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Let V be the volume enclosed by S.Then,by the divergence theorm \iint _{S}(F\times \nabla \phi )\cdot ndS=\iiint _{V}{\mathrm  {div}}(F\times \nabla \phi )dV\, --(1)

We shall now use the following vector identiry.

{\mathrm  {div}}(f\times g)=g\cdot {\mathrm  {curl}}f-f\cdot {\mathrm  {curl}}g\,

Replacing f by F and g by \nabla \phi \,,we have

{\mathrm  {div}}(F\times \nabla \phi )=\nabla \phi \cdot {\mathrm  {curl}}F-F\cdot {\mathrm  {curl}}(\nabla \phi )=\nabla \phi \cdot {\mathrm  {curl}}F+0\, [as {\mathrm  {curl}}\nabla \phi ={\mathrm  {curl}}{\mathrm  {grad}}\phi =0\,]

Thus,{\mathrm  {div}}(F\times \nabla \phi )=\nabla \phi \cdot {\mathrm  {curl}}F\, --(3)

But {\mathrm  {curl}}F={\begin{vmatrix}i&j&k\\{\frac  {\partial }{\partial x}}&{\frac  {\partial }{\partial y}}&{\frac  {\partial }{\partial z}}\\x^{2}&y^{2}&z^{2}\end{vmatrix}}\,

=[{\frac  {\partial z^{2}}{\partial y}}-{\frac  {\partial y^{2}}{\partial z}}]i-[{\frac  {\partial z^{2}}{\partial x}}-{\frac  {\partial x^{2}}{\partial z}}]j+[{\frac  {\partial y^{2}}{\partial x}}-{\frac  {\partial x^{2}}{\partial y}}]k=0\,

(3) implies {\mathrm  {div}}(F\times \nabla \phi )=\nabla \phi \cdot (0)=0\, --(4)

Using (4),(1) =\iint _{S}(F\times \nabla \phi )\cdot ndS=\iiint _{V}(0)dV=0\,

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