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Given F=xi+yj+z^{2}k\,

Therefore,\iint _{S}F\cdot ndS=\iiint _{V}{\mathrm  {div}}FdV=\iiint _{V}{\mathrm  {div}}(xi+yj+z^{2}k)dV\,,by divergence theorm.

=\iiint _{V}[{\frac  {\partial }{\partial }}(x)+{\frac  {\partial }{\partial y}}(y)+{\frac  {\partial }{\partial z}}(z^{2})]dV=\iiint _{V}(1+1+2z)dV=2\iiint _{V}dV+2\iiint _{V}zdV=2V+2Vz_{1}=2V(1+z_{1})\, --(1) where z1=z-coordinate of the centroid of the cone x^{2}+y^{2}=z^{2}\,

or z_{1}={\frac  {\iiint _{V}zdV}{\iiint _{V}dV}}={\frac  {\iiint _{V}zdV}{V}}\, so that \iiint _{V}zdV=Vz_{1}\,

Now we know that x^{2}+y^{2}=z^{2}\, represents a right circular cone whose vertex is (0,0,0) and whose axis is OZ,base is a circle with equation x^{2}+y^{2}=4^{2}\,

Here the radius of the cone =4 and height of the cone =4.If ZG be the centroid of the cone,then z_{1}=OG={\frac  {3}{4}}\times heightofthecone={\frac  {3}{4}}\times 4=3\, and V=volume of the cone ={\frac  {\pi }{3}}(radius)^{2}\times (height)={\frac  {\pi }{3}}(4)^{2}\times 4={\frac  {64\pi }{3}}\,

Therefore,(1)=\iiint _{S}F\cdot ndS=2{\frac  {64\pi }{3}}(1+3)={\frac  {256\pi }{3}}\,

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