Answer
$\dfrac{2}{33}$
Work Step by Step
The tangential form for Green Theorem is given as:
Work done $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy $ ...(1)
As per the question, we have $M=2xy^3 ; N=4x^2y^2$
Equation (1) gives: Work done is: $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial (4x^2y^2)}{\partial x}-\dfrac{\partial (2xy^3)}{\partial y}) dx dy=\iint_{R} 2xy^2 dx dy$
This implies that $\int_{0}^{1} \int_{0}^{x^3} 2xy^2 dx dy=\int_{0}^{1} [\dfrac{2xy^3}{3}]_{0}^{x^3} dx$
or, $\int_{0}^{1} [\dfrac{2x^{10}}{3}] dx=[\dfrac{2x^{11}}{33}]_0^1$
Thus, $[\dfrac{2x^{11}}{33}]_0^1=\dfrac{2}{33}$
