Be $D$ limited by the curve $\gamma(t)=(t-t^2, t^3)$ and the $y$ axis...
Calculate integral over $D$ of the function $y^3dxdy$.
So, it smells like Gauss-Green, i have a border and a surface but... ...first of all, if im not wrong, $y^3$ must be the divergence of Something $(S) | div(S)=y^3$. For example i can take $S=(0,0,\frac{y^4}{4})$, now? what im supposed to do? Im messing up things...
By using Stokes'es theorem we have:$$\int_S\nabla\times \vec A.\vec{ds}=\int_C \vec A.\vec {dl}$$here $\vec{ds}=dxdy\hat z$ , $\vec{dl}=dx\hat x+dy\hat y$ and take $A=-\dfrac{y^4}{4}\hat x$ therefore $\nabla\times \vec A=y^3\hat z$ and $$I=\int_S\nabla\times \vec A.\vec{ds}=\int_Sy^3{dxdy}$$also$$I=\int_C \vec A.\vec {dl}=\int_C -\dfrac{y^4}{4}{dx}$$by substituting $x=t-t^2$ and $y=t^3$ for $0\le t\le1$ we obtain:$$I=\int_{x=0} -\dfrac{y^4}{4}{dx}+\int_{0}^{1} -\dfrac{t^{12}}{4}(1-2t){dt}=\dfrac{3}{182}$$