Area of square with vertices: $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$

Question

If I know that:$$\int_C-ydx+xdy=\boxed{x_1y_2-x_2y_1}$$

So, why the area of square with vertices: $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4)$ is

$\frac{1}{2}[(x_1y_2-x_2y_1)+(x_2y_3-x_3y_2)+(x_3y_4-x_4y_3)+(x_4y_1-x_1y_4)]$

Answer 1

Well, if you evaluate that line integral over all four edges, you get the expression at the bottom, but without the factor of $1/2$.

On the other hand, Green's theorem tells you that $$ \int_{\partial R} P dx + Q dy = \int_R \frac{\partial Q}{\partial x} - \frac{\partial P} {\partial y} dA $$

(I might have a sign error here); in your case, that tells you your line integral is the same as the integral of the constant $2$ over the area enclosed by the edges, giving twice the area. Hence half the line integral gives the area.