VC5.61

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By divergence theorm,we have

\iint _{S}F\cdot ndS=\iiint _{V}{\mathrm  {div}}FdV\, --(1)

RHS of (1)=\iiint _{V}[{\frac  {\partial }{\partial x}}(4xz)+{\frac  {\partial }{\partial y}}(-y^{2})+{\frac  {\partial }{\partial z}}(yz)]dV\,

=\int _{{z=0}}^{{1}}\int _{{y=0}}^{{1}}\int _{{x=0}}^{{1}}(4z-2y+y)dxdydz\,

=\int _{{z=0}}^{{1}}\int _{{y=0}}^{{1}}(4z-y)[x]_{0}^{1}dydz\,

=\int _{{z=0}}^{{1}}\int _{{y=0}}^{{1}}(4z-y)dydz=\int _{{z=0}}^{{1}}[4zy-{\frac  {y^{2}}{2}}]_{0}^{1}dz\,

=\int _{{z=0}}^{{1}}[4z-{\frac  {1}{2}}]dz=[2z^{2}-{\frac  {z}{2}}]_{0}^{1}=2-{\frac  {1}{2}}={\frac  {3}{2}}\, --(2)

LHS=\iint _{{S_{1}}}F\cdot ndS+\iint _{{S_{2}}}F\cdot ndS+\iint _{{S_{3}}}F\cdot ndS+\iint _{{S_{4}}}F\cdot ndS+\iint _{{S_{5}}}F\cdot ndS+\iint _{{S_{6}}}F\cdot ndS\,

The six integrals varies as 2,0,-1,0,1/2,0.

Therefore,LHS=2+0-1+0+{\frac  {1}{2}}+0={\frac  {3}{2}}\, which is equal to RHS.

Hence divergence theorm is verified.

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