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
$\endgroup$ 01 Answer
$\begingroup$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.
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