Answer
$\dfrac{2}{33}$
Work Step by Step
The tangential form for Green's Theorem - work done can be defined as: $W=\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy $ $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy $
So, $M=2xy^3 ; N=4x^2y^2$
$W=\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial (4x^2y^2)}{\partial x}-\dfrac{\partial (2xy^3)}{\partial y}) dx dy=\iint_{R}$
or, $W= 2xy^2 dx dy=\int_{0}^{1} \int_{0}^{x^3} 2xy^2 dx dy$
So, work done: $W=\int_{0}^{1} [\dfrac{2xy^3}{3}]_{0}^{x^3} dx=\dfrac{2}{33}$
