- #1

PhantomPower

- 14

- 0

## 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):

[itex]\oint (x+y)^2 dx + (x-y)^2 dy [/itex]

## 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 : [itex]\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 [/itex]

Giving [itex] \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 [/itex]

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