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Greens theorem on a circle

  • Thread starter Liferider
  • Start date
  • #1
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Homework Statement


Use greens theorem to solve the closed curve line integral:
[itex]\oint[/itex](ydx-xdy)

The curve is a circle with its center at origin with a radius of 1.

Homework Equations


x^2 + y^2 = 1

The Attempt at a Solution


Greens theorem states that:
Given F=[P,Q]=[y, -x]=yi-xj

[itex]\oint[/itex]F*dr=[itex]\oint[/itex]Pdx+Qdy=[itex]\int[/itex][itex]\int[/itex]([itex]\frac{dQ}{dx}[/itex]-[itex]\frac{dP}{dy}[/itex])dA

From the circle equation i find:
x=[itex]\sqrt{1-y^2}[/itex]
y=[itex]\sqrt{1-x^2}[/itex]

Which means that:
[itex]\frac{dQ}{dx}[/itex]=0 and [itex]\frac{dP}{dy}[/itex]=0

Obviously, I am doing something wrong... but which rules am I breaking??

I did find the answer the "normal" way, without greens, which was -2[itex]\pi[/itex].

I think a lot of my difficulties originates from lack of knowledge about different coordinate systems and the conversion between these.
One could write:
x=cos t and y=sin t, but do I have to? (btw, i used that for solving the normal way).
 

Answers and Replies

  • #2
43
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Oh wait... I think I got the vector field F mixed up with the circle that I am integrating over.
I shouldn't insert for x and y before I have applied greens theorem... I think.
 
  • #3
557
1
It's [itex]Q = -x, P= y[/itex], so [itex]\frac{dQ}{dx} = -1[/itex], etc...
 
  • #4
43
0
Solved, used polar coordinates on the integral after greens theorem.
 

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