Q62P - Introduction to Electrodynamics

Question

The integral

                                       $\mathit{a}\mathbf{=}{\mathbf{\int }}_{\mathbf{s}}\mathit{d}\mathit{a}$

is sometimes called the vector area of the surface S. If S happens to be flat, then lal is the ordinary (scalar) area, obviously.

(a) Find the vector area of a hemispherical bowl of radius R.

(b) Show that a= 0 for any closed surface. [Hint: Use Prob. 1.6la.]

(c) Show that a is the same for all surfaces sharing the same boundary.

(d) Show that

 

where the integral is around the boundary line. [Hint: One way to do it is to draw the cone subtended by the loop at the origin. Divide the conical surface up into infinitesimal triangular wedges, each with vertex at the origin and opposite side dl, and exploit the geometrical interpretation of the cross product (Fig. 1.8).]

(e) Show that 

                                    $\mathbf{\oint }\mathbf{c}\mathbf{\cdot }\mathbf{r}\mathbf{=}\mathbf{a}\mathbf{\times }\mathbf{c}$

for any constant vector c. [Hint: Let T = c · r in Prob. 1.61e.] (

Step-by-Step Solution

Answer

(a)For hemispherical bowl, the area is $\mathrm{a}={\mathrm{\pi R}}^2 \hat{\mathrm{z}}$ .

(b)It is proved that for a sphere bowl ${\int }_{\mathrm{s}}\mathrm{da}=0$ .

(c)The area is same for the surfaces sharing same boundaries only as the difference between the areas is not equal to zero.

(d)For a conical surface the total area is obtained as$\mathrm{a}=\frac{1}{2}\oint \mathrm{r}\times \mathrm{dl}$  .

(e)The result  ${\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl}=\mathrm{a}\times \mathrm{c}$, has been shown.

Step 1: Describe the given information

It is given that the integral $\mathrm{a}={\int }_{\mathrm{s}}\mathrm{da}$, is called as the vector area of the surface S., where S is a flat area and, [a] is a ordinary (scalar) area.

Step 2: Define the Gauss divergence theorem and stokes theorem

According to the Gauss divergence theorem. The integral of divergence of a function $\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{ }$  over an closed surface area is equal to the surface integral of the function $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{v}\mathbf{)}\mathbf{\cdot }\mathbf{d\tau }\mathbf{=}{\mathbf{\oint }}_{\mathbf{s}}\mathbf{v}\mathbf{\cdot }\mathbf{da}$ . According to the stokes theorem the integral of divergence of a function $\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\mathbf{z}\mathbf{)}\mathbf{ }$over an open surface area is equal to the line integral of the function $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathbf{v}\mathbf{)}\mathbf{\cdot }\mathit{d}\mathit{a}\mathbf{=}{\mathbf{\oint }}_{\mathbf{l}}\mathbf{v}\mathbf{\cdot }\mathit{d}\mathit{l}$.

Step 3: Find the vector area in part (a)

The differential elemental area in a sphere is  $\mathrm{da}={\mathrm{R}}^2\mathrm{sin\theta } \mathrm{d\theta d\varphi }\hat{\mathrm{r}}$Thus, on integrating the differential area over northern hemisphere, we obtain as

 $$\begin{gathered} \mathrm{a}={\int }_{\mathrm{s}}\mathrm{da} \\ ={\int }_{\mathrm{s}}{\mathrm{R}}^2\mathrm{sin\theta cos\theta } \mathrm{d\theta d\varphi }\hat{\mathrm{z}} \\ =2{\mathrm{\pi R}}^2\hat{\mathrm{z}}\underset{0}{\overset{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\mathrm{sin\theta cos\theta d\theta } \\ =2\mathrm{\pi }{\mathrm{R}}^2\hat{\mathrm{z}}{\left(\frac{{\sin }^2 \mathrm{\theta }}{2}\right)}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

 

Solve further as, 

 $$\begin{gathered} \mathrm{a}=2\mathrm{\pi }{\mathrm{R}}^2\hat{\mathrm{z}}{\left(\frac{{\sin }^2\left(\mathrm{\pi }/2\right)-{\sin }^2\left(0\right)}{2}\right)}_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ =2\mathrm{\pi }{\mathrm{R}}^2\hat{\mathrm{z}}\left(\frac{1}{2}\right) \\ =\mathrm{\pi }{\mathrm{R}}^2\hat{\mathrm{z}} \end{gathered}$$

 

Thus, for another hemisphere, the area is $a=\pi R^2\hat{z}$.

Step 4: Prove expression in part (b)

It is known that the total surface function of any closed surface is $T=1$.

