Question

(a) Show that

$\underset{\mathbf{s}}{\mathbf{\int }}\mathit{f}\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathit{A}\mathbf{)}\mathbf{-}\mathit{d}\mathit{a}\mathbf{=}\underset{\mathbf{s}}{\mathbf{\int }}\mathbf{[}\mathbf{A}\mathbf{\times }\left(\nabla f\right)\mathbf{]}\mathbf{-}\mathit{d}\mathit{a}\mathbf{+}\underset{\mathbf{P}}{\mathbf{\int }}\mathit{f}\mathit{A}\mathbf{-}\mathit{d}\mathit{l}$

(b) Show that

$\underset{\mathbf{v}}{\mathbf{\int }}\mathit{B}\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathbf{A}\mathbf{)}\mathbf{-}\mathit{d}\mathit{\tau }\mathbf{=}\underset{\mathbf{v}}{\mathbf{\int }}\mathit{A}\mathbf{-}\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathbf{B}\mathbf{)}\mathbf{-}\mathit{d}\mathit{\tau }\mathbf{+}\underset{\mathbf{s}}{\mathbf{\int }}\mathit{A}\mathbf{\times }\mathit{B}\mathbf{-}\mathit{d}\mathit{a}$

Short answer

1. The expression in part (a) is proved.
2. The expression in part (b) is proved.

Step-by-step solution

Describe the given information.

The expressions $\underset{\mathrm{s}}{\int }f(\nabla \times A)·da=\underset{\mathrm{s}}{\int }\left\{\mathrm{A}\times (\nabla \mathrm{f})\right\}·da+\underset{\mathrm{P}}{\int }fA-dl$and $\underset{\mathrm{v}}{\int }B·(\nabla \times \mathrm{A})-d\tau =\underset{\mathrm{v}}{\int }A·(\nabla \times \mathrm{B})d\tau +\underset{\mathrm{s}}{\int }\left(A\times B\right)da$ have to be proved. Here, A, B are arbitrary vectors, and f is scalar function.

Define the product rule (i).

According to the product rule (i) $\mathbf{\nabla }\mathbf{\times }\left(fA\right)\mathbf{=}\mathit{f}\left(\nabla \times A\right)\mathbf{-}\mathit{A}\left(\nabla \times B\right)$.

Prove the expression in part (a).

(a)

According to stokes theorem, the surface integral of curl of a function is equal to the line integral of that function written mathematically as$\underset{s}{\int }\left(\nabla \times fA\right)·da=\underset{P}{\int }fA·dl$

Apply the product rule (i) into the right side of stokes theorem as,

$\begin{array}{rcl}\underset{P}{\int }fA·dl& =& \underset{s}{\int }\left\{f\left(\nabla \times A\right)\right\}·da-\underset{s}{\int }\left\{A\times \left(\nabla f\right)\right\}·da\\ \underset{s}{\int }\left\{f\left(\nabla \times A\right)\right\}·da& =& \underset{s}{\int }\left\{A\times \left(\nabla f\right)\right\}·da+\underset{P}{\int }fA·dl\end{array}$

Thus, the expression in part (a) is proved.

Prove the expression in part (b).

(b)

According to the formula of divergence of curl of two functions A, B

$\nabla ·\left(A\times B\right)=B·\left(\nabla \times A\right)-a·\left(\nabla \times B\right)$.

Apply the volume integral on the both sides of above mentioned formula as

role="math" localid="1658812206459" $\underset{v}{\int }\nabla ·\left(A\times B\right)d\tau =\underset{v}{\int }B·\left(\nabla \times A\right)d\tau -\underset{v}{\int }A·\left(\nabla \times B\right)d\tau$ …… (1)

According to Gauss divergence theorem, the volume integral of curl of a function is equal to the surface integral of that function written mathematically as$\underset{v}{\int }\left(\nabla ·A\right)·d\tau =\underset{s}{\int }A·da$

Apply the gauss divergence theorem on left side of equation (1)

$\begin{array}{rcl}\underset{v}{\int }\nabla ·\left(A\times B\right)d\tau & =& \underset{v}{\int }B·\left(\nabla \times A\right)d\tau -\underset{v}{\int }A·\left(\nabla \times B\right)d\tau \\ \underset{s}{\int }\left(A\times B\right)da& =& \underset{v}{\int }B·\left(\nabla \times A\right)d\tau -\underset{v}{\int }A·\left(\nabla \times B\right)d\tau \\ \underset{v}{\int }B·\left(\nabla \times A\right)d\tau & =& \underset{v}{\int }A·\left(\nabla \times B\right)d\tau +\underset{s}{\int }\left(A\times B\right)da\end{array}$

Thus, the expression in part (b) is proved.