VC5.46

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

$a\cdot\iint_S ndS=\iint_S a\cdot n dS=\iiint_V \mathrm{div}(a) dV\,$ ,by divergence theorm.

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

Thus, $a\cdot\iint_S n dS=0\,$ and a is any arbitrary vector.

Hence, $\iint_S ndS=0\,$

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

$a\cdot\iint_S r\times n dS=\iint_S a\cdot(r\times n)dS=\iint_S(a\times r)\cdot n dS\,$

= $\iiint_V\mathrm{div}(a\times r)dV\,$ ,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 $a\cdot\iint_S r\times n dS=0\,$ and a is an arbitrary vector.

Hence $\iint_S r\times n dS=0\,$

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

$a\cdot\iint_S(\nabla\phi)\times n dS=\iint_S a\cdot(\nabla\phi\times n)dS\,$

= $\iint_S(a\times\nabla\phi)\cdot n dS\,$

= $\iiint_V \mathrm{div}(a\times\nabla\phi)dV\,$ ,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, $a\cdot\iint_S(\nabla\phi)\times n dS=0\,$ and a is an arbitrary vector.

Therefore, $\iint_S(\nabla\phi)\times n dS=0\,$