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Here the given sphere x^{2}+y^{2}+z^{2}=a^{2}\, meets the xy plane in a circle given by x^{2}+y^{2}=a^{2}\,,z=0.Let R be the plane region of the circular disc,bounded by the circle.Let S1 denote the entire closed surface made up of S and R and let V be the volume enclosed by S1.Then,we have

\iint _{{S_{1}}}(\nabla \times F)\cdot ndS=\iiint _{{V_{1}}}{\mathrm  {div}}(\nabla \times F)dV\, by the divergence theorm

=0\, [as {\mathrm  {div}}(\nabla \times F)={\mathrm  {div}}{\mathrm  {curl}}F=0\,]

=\iint _{S}(\nabla \times F)\cdot ndS+\iint _{R}(\nabla \times F)\cdot ndS=0\, [as S1 consists of S and R]

=\iint _{S}(\nabla \times F)\cdot ndS+\iint _{R}(\nabla \times F)\cdot (-k)dS=0\,,as on R,n=-k.

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

Now {\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\0&0&xye^{z}+\log(z+1)-\sin x\end{vmatrix}}\,

=[{\frac  {\partial }{\partial y}}(xye^{z}+\log(z+1)-\sin x]i-[{\frac  {\partial }{\partial x}}(xye^{z}+\log(z+1)-\sin x)]j+0k=[xe^{z}i-(ye^{z}-\cos x)j]\,

=(\nabla \times F)\cdot k=[xe^{z}i-(xe^{z}-\cos x)j]\cdot k=0\,

Then, (1) becomes \iint _{S}(\nabla \times F)\cdot ndS=\iint _{R}0dS=0\,

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