# Evaluate the line integral of a vector field around a square

I am asking this question because I believe the answer sheet I was given has an incorrect solution.

The task is to evaluate (by hand!) the line integral of the vector field $$\mathbf{F}(x,y) = x^2y^2 \mathbf{\hat{i}} + x^3y \mathbf{\hat{j}}$$ over the square given by the vertices (0,0), (1,0), (1,1), (0,1) in the counterclockwise direction. This vector field is not conservative by the way.

The answer I was given is as follows:

Now the part I believe to be incorrect is the parametrization of the third curve $$\mathcal{C}_3$$. I think it is wrong due to the direction: the given parametrization is the curve going up instead of down. Since we are going in the counterclockwise direction, I believe the parametrization should be $$\mathbf{r}(t) = 1-t\mathbf{\hat{i}} + \mathbf{\hat{j}}$$ giving: $$\int_{\mathcal{C}_3} x^2y^2 dx + x^3ydy = \int^1_0(1-t)^2dt=\int^1_0 1-2t+t^2=\frac{1}{3}$$

Hopefully someone can confirm my suspicion or tell me why I am wrong, thank you

Notice that you forgot to parameterize $${\rm d}x$$. By choosing $$x(t) = 1-t$$, you would then have $${\rm d}x = -{\rm d}t$$, giving you the same answer as the solution.