# VC5.46

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