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The equation of the given surface is z=2-{\sqrt  {(x^{2}+y^{2})}}\,. It meets the xy-plane in the circle C whose equations are x^{2}+y^{2}=4,z=0\,.Hence the boundary of the surface S is the circle C. The parametric equations of the curve C are

x=2\cos \theta ,y=2\sin \theta ,z=0,0\leq \theta \leq 2\pi \, --(1)

Hence,by Stokes'theorm,we have \iint _{S}(\nabla \times A)\cdot ndS\,

=\int _{C}A\cdot dr=\int _{C}[(x-z)i+(x^{3}+yz)j-3xy^{2}k]\cdot (dxi+dyj+dzk)\,

=\int _{C}[(x-z)dx+(x^{3}+yz)dy-3xy^{2}dz]=\int _{C}(xdx+x^{3}dy)\,

=\int _{{\theta =0}}^{{2\pi }}[x{\frac  {dx}{d\theta }}+x^{3}{\frac  {dy}{d\theta }}]d\theta \,

=\int _{{0}}^{{2\pi }}[(2\cos \theta )(-2\sin \theta )+(8\sin ^{3}\theta )(2\cos \theta )]d\theta \,

=-2\int _{{0}}^{{2\pi }}\sin 2\theta d\theta +16\int _{{0}}^{{2\pi }}\cos ^{{16}}\theta d\theta \,

=-2[-{\frac  {\cos 2\theta }{2}}]_{{0}}^{{2\pi }}+16\cdot 2\int _{{0}}^{{\pi }}\cos ^{4}\theta d\theta \,

=0+32\cdot 2\int _{{0}}^{{{\frac  {\pi }{2}}}}\cos ^{4}\theta d\theta \,

=64{\frac  {3}{4}}\cdot {\frac  {1}{2}}\cdot {\frac  {\pi }{2}}=12\pi \,

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