Line integral problem. HELP PLEASE!!

[Faux Carnival]

Is the answer 1 or 0?

I get different answers by doing the line integral and using the path-independency theorem. (Because the vector field is conservative)

Any help will be appreciated.

 

[galactus]

Wow, I completely overlooked this problem.

I get 0.

Using $\displaystyle r(t)=(1-t)r_{0}+tr, \;\ 0\leq t\leq 1$

$\displaystyle C_{1}:r(t)=(1-t)[0,-1,0]+t[0,0,0]=[0,t-1,0]$

$\displaystyle C_{2}:r(t)=(1-t)[0,0,0]+t[1,0,0]=[t,0,0]$

$\displaystyle C_{3}:r(t)=(1-t)[1,0,0]+t[2,1,0]=[t+1,t,0]$

$\displaystyle C_{1}: \;\ \int_{0}^{1}\left[2(0)(t-1)(0)-(t-1)\right](0)dt+\int_{0}^{1}\left[(0)^{2}(0)+3(t-1)^{2}-0\right](1)dt+\int_{0}^{1}\left[(0)^{2}(t-1)^{2}\right](0)dt$$\displaystyle =3\int_{0}^{1}(t-1)^{2}dt=\fbox{1}$

$\displaystyle C_{2}: \;\ \int_{0}^{1}\left[2(t)(0)(0)-(0)\right](1)dt+\int_{0}^{1}\left[t^{2}(0)+3(0)^{2}-t^{2}\right](0)dt+\int_{0}^{1}t^{2}(0)(0)dt=\fbox{0}$

$\displaystyle C_{3}: \;\ \int_{0}^{1}\left[2(t+1)(t)(0)-t\right](0)dt+\int_{0}^{1}\left[(t+1)^{2}(0)+3t^{2}-(2t+1)\right](1)dt+\int_{0}^{1}(t+1)^{2}(t)(0)dt$
$\displaystyle =\int_{0}^{1}\left[3t^{2}-2t-1\right]dt=\fbox{-1}$

$\displaystyle C_{1}+C_{2}+C_{3}=\fbox{1+0+(-1)=0}$