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Here the given cone z=2-{\sqrt  {(x^{2}-y^{2})}}\, meets the xy plane in a circle C given by x^{2}+y^{2}=4,z=0\,.Let R be the plane region of the circular disc bounded by C.Let S1 denote the entire closed surface made up of S and R and Let V1 be the volume enclosed by S1.

Then,we have \iint _{{S_{1}}}(\nabla \times F)\cdot ndS=\iint _{{S_{1}}}{\mathrm  {curl}}F\cdot ndS\,

={\mathrm  {div}}{\mathrm  {curl}}FdV\, by divergence theorm.

=0\, [As {\mathrm  {div}}{\mathrm  {curl}}F=0\,]

Therefore,\iint _{S}(\nabla \times F)\cdot ndS+\iint _{R}(\nabla \times F)\cdot ndS=0\,

<\iint _{S}(\nabla \times F)\cdot ndS=-\iint _{R}{\mathrm  {curl}}F\cdot ndS\, --(1)

Here,{\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\(x-z)&(x^{3}+yz)&-3xy^{2}\end{vmatrix}}\,

=(-6xy-y)i+(-1+3y^{2})j+(3x^{2}-0)k\, --(2)

Since on R,n=-k,therefore,we have

{\mathrm  {curl}}F\cdot n={\mathrm  {curl}}F\cdot (-k)=-3x^{2}\,,using (2)

Therefore,from (1),\iint _{S}(\nabla \times f)\cdot ndS\,

=\iint _{R}3x^{2}dS=3\int _{{r=0}}^{{2}}\int _{{\theta =0}}^{{2\pi }}(r\cos \theta )^{2}\times rd\theta dr\,,changing to polars

=3\int _{{r=0}}^{{2}}\int _{{\theta =0}}^{{2\pi }}r^{3}[{\frac  {1+\cos 2\theta }{2}}]drd\theta \,

=3\int _{0}^{2}r^{3}[{\frac  {\theta }{2}}+{\frac  {\sin 2\theta }{4}}]_{{0}}^{{2\pi }}dr\,

=3\pi \int _{0}^{2}r^{3}dr=3\pi [{\frac  {r^{4}}{4}}]_{0}^{2}=12\pi \,

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