Answer
$$\dfrac{1}{24}$$
Work Step by Step
Stokes' Theorem states that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr$
where, $C$ corresponds to the boundary of the surface oriented counter-clockwise.
Now $curl \ F=(x-y) i-yj +k$
Since, $D$ 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 intersection points for $(1/3,0)$ and $0,1/2)$ he $\dfrac{x}{1/3}+\dfrac{y}{1/2}=1$
or, $ 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 $y=\dfrac{1-3x}{2}$
$$\iint_{D} 3x-5y+1 dA=\int_{0}^{1/3} \dfrac{3x-9x^2}{2} -\dfrac{45x^2}{8}+\dfrac{3x}{2} dx=\dfrac{1}{8} \times \int_0^{1/3} [24x -81x^2] \ dx \\=\dfrac{1}{8}[12x^2-27 x^3]_0^{(1/3)} \\=\dfrac{1}{8} \times (\dfrac{4}{3}-1) \\=\dfrac{1}{24}$$
