VC5.46

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i).Let a be an arbitrary constant vector.Then,we have

a\cdot \iint _{S}ndS=\iint _{S}a\cdot ndS=\iiint _{V}{\mathrm  {div}}(a)dV\,,by divergence theorm.

=0\, [Since {\mathrm  {div}}a=0\, as a is a constant vector]

Thus,a\cdot \iint _{S}ndS=0\, and a is any arbitrary vector.

Hence,\iint _{S}ndS=0\,

ii). Let a be an arbitrary vector.Then,we have

a\cdot \iint _{S}r\times ndS=\iint _{S}a\cdot (r\times n)dS=\iint _{S}(a\times r)\cdot ndS\,

=\iiint _{V}{\mathrm  {div}}(a\times r)dV\,,by divergence theorm.

=\iiint _{V}[r\cdot {\mathrm  {curl}}a-a\cdot {\mathrm  {curl}}r]dV\,,by a vector identity

=0\, [Since {\mathrm  {curl}}r=0\,,{\mathrm  {curl}}a=0\, as a is a constant vector]

Thus a\cdot \iint _{S}r\times ndS=0\, and a is an arbitrary vector.

Hence \iint _{S}r\times ndS=0\,

iii). Let a be an arbitrary vector. Then,we have

a\cdot \iint _{S}(\nabla \phi )\times ndS=\iint _{S}a\cdot (\nabla \phi \times n)dS\,

=\iint _{S}(a\times \nabla \phi )\cdot ndS\,

=\iiint _{V}{\mathrm  {div}}(a\times \nabla \phi )dV\,,by divergence theorm.

=\iiint _{V}[\nabla \phi {\mathrm  {curl}}a-a\cdot {\mathrm  {curl}}\nabla \phi ]dV\,

=0\, [Since curl a=0 and {\mathrm  {curl}}\nabla \phi =0\,]

Thus,a\cdot \iint _{S}(\nabla \phi )\times ndS=0\, and a is an arbitrary vector.

Therefore,\iint _{S}(\nabla \phi )\times ndS=0\,

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