Originally Posted by

**tammyl** Thanks for the response.

So for $\displaystyle C_1$,

Parameterisation of straight line from (0,0) to (1,1)

$\displaystyle x = a(1-t) + ct = t$ where $\displaystyle a=0, c=1$

$\displaystyle y = b(1-t) + dt = t$ where $\displaystyle b=0, d=1$

Thus

$\displaystyle dx=dt$ and $\displaystyle dy=dt$ where $\displaystyle 0\leq t \leq1$

So,

$\displaystyle \int_{C_1} (sin y + 2x)dx + (2y + x cos y)dt$

$\displaystyle = \int_0^1 (sin t + 2t)dt + (2t + t cos t)dt$

$\displaystyle = \int_0^1 [sin t + 4t + t cos t] dt$

$\displaystyle = [-cost + 2t^2 + t sin t + cos t]_0^1$

$\displaystyle = [2t^2 + t sin t]_0^1$

$\displaystyle = (2.(1)^2 - 2.(0)^2) + (1.sin (1) - 0.sin(0))$

$\displaystyle = 2 + sin (1)$

Repeating the procedure for the other two straight lines, I got

$\displaystyle \int_{Z} = \int_{C_1} + \int_{C_2} + \int_{C_3} = (2 + sin (1)) + (-16 - 4sin (1))+(7+2cos(1))= 2 cos(1) - 3 sin(1) -7$