Q61P - Introduction to Electrodynamics

Question

Although the gradient, divergence, and curl theorems are the fundamental integral theorems of vector calculus, it is possible to derive a number of corollaries from them. Show that:

(a) $\underset{v}{\int }\left(\nabla T\right)d\tau =\underset{s}{\oint }Tda$. [Hint: Let v = cT, where c is a constant, in the divergence theorem; use the product rules.]

(b) $\underset{v}{\int }\left(\nabla \times v\right)d\tau =\underset{s}{\oint }v\times da$. [Hint: Replace v by (v x c) in the divergence

theorem.]

(c) $\underset{v}{\int }\left(T{\nabla }^{2}U+\nabla T\cdot \nabla U\right)d\tau =\underset{s}{\oint }T\nabla U\cdot da$ . [Hint: Let  in the

divergence theorem.]

(d) $\underset{v}{\int }\left(U{\nabla }^{2}T+\nabla U\cdot \nabla V\right)d\tau =\underset{s}{\oint }U\nabla T\cdot da$. [Comment: This is sometimes

called Green's second identity; it follows from (c), which is known as

Green's identity.]

(e)  $\underset{S}{\int }\nabla T\times da=\underset{P}{\oint }T\cdot dl$ [Hint: Let v = cT in Stokes' theorem.]

Step-by-Step Solution

Answer

1. The result, $\int \nabla T\cdot d\tau =\oint T\cdot da$ , has been shown.
2. The result $\underset{}{\int }\left(\nabla \times v\right)\cdot d\tau =-\underset{}{\int }\left(v\times da\right)$ has been shown.
3. The result $\int T{\nabla }^{2}U+\nabla U\cdot \nabla T=\int \left(T\nabla U\right)da$ has been shown.
4. The result $\int U{\nabla }^{2}T+\nabla T\cdot \nabla U=\int \left(U\nabla T\right)da$ has been shown.
5. The result $\int \nabla T\times da=-\oint T\cdot dl$, has been shown.

 

Step 1: Describe the given information

The identities $\int \nabla T\cdot d\tau =\oint T\cdot da$ , $\underset{}{\int }\left(\nabla \times v\right)\cdot d\tau =-\underset{}{\int }\left(v\times da\right)$ , $\int T{\nabla }^{2}U+\nabla U\cdot \nabla T=\int \left(T\nabla U\right)da$ ,$\int U{\nabla }^{2}T+\nabla T\cdot \nabla U=\int \left(U\nabla T\right)da$

and $\int \nabla T\times da=-\oint T\cdot dl$ have to be proved. Here  $T,U,V$ are the vector and c is a constant.

Step 2: Define the Gauss divergence theorem and stokes theorem

According to the Gauss divergence theorem The integral of divergenceof a function $f\left(x,y,z\right)$  over an closed surface area is equal to the surface integral of the function $\int \left(\nabla v\right)\cdot d\tau =\underset{s}{\oint }v\cdot da$ . According to the stokestheorem, 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 $\int \left(\nabla \times v\right)\cdot da=\underset{l}{\oint }v\cdot dl$ .

Step 3: Prove expression in part (a).

The divergence theorem is defined as $\int \left(\nabla v\right)\cdot d\tau =\oint v\cdot da$ . Substitute $cT$  for v into $\int \left(\nabla v\right)\cdot d\tau =\oint v\cdot da$ as follows:

  $\int \left(\nabla \left(cT\right)\right)\cdot d\tau =\oint cT\cdot da$             ……….. (1)

 

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

 

 $\begin{array}{c}\int \left(\nabla \left(cT\right)\right)\cdot d\tau =\oint cT\cdot da\\ \int \left(T\left(\nabla \cdot c\right)+c\cdot \left(\nabla T\right)\right)\cdot d\tau =\oint cT\cdot da\end{array}$             ……….. (2)

 

As c is a constant, so its divergence is 0, that is, $\nabla \cdot c=0$ .

 

Substitute 0 for $\nabla \cdot c$  into equation (2)

$\begin{array}{c}\int \left(T\left(0\right)+c\cdot \left(\nabla T\right)\right)\cdot d\tau =\oint cT\cdot da\\ \int \left(c\cdot \left(\nabla T\right)\right)\cdot d\tau =\oint cT\cdot da\\ c\int \nabla T\cdot d\tau =c\oint T\cdot da\\ \int \nabla T\cdot d\tau =\oint T\cdot da\end{array}$ 

 

Thus, $\int \nabla T\cdot d\tau =\oint T\cdot da$, has been shown.

Step 4: Prove expression in part (b).

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, $\underset{v}{\int }\nabla v\cdot d\tau =\underset{s}{\int }vda$ 

Substitute  $v\times c$ for v into $\underset{v}{\int }\nabla v\cdot d\tau =\underset{s}{\int }vda$.

 $\underset{v}{\int }\nabla \left(v\times c\right)\cdot d\tau =\underset{s}{\int }\left(v\times c\right)da$…… (3)

 

Apply the rule  $\nabla \left(\text{A}\times \text{B}\right)=\text{B}\left(\nabla \times \text{A}\right)-\text{A}\left(\nabla \times \text{B}\right)$ into equation (3)

 $\underset{v}{\int }\left(c\left(\nabla \times v\right)-v\left(\nabla \times c\right)\right)\cdot d\tau =\underset{s}{\int }\left(v\times c\right)da$

As c is a constant, so it’s curl is 0, that is, $\nabla \times c=0$ .

