Answer
$16 \pi$
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 $M=4x-2y ; N=2x-4y$
Equation (1) becomes: Work done is: $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial (2x-4y)}{\partial x}-\dfrac{\partial (4x-2y)}{\partial y}) dx dy=\iint_{R} 2-(-2) dx dy$
This implies that $\int (4) dx dy=4 \times$ Area of a circle having radius 2
Thus, work done is $4 (\pi) (2)^2=16 \pi$
