By green's theorem $\int\limits_{c} Pdx + Qdy = \int\int\limits_{R} \bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dx dy$

Here,

$P = x^2y \\ Q = y^3$

Now $\frac{\partial P}{\partial y} = x^2$ and $\frac{\partial Q}{\partial x} = 0$

Now defining the limits

Given, y = x, y = $x^2 \\ \therefore x = x^2$

i.e.

$x^2 - x = 0 \\ \therefore x(x - 1) = 0 \\ \therefore x = 0, x = 1$

Hence limit of x : 0 to 1 [Here we have defined strip parallel to y-axis]

y: $x^2$ to x

Hence

$\int\limits_{x = 0}^{1}\int\limits_{y = x^2}^{x} -x dxdy = \int\limits_{x = 0}^{1} - x^2 dx \int\limits_{y = x^2}^{x}dy \\ = \int\limits_{x = 0}^{1} - x^2 [y]_{x^2}^{x}dx \\ = \int\limits_{x = 0}^{1} - x^2 [ x - x^2]dx \\ = - \int\limits_{x = 0}^{1} x^3 - x^4 dx \\ = -\bigg[\frac{x^4}{4} - \frac{x^5}{5}\bigg]_{0}^{1} \\ = - \bigg(\frac{1}{4} - \frac{1}{5}\bigg) \\ = \frac{-1}{20}$

Here green's theorem

$\oint\limits_{c} x^2 ydx + y^3 dy = \frac{-1}{20}$