$y=x \ and \ y=x^2 \ \cdots (Given)$ * Solving simultaneously we get, $x=x^2;\ Hence, x-x^2=0 $ $x(1-x)=0;\ hence, \ x=0 \ or\ x=1$ * Since, y=x, y=0 or y=1. * The two curves intersect at (0,0) and (1,1) $Let\ P=x^2y \ and\ Q=y^3$ $Hence, \dfrac {\delta P} {\delta y}=x^2 \ and \dfrac {\delta Q} {\delta x}=0$ * By Green's Thm,$\int_cPdx+Qdy=\iint_R \bigg[\dfrac {\delta Q} {\delta x}-\dfrac{\delta P} {\delta y}\bigg]dx.dy$ $\int_c(x^2ydx+y^3dy)=\iint_R[0-x^2]dxdy$ $=\int \limits_0^1\int \limits_{x^2}^x-x^2dydx$ $=\int \limits_0^1-x^2[y]_{x^2}^xdx$ $=\int \limits_0^1-x^2[x-x^2]dx$ $=\int \limits_0^1[-x^3+x^4]dx$ $=\bigg[-\dfrac {x^4}4+\dfrac {x^5} 5\bigg]_0^1$ -$=\bigg(\dfrac {-1} 4+\dfrac 1 5\bigg)-(-0+0) $ $=\dfrac {-1} {20}$