Given two scalars fields $u$ and $v$ that are continiously differentiable on an open set containing the circular disk $R$ whose boundary is the circle $x^2 + y^2=1 $. Define the two vector fields $f$ and $g$ as follows $f(x,y) = v(x,y) i + u(x,y) j $
$$g(x,y) = \biggl(\frac {\partial u}{\partial x}- \frac{\partial u} { \partial y}\biggr) i + \biggl(\frac {\partial v}{\partial x} - \frac{\partial v }{\partial y}\biggr) j$$
Find the value of the double integral $\iint_{R} f.g\,dx\,dy $ if it is known that on the boundary of $R $ we have $u(x,y) =1$ and $v(x,y) =y$
My attempt : By using Green's theorem , \begin{align} \iint_{R} f.g\:dx\, dy &= \iint_{R} \biggl(v\frac{\partial u}{\partial x}- v\frac{\partial u} {\partial y}\biggr) + \biggl(u\frac{\partial v}{\partial x} - u\frac{\partial v }{\partial y}\biggr)\,dx\,dy\\ &=\iint_{R} \frac {\partial( {uv})}{\partial x}-\frac {\partial( {uv})}{\partial y}\,dx\,dy \end{align} according to question it is given that $ u=1 , v=y$
so $$\iint_{R} f.g\,dx\, dy= \oint_{C} y\,dx + y\,dy$$
After that I'm not able to proceed further
You're nearly there. You found
$$ \iint_R f(x,y) \cdot g(x,y) \; d(x,y) = \oint_C y \; dx + y \; dy.$$
But this last integral is by definition the work of the vector field $F(x,y) = (y,y)$ along the curve $C$. Using the anticlockwise parametrisation $\gamma :[0,2\pi] \rightarrow C : t \mapsto (\cos(t),\sin(t))$ of $C$ the last integral is by definition
$$ \oint_C y \; dx + y \; dy := \int_0^{2\pi} F(\gamma(t)) \cdot \gamma'(t) dt = \int_0^{2\pi} \sin(t)\cos'(t) + \sin(t)\sin'(t)dt.$$
Notice that$$ \int_0^{2\pi} \sin(t) \sin'(t)= \int_{\sin(0)}^{\sin(2\pi)} u \; du = 0$$ so $$ \iint_Rf(x,y)\cdot g(x,y) \; d(x,y) = \int_0^{2\pi} -\sin^2(t) =- \pi.$$