Calculating Curve Integral: \int_{\gamma}(x^2+xy)dx+(y^2-xy)dy

[Hannisch]

Homework Statement

Calculate the curve integral

\int_{\gamma}(x^2+xy)dx+(y^2-xy)dy

where \gamma is the line segments from (0,0) to (2,0) and from (2,0) to (2,2).

Homework Equations

\int_{\gamma} P(x,y)dx+Q(x,y)dy= \int^{\beta}_{\alpha}(P(g(t),h(t))g'(t)+Q(g(t),h(t))h'(t))dt

where

x=g(t)
y=h(t)

The Attempt at a Solution

I started by splitting the line up in two parts,

\gamma _{1} = the segment from (0,0) to (2,0)
\gamma _{2} = the segment from (2,0) to (2,2)

I then stated that for the first segment, the parametric equation would be

(x,y)=(t,0) and 0 \leq t \leq 2

and for the other segment:

(x,y)=(0,t) and 0 \leq t \leq 2

Both of these segments are in the positive direction, so

\int _{\gamma}(x^2+xy)dx+(y^2-xy)dy = \int _{\gamma_{1}}(x^2+xy)dx+(y^2-xy)dy + \int _{\gamma{2}}(x^2+xy)dx+(y^2-xy)dy = \int^{2}_{0} t^2 dt + \int^{2}_{0} t^2 dt = 16/3

But this is incorrect and I haven't the faintest idea where and why I went wrong. It's driving me crazy (it should be such a simple problem) but I can't figure it out. I thought this was what the did in the book, but alas.. According to the book the answer is 4/3.

 

[Hannisch]

Okay, never mind, I realized my mistake. I'd stared myself blind in my tiredness x)