By green's theorem,

$$\therefore \int Pdx + Qdy = \int\int\limits_{s} \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}dxdy$$

Here $P = 2x^2 - y^2 \\ Q = x^2 - y^2$

$\therefore \frac{\partial P}{\partial y} = -2y \\ \frac{\partial Q}{\partial x} = 2x$

Now considering the horizontal strip which will slide from 0 to 2 in x and thus will become the outer integration units.

However, for inner limits the upper limits will be line i.e. y = 2 and lower limit is y = 0

$\therefore A = \int\limits_{x = 0}^{2}\int\limits_{y = 0}^{2} (2x + 2y)dxdy$

Integrating first w.r.t. y

$\therefore A = \int\limits_{x = 0}^{2} \bigg[2xy + \frac{2y^2}{2}\bigg]_{0}^{2}dx = \int\limits_{0}^{2}[4x + 4]dx$

Integrating w.r.t. x we get

$ = \bigg[\frac{4x^2}{2} + 4x \bigg]_{0}^{2} = [2 \times 2^2 + 8] = 16$