VC5.39

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Here {\mathrm  {div}}F={\frac  {\partial }{\partial x}}(x)+{\frac  {\partial }{\partial y}}(-y)+{\frac  {\partial }{\partial z}}(z^{2}-1)=1-1+2z=2z\, --(1)

Now,by divergence theorm,we have

\iint _{S}F\cdot ndS=\iiint _{V}{\mathrm  {div}}FdV\,,V being the volume enclosed by S.

=\iiint _{V}2zdV\, using(1)

=2\int _{{z=0}}^{{1}}\int _{{y=-2}}^{{2}}\int _{{x=-{\sqrt  {4-y^{2}}}}}^{{{\sqrt  {4-y^{2}}}}}zdxdydz\,

=2\int _{{z=0}}^{{1}}\int _{{y=-2}}^{{2}}[zx]_{{x=-{\sqrt  {4-y^{2}}}}}^{{{\sqrt  {4-y^{2}}}}}dydz\,

=2\int _{{z=0}}^{{1}}\int _{{y=-2}}^{{2}}2z{\sqrt  {2^{2}-y^{2}}}dydz\,

=8\int _{0}^{1}\int _{{y=-2}}^{{2}}z{\sqrt  {2^{2}-y^{2}}}dydz\, Since the square root is an even function of y.

=8\int _{0}^{1}z[{\frac  {y}{2}}{\sqrt  {2^{2}-y^{2}}}+{\frac  {2^{2}}{2}}\arcsin {\frac  {y}{2}}]_{{y=-2}}^{{2}}dz\,

=8\int _{0}^{1}z[2\arcsin 1]dz=8\times 2\times {\frac  {\pi }{2}}\int _{0}^{1}zdz=8\pi [{\frac  {z^{2}}{2}}]_{0}^{1}=4\pi \,

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