#### Answer

$16 \pi$

#### 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, $M=4x-2y ; N=2x-4y$
$W=\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial (2x-4y)}{\partial x}-\dfrac{\partial (4x-2y)}{\partial y}) dx dy$
or, $W=\iint_{R} 2-(-2) dx dy=\int (4) dx dy$
Since, $\int (4) dx dy=4 \times$ Area of a circle having radius 2
So, work done: $W=4 \times \pi \times (2)^2=16 \pi$