 

Substitute 1 for T  into gauss divergence theorem ${\int }_{\mathrm{v}}\nabla \mathrm{v}\cdot \mathrm{d\tau }={\int }_{\mathrm{s}}\mathrm{vda}$  .

 $$\begin{gathered} {\int }_{\mathrm{v}}\nabla \left(1\right).\mathrm{d\tau }={\int }_{\mathrm{s}}\left(1\right)\mathrm{da} \\ 0={\int }_{\mathrm{s}}\mathrm{da} \\ {\int }_{\mathrm{s}}\mathrm{da}=0 \end{gathered}$$                    …… (3)

 

Thus, it is proved that for a sphere bowl ${\int }_{\mathrm{s}}\mathrm{da}=0$ .

Step 5: Prove expression in part (c)

For the closed surfaces sharing same boundary, having different area, say, $a_1\ne a_2$, such that  $a_1-a_2\ne 0$. The total integral of the difference of the areas can be obtained as,

 

 $$\begin{gathered} {\int }_{s}d\left(a_1-a_2\right)=a_1-a_2 \\ \ne 0 \end{gathered}$$

 

Thus, the area is same for the surfaces sharing same boundaries only as the difference between the areas is not equal to zero.

Step 6: Prove expression in part (d)

Let us consider that conical surface is broke down into infinitesimal triangular shapes. The differential area of the triangular element is written as$da=\frac{1}{2} (r\times dl)$ , where r  is the base of the triangle and dl is the height of the triangle.

 

From the expression $r\times dl$, it can be inferred that it also represents the area of a parallelogram, having direction perpendicular to the surface of parallelogram.

 

Thus, for a conical the total area can be written as

 $$\begin{gathered} a={\int }_{s}da \\ =\frac{1}{2}\oint r\times dl \end{gathered}$$

 

Therefore, for a conical surface the total area is obtained as $a=\frac{1}{2}\oint r\times dl$ .

Step 7: Prove expression in part (e)

(e)

The Stokes theorem is defined as $\int (\nabla \mathrm{T})\times \mathrm{da}={\oint }_{\mathrm{P}}\mathrm{T}\cdot \mathrm{dl}$ . Substitute $c\cdot r$   for T into $\int (\nabla \mathrm{T})\times \mathrm{da}={\oint }_{\mathrm{P}}\mathrm{T}\cdot \mathrm{dl}$  as follows:

 $\int (\nabla (\mathrm{c}\cdot \mathrm{r}\left)\right)\times \mathrm{da}={\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl}$

 

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

 $\int (c\times (\nabla \times r)+(c\cdot \nabla \left)r\right)\times da={\oint }_P(c\cdot r)\cdot dl$

As  r is a position vector, so its curl is 0, that is,$\nabla \times r=0$  .

 

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

 $$\begin{gathered} \int \left(\mathrm{C}\times \left(0\right)+\left( \mathrm{c}.\nabla \right)\mathrm{r}\right)\times \mathrm{da}={\oint }_{\mathrm{P}}\left(\mathrm{c}.\mathrm{r}\right).\mathrm{dl} \\ \int \left( \mathrm{c}.\nabla \right)\mathrm{r}\times \mathrm{da}={\oint }_{\mathrm{P}}\left(\mathrm{c}.\mathrm{r}\right).\mathrm{dl} \end{gathered}$$

 

Substitute$c_x\frac{\partial }{\partial x}+c_y\frac{\partial }{\partial y}+c_z\frac{\partial }{\partial x}$   for  $c\cdot \nabla$,$\mathrm{xi}+\mathrm{yj}+\mathrm{zk}$ for  into equation (3)

 $$\begin{gathered} \int ({\mathrm{c}}_{\mathrm{x}}\partial /\partial \mathrm{x}+{\mathrm{c}}_{\mathrm{y}}\partial /\partial \mathrm{y}+{\mathrm{c}}_{\mathrm{z}}\partial /\partial \mathrm{x})(\mathrm{xi}+\mathrm{yj}+\mathrm{zk})\times \mathrm{da}={\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl} \\ \int ({\mathrm{c}}_{\mathrm{x}}\mathrm{i}+{\mathrm{c}}_{\mathrm{y}}\mathrm{j}+{\mathrm{c}}_{\mathrm{z}}\mathrm{k})\times \mathrm{da}={\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl} \\ -{\int }_{\mathrm{s}}\left(\mathrm{c}\right)\times \mathrm{da}={\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl} \\ -\mathrm{c}\times {\int }_{\mathrm{s}}\mathrm{da}={\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl} \end{gathered}$$

 

 Simplify further as,

 $$\begin{gathered} -\mathrm{c}\times {\int }_{\mathrm{s}}\mathrm{da}={\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl} \\ -\mathrm{c}\times \mathrm{a}={\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl} \\ {\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl}=-\mathrm{c}\times \mathrm{a} \\ {\oint }_{\mathrm{P}}(\mathrm{c}\cdot \mathrm{r})\cdot \mathrm{dl}=\mathrm{a}\times \mathrm{c} \end{gathered}$$

 

Thus, ${\oint }_P(c\cdot r)\cdot dl=a\times c$ , has been shown.