91Ó°ÊÓ

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

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

(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$\underset{s}{∫}\left(∇×fA\right)·da=\underset{P}{∫}fA·dl$

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

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

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

(b)

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

$∇·\left(A×B\right)=B·\left(∇×A\right)-a·\left(∇×B\right)$.

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

role="math" localid="1658812206459" $\underset{v}{∫}∇·\left(A×B\right)d\mathrm{τ}=\underset{v}{∫}B·\left(∇×A\right)d\mathrm{τ}-\underset{v}{∫}A·\left(∇×B\right)d\mathrm{τ}$ …… (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$\underset{v}{∫}\left(∇·A\right)·d\mathrm{τ}=\underset{s}{∫}A·da$

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

$\begin{array}{rcl}\underset{v}{∫}∇·\left(A×B\right)d\mathrm{τ}& =& \underset{v}{∫}B·\left(∇×A\right)d\mathrm{τ}-\underset{v}{∫}A·\left(∇×B\right)d\mathrm{τ}\\ \underset{s}{∫}\left(A×B\right)da& =& \underset{v}{∫}B·\left(∇×A\right)d\mathrm{τ}-\underset{v}{∫}A·\left(∇×B\right)d\mathrm{τ}\\ \underset{v}{∫}B·\left(∇×A\right)d\mathrm{τ}& =& \underset{v}{∫}A·\left(∇×B\right)d\mathrm{τ}+\underset{s}{∫}\left(A×B\right)da\end{array}$

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