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Project the given sphere S on the xy-plane in the region R bounded by the circle x^{2}+y^{2}=1,z=0\,

Hence {\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\y&z&x\end{vmatrix}}\,

=[{\frac  {\partial x}{\partial y}}-{\frac  {\partial z}{\partial z}}]i-[{\frac  {\partial x}{\partial x}}-{\frac  {\partial y}{\partial z}}]j+[{\frac  {\partial z}{\partial x}}-{\frac  {\partial y}{\partial y}}]k=-(i+j+k)\, --(1)

Given surface S is \phi (x,y,z)=x^{2}+y^{2}+z^{2}=1\, --(2)

From(2), \nabla \phi =[i{\frac  {\partial }{\partial x}}+j{\frac  {\partial }{\partial y}}+k{\frac  {\partial }{\partial z}}](x^{2}+y^{2}+z^{2})=2(xi+yj+zk)\,

Therefore,n=unit normal vector to surface(2)={\frac  {\nabla \phi }{|\nabla \phi |}}={\frac  {2(xi+yj+zk)}{{\sqrt  {(4x^{2}+4y^{2}+4z^{2})}}}}=xi+yj+zk\, from (2)

Now,n\cdot k=(xi+yj+zk)\cdot k=z\, and {\mathrm  {curl}}F\cdot n=-(i+j+k)\cdot (xi+yj+zk)=-(x+y+z)\,

Therefore,\iint _{S}({\mathrm  {curl}}F)\cdot ndS=\iint _{R}({\mathrm  {curl}}F)\cdot n{\frac  {dxdy}{n\cdot k}}=\iint _{R}(-1)(x+y-z){\frac  {dxdy}{z}}=-\iint _{R}{\frac  {x+y+{\sqrt  {(1-x^{2}-y^{2})}}}{{\sqrt  {(1-x^{2}-y^{2})}}}}dxdy\, --(3)

We now transform the double integral(3) into polar coordinates by writing x=r\cos \theta ,y=r\sin \theta ,x^{2}+y^{2}=r^{2},dxdy=rd\theta dr\,,Again for the circular region,r varies from 0 to 1 and theta varies from 0 to 2\pi \,.Then (3) may be rewritten as

=-\int _{{\theta =0}}^{{2\pi }}\int _{{r=0}}^{{1}}{\frac  {r\cos \theta +r\sin \theta +{\sqrt  {1-r^{2}}}}{{\sqrt  {1-r^{2}}}}}rd\theta dr\,

=-\int _{{\theta =0}}^{{2\pi }}\int _{{r=0}}^{{1}}{\frac  {r^{2}}{{\sqrt  {1-r^{2}}}}}(\sin \theta +\cos \theta )d\theta dr=-\int _{{0}}^{{1}}{\frac  {r^{2}}{{\sqrt  {1-r^{2}}}}}(-\cos \theta +\sin \theta )_{{0}}^{{2\pi }}dr-\int _{{r=0}}^{{1}}r[\theta ]_{{0}}^{{2\pi }}dr\,

=-\int _{{0}}^{{1}}{\frac  {r^{2}}{{\sqrt  {1-r^{2}}}}}(0)dr-\int _{{0}}^{{1}}r(2\pi )dr=0-2\pi [{\frac  {r^{2}}{2}}]_{{0}}^{{1}}=-2\pi [{\frac  {1}{2}}-0]=-\pi \,

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