 Solution

For an oriented piecewise smooth surface $S$, a continuously differentiable vector field $F$ on a neighborhood of it, and its compatibly oriented boundary $C=\partial S$, Stokes theorem statesHere $n$ is the chosen unit normal vector, $dS$ is the unsigned surface area element, and $dr$ is the tangent displacement along $C$. The right-hand rule relates the boundary direction to $n$.

The supplied field has curl $\nabla \times F=(2z,2x,2y)$. Parametrize the plane by $r(x,y)=(x,y,1-x-y)$ over the triangle $D=\{x,y\ge 0:x+y\le 1\}$. Its upward oriented surface area element is $(1,1,1)dxdy$, givingFor the boundary line integral, take the positively oriented vertices $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$. Their projections run counterclockwise in the horizontal plane. With $0\le t\le 1$, the three edges areOn each edge, direct substitution gives $F(r(t))\cdot r^{\prime }(t)=t^2+2t-1$. HenceThe two independent computations agree, including the orientation sign.