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Let \phi (x,y,z)=x^{2}+y^{2}=9\, --(1)

Therefore \nabla \phi =i{\frac  {\partial \phi }{\partial x}}+j{\frac  {\partial \phi }{\partial y}}+k{\frac  {\partial \phi }{\partial z}}=2xi+2yj\,

n=unit normal vector at any point (x,y,z) of S ={\frac  {2xi+2yj}{{\sqrt  {4x^{2}+4y^{2}}}}}={\frac  {xi+yj}{3}}\,

Therefore,F\cdot n=(zi+xj-yzk)\cdot {\frac  {xi+yj}{3}}={\frac  {zx+xy}{3}}\, -(2)

While solving a problem connected with a cylindrical surface we should use cylindrical coordinates r,theta,z for better solution.on the surface of the cylinder(1) we have x=3\cos \theta ,y=3\sin \theta ,z=z\, --(3)

dS=elementaty area on the cylindrical surface x^{2}+y^{2}=9,dS=3d\theta dz\, --(4)

\iint _{S}F\cdot ndS=\iint _{S}{\frac  {1}{3}}x(z+y)3d\theta dz\, using (2),(4)

=\int _{{z=0}}^{{4}}\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}(3\cos \theta )(z+3\sin \theta )d\theta dz\, using (3)

=3\int _{{\theta =0}}^{{{\frac  {\pi }{2}}}}\cos \theta [{\frac  {1}{2}}z^{2}+3z\sin \theta ]_{{z=0}}^{{4}}d\theta \,

=3\int _{{0}}^{{{\frac  {\pi }{2}}}}\cos \theta [8+12\sin \theta ]d\theta \,

=3\int _{{0}}^{{{\frac  {\pi }{2}}}}[8\cos \theta +6\sin 2\theta ]d\theta \,

=3[8\sin \theta -3\cos 2\theta ]_{{0}}^{{{\frac  {\pi }{2}}}}=3[8+3+3]=42\,

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