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The identities $鈭�鈭�T鈰�d蟿=鈭�T鈰�da$, $\underset{}{鈭�}\text{鈥夆赌�}\left(鈭�脳v\right)鈰�d蟿=鈭�\underset{}{鈭�}\left(v脳da\right)\text{鈥夆赌�}$, $鈭�T{鈭�}^{2}U+鈭�U鈰�鈭�T=鈭�\left(T鈭�U\right)\text{鈥�}da\text{鈥�}$,$鈭�U{鈭�}^{2}T+鈭�T鈰�鈭�U=鈭�\left(U鈭�T\right)\text{鈥�}da\text{鈥�}$

and $鈭�鈭�T脳da=鈭�鈭�T鈰�dl$have to be proved. Here $T,\text{鈥�}U,\text{鈥夆赌�}V$are the vector and c is a constant.

According to the Gauss divergence theorem The integral ofdivergenceof a function$f\left(x,y,z\right)$ over an closed surface area is equal to the surfaceintegral of the function $鈭�\left(鈭�v\right)鈰�\text{鈥�}d蟿=\underset{s}{鈭�}v鈰�\text{鈥�}da$. According to thestokestheorem, the integral of divergence of a function $f\left(x,y,z\right)$over an open surface area is equal to the line integral of the function $鈭�\left(鈭�脳v\right)鈰�\text{鈥�}da=\underset{l}{鈭�}v鈰�\text{鈥�}dl$.

The divergence theorem is defined as $鈭�\left(鈭�v\right)鈰�d蟿=鈭�v鈰�da$.Substitute $cT$ for v into $鈭�\left(鈭�v\right)鈰�d蟿=鈭�v鈰�da$as follows:

$鈭�\left(鈭�\left(cT\right)\right)鈰�d蟿=鈭�cT鈰�da$ 鈥︹�︹��.. (1)

Apply the product rule (i) ,$鈭�\left(fA\right)=f\left(鈭�鈰�A\right)+A鈰�\left(鈭�f\right)$ in equation (1),

$\begin{array}{c}鈭�\left(鈭�\left(cT\right)\right)鈰�d蟿=鈭�cT鈰�da\\ 鈭�\left(T\left(鈭�鈰�c\right)+c鈰�\left(鈭�T\right)\right)鈰�d蟿=鈭�cT鈰�da\end{array}$ 鈥︹�︹��.. (2)

As c is a constant, so its divergence is 0, that is, $鈭�鈰�c=0$.

Substitute 0 for $鈭�鈰�c$ into equation (2)

$\begin{array}{c}鈭�\left(T\left(0\right)+c鈰�\left(鈭�T\right)\right)鈰�d蟿=鈭�cT鈰�da\\ 鈭�\left(c鈰�\left(鈭�T\right)\right)鈰�d蟿=鈭�cT鈰�da\\ c鈭�鈭�T鈰�d蟿=c鈭�T鈰�da\\ 鈭�鈭�T鈰�d蟿=鈭�T鈰�da\end{array}$

Thus, $鈭�鈭�T鈰�d蟿=鈭�T鈰�da$, has been shown.

The gauss divergence theorem states that the volume integral of the divergence of a function v is equal to the surface integral of the function v, that is,$\underset{v}{鈭�}\text{鈥�}鈭�v鈰�d蟿=\underset{s}{鈭�}v\text{鈥�}da\text{鈥�}$

Substitute $v脳c$for v into $\underset{v}{鈭�}\text{鈥�}鈭�v鈰�d蟿=\underset{s}{鈭�}v\text{鈥�}da\text{鈥�}$.

$\underset{v}{鈭�}\text{鈥�}鈭�\left(v脳c\right)鈰�d蟿=\underset{s}{鈭�}\left(v脳c\right)\text{鈥�}da\text{鈥�}$鈥︹�� (3)

Apply the rule $\text{鈥�}鈭�\left(\text{A}脳\text{B}\right)=\text{B}\left(鈭�脳\text{A}\right)鈭�\text{A}\left(鈭�脳\text{B}\right)$ into equation (3)

$\underset{v}{鈭�}\text{鈥夆赌�}\left(c\left(鈭�脳v\right)鈭�v\left(鈭�脳c\right)\right)鈰�d蟿=\underset{s}{鈭�}\left(v脳c\right)\text{鈥�}da\text{鈥�}$

As c is a constant, so it鈥檚 curl is 0, that is, $鈭�脳c=0$.

Substitute 0 for $鈭�脳c$ into equation$\underset{v}{鈭�}\text{鈥夆赌�}\left(c\left(鈭�脳v\right)鈭�v\left(鈭�脳c\right)\right)鈰�d蟿=\underset{s}{鈭�}\left(v脳c\right)\text{鈥�}da\text{鈥�}$

$\begin{array}{l}\underset{v}{鈭�}\text{鈥夆赌�}\left(c\left(鈭�脳v\right)鈭�v\left(0\right)\right)鈰�d蟿=\underset{s}{鈭�}\left(v脳c\right)\text{鈥�}da\text{鈥�}\\ \underset{v}{鈭�}\text{鈥夆赌�}c\left(鈭�脳v\right)鈰�d蟿=\underset{s}{鈭�}c\left(鈭�脳da\right)\text{鈥夆赌�}\\ c\underset{v}{鈭�}\text{鈥夆赌�}\left(鈭�脳v\right)鈰�d蟿=\underset{s}{鈭�}c\left(da脳v\right)\text{鈥夆赌�}\\ c\underset{v}{鈭�}\text{鈥夆赌�}\left(鈭�脳v\right)鈰�d蟿=鈭�\underset{s}{鈭�}c\left(v脳da\right)\text{鈥夆赌�}\end{array}$

