Proof of Line Integral Using ∇f & ∇g: R Region, C Curve

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 \mathbf{\nabla}\left[f(x,y)g(x,y)\right]? 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.

\oint \mathbf{\nabla}(fg)\cdot d\textbf{r}=0

You can also express the integrand in terms of f, g, \mathbf{\nabla}f and \mathbf{\nabla}g 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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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