How to use the normal form of the Green's Theorem?

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randomcat
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Homework Statement


Suppose that F = ∇f for some scalar potential function f(x, y) = 1/2(x2 + y2)
Let C denote the positively oriented unit circle, parametrized by r(t) = (cos t, sin t), 0 ≤ t ≤ 2∏. Compute the flux integral of [itex]\oint[/itex]F[itex]\bullet[/itex]N ds, where N is the outward unit normal to C.

Homework Equations


I know that the theorem is basically [itex]\oint[/itex]F[itex]\bullet[/itex]N ds = double integral of ([itex]\frac{\partial M}{\partial x}[/itex] + [itex]\frac{\partial N}{\partial y}[/itex]) dx dy

My question is, how do you find F(x, y)? I'm given f(x, y) in the problem, but do I need to use it to solve for the answer?

The Attempt at a Solution


What I did (which was wrong) was that I set M = 1/2x2 and N = 1/2y2 and took the partial derivatives of those. That gives me (x, y). With the parametrization, I get (cos t, sin t). Then I integrate them over the unit circle and got 0, which was wrong. The correct answer is 2pi.

My professor wrote that F = ∇f = F(x, y) in the solutions, but where did F(x, y) come from?
 
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randomcat said:

Homework Statement


Suppose that F = ∇f for some scalar potential function f(x, y) = 1/2(x2 + y2)
Let C denote the positively oriented unit circle, parametrized by r(t) = (cos t, sin t), 0 ≤ t ≤ 2∏. Compute the flux integral of [itex]\oint[/itex]F[itex]\bullet[/itex]N ds, where N is the outward unit normal to C.


Homework Equations


I know that the theorem is basically [itex]\oint[/itex]F[itex]\bullet[/itex]N ds = double integral of ([itex]\frac{\partial M}{\partial x}[/itex] + [itex]\frac{\partial N}{\partial y}[/itex]) dx dy

My question is, how do you find F(x, y)? I'm given f(x, y) in the problem, but do I need to use it to solve for the answer?

The Attempt at a Solution


What I did (which was wrong) was that I set M = 1/2x2 and N = 1/2y2 and took the partial derivatives of those. That gives me (x, y). With the parametrization, I get (cos t, sin t). Then I integrate them over the unit circle and got 0, which was wrong. The correct answer is 2pi.

My professor wrote that F = ∇f = F(x, y) in the solutions, but where did F(x, y) come from?

You are given ##f=(x^2+y^2)/2## and ##F=\nabla f##. So calculate ##\nabla f## to get ##F##.
 
Perhaps the problem is that you don't know what [itex]\nabla f[/itex] means? If f is as function of two variables, x and y, the [itex]\nabla f[/itex] is the vector function
[tex]\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}[/tex]

That should be easy to calculate for this f.