Question

1-25 By applying the divergence theorem to the special case in which A is a constant but other- wise arbitrary vector, show that the total vector area of a closed surface is zero, that is, $da = 0. Similarly, show that $ds = 0. Do these results surprise you? 1-26 Verify (1-122). (1-123), and (1-124 by using

Answer 1

Using Gauss divergence theorem

$\int_V( \nabla .\vec A )dV=\oint \vec A .d\vec a$

When A is a constant vector, its divergence vanishes.

$\nabla .\vec A = 0$

Hence the left-hand side is zero. And A can be pulled out of the integral on the right-hand side. Therefore

$\vec A.\oint d\vec a=0$

Since A is not zero,

$\oint d\vec a=0$

Stokes' theorem states that

$\int (\nabla \times \vec A) .d\vec a= \oint \vec A .d\vec s$

Since A is a constant vector, its curl vanishes

$(\nabla \times \vec A) =0$

Hence the left-hand side is zero. And A can be pulled out of the integral on the right-hand side. Therefore$\oint \vec A .d\vec s=\vec A.\oint d\vec s=0$

Since A is not zero,

$\oint d\vec s=0$

These results are expected because integral on a closed surface or a closed means you start from a point on the surface or the line and return back to the same point. The net vector magnitude of such an operation is zero. Therefore the closed surface and line integrals are zero.