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Since the z-coordinates of each vertex of the given triangle is zero,hence the given triangle must lie in the xy-plane. In the present problem A is (1,0) and B is(1,1) and so the given triangle is OAB.Also,we have

{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\y^{2}&x^{2}&-(x+z)\end{vmatrix}}=j+2(x-y)k\, --(1)

Equation of OB is y-0={\frac  {1-0}{1-0}}(x-0)\, i.e y=x.

Since here, n=k,so by Stokes'theorm, we have

\oint _{C}F\cdot dr=\iint _{S}{\mathrm  {curl}}F\cdot ndS\,

=\iint _{S}[j+2(x-y)k]\cdot kdxdy\,

=\int _{{x=0}}^{{1}}\int _{{y=0}}^{{x}}2(x-y)dxdy\,

=2\int _{0}^{1}[xy-{\frac  {y^{2}}{2}}]_{0}^{x}dx=2\int _{0}^{1}{\frac  {x^{2}}{2}}dx={\frac  {1}{3}}\,

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