how to find the line integral of a square with vertices $(-1,0) (0,1) (1,0) $ and $ (0,-1)$ when $F=xyi+xy^2j$

Question

I know how to usually do these kinds of questions but usually there is something given to substitute for x or y in F or parametric equation is given.

Answer 1

One way: if $S$ is the given square and $\partial S$ its boundary, $$\int_{\partial S}xydx+xy^2dy\underbrace{=}_{\text{Green's Theorem}}\iint_{S}(y^2-x)dxdy=\ldots$$

Answer 2

If you don't want to use Green's Theorem, then you need to concoct your own parametrization. In this case, the curve is made of 4 pieces, so the integral breaks into 4 parts. The segment from $(-1,0)$ to $(0,1)$ is given by $x=t-1$, $y=t$ for $0\leq t \leq 1$. Work out that integral and then do the other 3 segments similarly. Then add up the 4 results.

The point of such exercises is to make you appreciate Green's Theorem.