nabla}f[/itex] and [itex]\mathbf{\nabla}g[/itex] using the product rule I mentioned above. Doing so, then splitting up the integral and using the fact that the result is zero will allow you to show the desired result.