Answer
$\dfrac{1}{24}$
Work Step by Step
Stokes' Theorem states that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr $
curl $F=(x-y) i-yj +k$
The region $D$ is the triangle formed by the vertices $(0,0), (1/3,0)$ and $0,1/2)$
So, $\iint_{S} curl \space F \cdot dS=\iint_{D} 3x-5y+1 dA$
The equation of joining points $(1/3,0)$ and $(0,1/2)$ is: $\dfrac{x}{1/3}+\dfrac{y}{1/2}=1 \implies y=\dfrac{1-3x}{2}$
So, $\iint_{D} 3x-5y+1 dA=\int_{0}^{1/3} [3xy -\dfrac{5y^2}{2}+y]_0^{(1-3x)/2} dx$
Plug in $y=\dfrac{1-3x}{2}$
$\iint_{D} 3x-5y+1 dA=\int_{0}^{1/3} \dfrac{3x-9x^2}{2} -\dfrac{45x^2}{8}+\dfrac{3x}{2} dx=\dfrac{1}{8} \int_0^{1/3} 24x -81x^2 dx$
or, $=\dfrac{1}{8}[12x^2-27 x^3]_0^{1/3}$
or, $=\dfrac{1}{8} (\dfrac{4}{3}-1)$
or, $=\dfrac{1}{24}$