Solve further as,

$\begin{array}{c}c\underset{v}{鈭�}\text{鈥夆赌�}\left(鈭�脳v\right)鈰�d蟿=鈭�c\underset{s}{鈭�}\left(v脳da\right)\text{鈥夆赌�}\\ \underset{}{鈭�}\text{鈥夆赌�}\left(鈭�脳v\right)鈰�d蟿=鈭�\underset{}{鈭�}\left(v脳da\right)\text{鈥夆赌�}\end{array}$

Thus, the result $\underset{}{鈭�}\text{鈥夆赌�}\left(鈭�脳v\right)鈰�d蟿=鈭�\underset{}{鈭�}\left(v脳da\right)\text{鈥夆赌�}$has been shown.

Let a function V is defined as, $v=T鈭�U$and the divergence theorem is defined as $\underset{v}{鈭�}\text{鈥�}鈭�v鈰�d蟿=\underset{s}{鈭�}v\text{鈥�}da\text{鈥�}$.

Substitute for into $\underset{v}{鈭�}\text{鈥�}鈭�v鈰�d蟿=\underset{s}{鈭�}v\text{鈥�}da\text{鈥�}$.

$\underset{v}{鈭�}\text{鈥�}鈭�\left(T鈭�U\right)鈰�d蟿=\underset{s}{鈭�}\left(T鈭�U\right)\text{鈥�}da\text{鈥�}$鈥︹�� (4)

Apply the product rule $鈭�\left(fA\right)=f\left(鈭�鈰�A\right)+A鈰�\left(鈭�f\right)$ into equation (4)

$\begin{array}{c}鈭�\text{鈥夆赌�}\left(T鈭�鈰�鈭�U鈭�鈭�U鈰�鈭�T\right)鈰�d蟿=鈭�\left(T鈭�U\right)\text{鈥�}da\text{鈥�}\\ 鈭�T{鈭�}^{2}U+鈭�U鈰�鈭�T=鈭�\left(T鈭�U\right)\text{鈥�}da\text{鈥�}\end{array}$

Thus, the result $鈭�T{鈭�}^{2}U+鈭�U鈰�鈭�T=鈭�\left(T鈭�U\right)\text{鈥�}da\text{鈥�}$has been shown.

Swap the variables $T鈫�U$in the result of part (c)$鈭�T{鈭�}^{2}U+鈭�U鈰�鈭�T=鈭�\left(T鈭�U\right)\text{鈥�}da\text{鈥�}$, as shown below:

$鈭�U{鈭�}^{2}T+鈭�T鈰�鈭�U=鈭�\left(U鈭�T\right)\text{鈥�}da\text{鈥�}$

Subtract the resulting equation from $鈭�T{鈭�}^{2}U+鈭�U鈰�鈭�T=鈭�\left(T鈭�U\right)\text{鈥�}da\text{鈥�}$,as,

$\begin{array}{c}鈭�\left(T{鈭�}^{2}U+鈭�U鈰�鈭�T\right)鈭�\left(U{鈭�}^{2}T+鈭�T鈰�鈭�U\right)d蟿=鈭�\left(T鈭�U\text{鈥夆赌�}鈭�U鈭�T\right)\text{鈥�}da\text{鈥�}\\ 鈭�\left(T{鈭�}^{2}U鈭�U{鈭�}^{2}T\right)d蟿=鈭�\left(T鈭�U\text{鈥夆赌�}鈭�U鈭�T\right)\text{鈥�}da\text{鈥�}\end{array}$

Thus, the result $鈭�\left(T{鈭�}^{2}U鈭�U{鈭�}^{2}T\right)d蟿=鈭�\left(T鈭�U\text{鈥夆赌�}鈭�U鈭�T\right)\text{鈥�}da\text{鈥�}$has been shown.

Sokes theorem is defined as $鈭�\left(鈭�脳v\right)鈰�da=鈭�v鈰�dl$.Substitute $cT$for v into $鈭�\left(鈭�脳v\right)鈰�da=鈭�v鈰�dl$as follows:

$鈭�\left(鈭�脳\left(cT\right)\right)鈰�da=鈭�\left(cT\right)鈰�dl$ 鈥︹�︹��.. (5)

Apply the product rule (ii) ,$鈭�脳\left(fA\right)=f\left(鈭�脳A\right)鈭�A脳\left(鈭�f\right)$in equation (5),

$\begin{array}{c}鈭�\left(鈭�脳\left(cT\right)\right)鈰�da=鈭�cT鈰�dl\\ 鈭�\left(T\left(鈭�鈰�脳c\right)鈭�c脳\left(鈭�T\right)\right)鈰�da=鈭�cT鈰�dl\end{array}$鈥︹�︹��.. (6)

As is a constant, so its curl is 0, that is $鈭�脳c=0$.

Substitute 0 for $鈭�脳c$into equation (6)

$\begin{array}{l}鈭�\left(T\left(0\right)鈭�c脳\left(鈭�T\right)\right)鈰�da=鈭�cT鈰�dl\\ 鈭�鈭�\left(c脳\left(鈭�T\right)\right)鈰�da=鈭�cT鈰�da\\ c鈭�\left(鈭�T脳da\right)=鈭�c鈭�T鈰�dl\\ 鈭�鈭�T脳da=鈭�鈭�T鈰�dl\end{array}$

Thus, $鈭�鈭�T脳da=鈭�鈭�T鈰�dl$has been shown.