$$\int\limits_{C} (Pdx + Qdy) = \int\int\limits_{R} \bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \bigg) dxdy$$

Here $P = x^2 - y$ and $Q = 2y^2 + x$

$$\frac{\partial P}{\partial y} = -1 , \frac{\partial Q}{\partial x} = 1$$

Along arc ADB $y = x^2$ -> dy = 2xdx

$$\int\limits_{C_{1}}(Pdx + Qdy) = \int\limits_{0}^{2} (x^2 - x^2) + (2x4 + x)2xdx$$ $$\therefore \int\limits_{C_{1}}(Pdx + Qdy)= \int\limits_{0}^{2}0 + (4x^5 + 2x^2)dx$$ $$\therefore \int\limits_{C_{1}}(Pdx + Qdy) = \bigg[\frac{4x^6}{6} + \frac{2x^3}{3} \bigg]_{0}^{2}$$ $$\therefore \int\limits_{C_{1}}(Pdx + Qdy) = \frac{128}{3} + \frac{16}{3} = \frac{144}{3}$$

Along $C_{2}$, y = 4, dy = 0

$$\therefore \int\limits_{C_{2}} Pdx + Qdy = \int\limits_{2}^{0}(x^2 - 4)dx$$ $$\therefore \int\limits_{C_{2}} Pdx + Qdy = \bigg[\frac{x^3}{3} - 4x\bigg]_{2}^{0}$$ $$\therefore \int\limits_{C_{2}} Pdx + Qdy = \frac{-8}{3} + 8$$ $$\therefore \int\limits_{C_{2}} Pdx + Qdy = \frac{16}{3}$$

Along $C_{3}$, x = 0, dx = 0

$$\int\limits_{C} Pdx + Qdy = \int\limits_{4}^{0}2y^2 dy = \bigg[\frac{2y^3}{3}\bigg]_{4}^{0}$$ $$= \frac{-2 * 64}{3}$$ $$=\frac{-128}{3}$$ $$\therefore$ Total i.e. $\int\limits_{C} Pdx + Qdy = \frac{144}{3} + \frac{16}{3} - \frac{128}{3} = \frac{32}{3}$$

Now outer limits will extend from 0 to 2 in horizontal direction thus x -> 0 to 2

For inner limits

Upper limit is line equation i.e. J = 4

And lower limit is curve of parabola i.e. $y = x^2$

$$\therefore \int\int \bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \bigg) dxdy = \int\limits_{0}^{2} \int\limits_{x^2}{4} (1 + 1)dxdy$$ $$\therefore \int\limits_{0}^{2} [2y]_{x^2}^{4}dx = \int\limits_{0}^{2} 2(4 - x^2)dx$$ $$= \bigg[4x - \frac{x^3}{3}\bigg]_{0}^{2}$$ $$= 2\bigg[8 - \frac{8}{3} \bigg]$$ $$= 2 * \frac{16}{3}$$ $$= \frac{32}{3}$$

Hence verified