 

Substitute 0 for  $\nabla \times c$ into equation$\underset{v}{\int }\left(c\left(\nabla \times v\right)-v\left(\nabla \times c\right)\right)\cdot d\tau =\underset{s}{\int }\left(v\times c\right)da$ 

 $\begin{array}{l}\underset{v}{\int }\left(c\left(\nabla \times v\right)-v\left(0\right)\right)\cdot d\tau =\underset{s}{\int }\left(v\times c\right)da\\ \underset{v}{\int }c\left(\nabla \times v\right)\cdot d\tau =\underset{s}{\int }c\left(\nabla \times da\right)\\ c\underset{v}{\int }\left(\nabla \times v\right)\cdot d\tau =\underset{s}{\int }c\left(da\times v\right)\\ c\underset{v}{\int }\left(\nabla \times v\right)\cdot d\tau =-\underset{s}{\int }c\left(v\times da\right)\end{array}$

 

Solve further as,

 $\begin{array}{c}c\underset{v}{\int }\left(\nabla \times v\right)\cdot d\tau =-c\underset{s}{\int }\left(v\times da\right)\\ \underset{}{\int }\left(\nabla \times v\right)\cdot d\tau =-\underset{}{\int }\left(v\times da\right)\end{array}$

 

Thus, the result $\underset{}{\int }\left(\nabla \times v\right)\cdot d\tau =-\underset{}{\int }\left(v\times da\right)$ has been shown.

Step 5: Prove expression in part (c)

Let a function  V is defined as, $v=T\nabla U$ and the divergence theorem is defined as $\underset{v}{\int }\nabla v\cdot d\tau =\underset{s}{\int }vda$ .

Substitute   for  into $\underset{v}{\int }\nabla v\cdot d\tau =\underset{s}{\int }vda$ .

$\underset{v}{\int }\nabla \left(T\nabla U\right)\cdot d\tau =\underset{s}{\int }\left(T\nabla U\right)da$ …… (4)

 

Apply the product rule  $\nabla \left(fA\right)=f\left(\nabla \cdot A\right)+A\cdot \left(\nabla f\right)$ into equation (4)

 $\begin{array}{c}\int \left(T\nabla \cdot \nabla U-\nabla U\cdot \nabla T\right)\cdot d\tau =\int \left(T\nabla U\right)da\\ \int T{\nabla }^{2}U+\nabla U\cdot \nabla T=\int \left(T\nabla U\right)da\end{array}$

 

Thus, the result $\int T{\nabla }^{2}U+\nabla U\cdot \nabla T=\int \left(T\nabla U\right)da$ has been shown. 

Step 6: Prove expression in part (d)

Swap the variables $T\leftrightarrow U$ in the result of part (c)  $\int T{\nabla }^{2}U+\nabla U\cdot \nabla T=\int \left(T\nabla U\right)da$, as shown below:

 

 

 $\int U{\nabla }^{2}T+\nabla T\cdot \nabla U=\int \left(U\nabla T\right)da$

Subtract the resulting equation from $\int T{\nabla }^{2}U+\nabla U\cdot \nabla T=\int \left(T\nabla U\right)da$, as,

 $\begin{array}{c}\int \left(T{\nabla }^{2}U+\nabla U\cdot \nabla T\right)-\left(U{\nabla }^{2}T+\nabla T\cdot \nabla U\right)d\tau =\int \left(T\nabla U-U\nabla T\right)da\\ \int \left(T{\nabla }^{2}U-U{\nabla }^{2}T\right)d\tau =\int \left(T\nabla U-U\nabla T\right)da\end{array}$

 

Thus, the result $\int \left(T{\nabla }^{2}U-U{\nabla }^{2}T\right)d\tau =\int \left(T\nabla U-U\nabla T\right)da$ has been shown.

Step 7: Prove expression in part (e)

Sokes theorem is defined as $\int \left(\nabla \times v\right)\cdot da=\oint v\cdot dl$. Substitute $cT$ for v into $\int \left(\nabla \times v\right)\cdot da=\oint v\cdot dl$ as follows:

$\int \left(\nabla \times \left(cT\right)\right)\cdot da=\oint \left(cT\right)\cdot dl$               ……….. (5)

 

Apply the product rule (ii) , $\nabla \times \left(fA\right)=f\left(\nabla \times A\right)-A\times \left(\nabla f\right)$ in equation (5),

 

    $\begin{array}{c}\int \left(\nabla \times \left(cT\right)\right)\cdot da=\oint cT\cdot dl\\ \int \left(T\left(\nabla \cdot \times c\right)-c\times \left(\nabla T\right)\right)\cdot da=\oint cT\cdot dl\end{array}$……….. (6)

 

As   is a constant, so its curl is 0, that is $\nabla \times c=0$.

 

Substitute 0 for  $\nabla \times c$ into equation (6)

 $\begin{array}{l}\int \left(T\left(0\right)-c\times \left(\nabla T\right)\right)\cdot da=\oint cT\cdot dl\\ -\int \left(c\times \left(\nabla T\right)\right)\cdot da=\oint cT\cdot da\\ c\int \left(\nabla T\times da\right)=-c\oint T\cdot dl\\ \int \nabla T\times da=-\oint T\cdot dl\end{array}$

 

Thus, $\int \nabla T\times da=-\oint T\cdot dl$ has been shown.