VC4.28

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Given surface \phi (x,y,z)=x^{2}+y^{2}-(2-z)^{2}=0\, --(1) Therefore \nabla \phi =2(xi+yj+(2-z)k,n={\frac  {xi+yj+(2-z)^{2}}{{\sqrt  {(x^{2}+y^{2}+(2-z)^{2})}}}}\,

n\cdot k={\frac  {(2-z)}{{\sqrt  {(x^{2}+y^{2}+(2-z)^{2})}}}}\,

\iint F\cdot ndS=\iint [x(x-z)+y(x^{3}+yz)-(2-z)3xy^{2}]{\frac  {dxdy}{2-z}}\,

=\iint [(x^{2}+yx^{3}-2x+2y^{3})-(2-z)(-x+y^{2}+3xy^{2})]{\frac  {dxdy}{2-z}}\,

=\iint {\frac  {x^{2}+yx^{3}-2x+2y^{2}-{\sqrt  {x^{3}+y^{2}}}(-x+y^{2}+3xy^{2})}{{\sqrt  {x^{2}+y^{2}}}}}dxdy\,, by (1)

=\int _{{r=0}}^{{2}}\int _{{\theta =0}}^{{2\pi }}[(r^{2}\cos ^{2}\theta +r^{4}\cos ^{3}\theta \sin \theta -2r\cos \theta +2r^{2}\sin ^{2}\theta )+r(r\cos \theta -r^{2}\sin ^{2}\theta -3r^{3}\cos \theta \sin ^{2}\theta )]drd\theta \, [on changing to polar coordinates]

=4\int _{{r=0}}^{{2}}\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}(r^{2}\cos ^{2}\theta +2r^{2}\sin ^{2}\theta -r^{3}\sin ^{2}\theta )drd\theta )drd\theta =4\pi \, [on simplification]

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