Green's Theorem?

Question

Use Green's Theorem to evaluate
I=∫_Cy²dx+xdy, where C is the boundary of the square with vertices (0,0),(2,0),(2,2),(0,2).

Answer 1

I=∫_Cy²dx+xdy = ∫∫_R 1-2y dA
= ∫_0 ^2 ∫_0 ^2 1-2y dxdy
here the symbol ∫_0 ^2 means integral from 0 to 2.
= ∫_0 ^2 (1-2y)x from 0 to 2 dy
= ∫_0 ^2 (1-2y)2 dy = ∫_0 ^2 2-4y dy
= 2y-4y^2/2 from 0 to 2
= (2(2) -2(4)) = 4-8=-4 .

Answer 2

Agree if C is *positively* oriented: then the contour integral equals

∫∫_Square [dx/dx-d(y^2)/dy]dxdy = ∫_0^2∫_0^2 (1-2y)dxdy = -4.

(The deriv. symbols should be of course partial.)

Answer 3

So what's the problem? Do the double integral of
(1-2y)dydx evaluated from (0,0) to (0,2), and (0,0) to (2,0).


Doug