Line Integral does not match Greens Theorem?

[PhantomPower]

Homework Statement

To evaluate the following line integral where the curve C is given by the boundary of the square 0 < x < 2 and 0 < y < 2 (In the anti clockwise sense):

\oint (x+y)^2 dx + (x-y)^2 dy

The Attempt at a Solution

Firstly it is noted that for a square ABDE :
Between AB, dy = 0 ,y = 0
Between BD, dx = 0 ,x = 2
Between DE, dy = 0 ,y = 2
Between EA, dx = 0 ,x =0

Thus : \int^2_0 x^2 dx + \int^2_0 (2-y)^2 dy + \int^0_2 (x+2)^2 dx + \int^0_2 -y^2 dy
Giving \frac{x^3}{3} |^2_0 + \frac{-(2-y)^3}{3} |^2_0 + \frac{(x+2)^3}{3} |^0_2 + \frac{-y^3}{3} |^0_2

Evaluating gives 10.6? but applying greens theorem gives -16. Can anyone spot my mistake - Probaly a negative sign?

Thanks very much.
P.s sorry for typos this keyboard is broken

 

[tiny-tim]

Hi PhantomPower!

erm …

∫ (2 - y)² is the same as ∫ (y - 2)² ! ​

(similary for (-y)²)

 

[PhantomPower]

Ooops.

Thanks very much - been a long day...