Q35P - Introduction to Electrodynamics

Question

Question:  Check Corollary 1 by using the same function and boundary line as in Ex. 1.11, but integrating over the five faces of the cube in Fig. 1.35. The back of the cube is open.

Step-by-Step Solution

Answer

The corollary 1 says that $\int (\nabla \times V)·da$ depends on the boundary line of the open surface but not on the surface itself. 

Step 1: Define the Gauss divergence theorem

The integral of derivative of a function  $\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{)}$ over an open surface area is equal to the volume integral of the function, $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{\times }\mathit{V}\mathbf{)}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}{\mathbf{\int }}_{\mathbf{s}}\mathit{v}\mathbf{·}\mathit{d}\mathit{s}\mathbf{.}$ . 

The right side of the gauss divergence theorem is the surface integral, that is, ${\int }_{s}v·da$

 

The diagram of the closed volume possessed by cube of 2 units is shown below:

Step: 2 Compute the left side of gauss divergence theorem

The corollary1 says that the integral vector $\int (\nabla \times v)·da$ depends on the boundary line of any open surface used. The given function is $v=(2xz+3y^2)i+\left(4y{z}^2\right)$ and the del operate is defined as $\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$ . 

 

The curl of vector v is computed a s follows:

$$\begin{gathered} \nabla \times v= \left| i j k \\ \frac{\partial }{\partial x} \frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{y}} \frac{\mathbf{\partial }}{\mathbf{\partial }\mathit{z}} \\ 0 2xz+3y^2 4y{z}^2\right| \\ =\left[\left(\frac{\partial }{\partial y}\left(4y{z}^2\right)-\frac{\partial }{\partial z}\left(2xz+3y^2\right)\right)i-{\left(\frac{\partial }{\partial x}\left(4y{z}^2\right)-\frac{\partial }{\partial z}\left(0\right)\right)}^{°}j \\ +\left(\frac{\partial }{\partial x}(2xz+3y^2)-\frac{\partial }{\partial z}\left(0\right)\right)k\right] \\ =(4z^2-2x)i+\left(2z\right)j \end{gathered}$$

 

 

Step 3: Compute the surface integral of the vector v along path (i)

The path (i) goes along the plane yz, where y and z from 0 to 1 and x = 1 . The area vector becomes $\overrightarrow{da}=dydz$ .

The surface integral of vector $\stackrel{\rightharpoonup }{v}$ , along the path (i) is computed as: 

$$\begin{gathered} \int (\nabla \times v)·da={\int }_0^1{\int }_0^1\left(\left(4z^2-2x\right)i+\left(2z\right)j\right)dydz i \\ ={\int }_0^1{\int }_0^{1}4z^2-2x dydz \\ =\left(\frac{4z^3}{3}\right)-{\left(2\left(1\right)y\right)}_0^1 \\ =2x{\left(\frac{y^2}{2}\right)}_0^2 \end{gathered}$$ 

 Solve further as,

$$\begin{gathered} \int (\nabla \times v)·da=\frac{4}{3}-2 \\ =-\frac{2}{3} \end{gathered}$$

Step 4: Compute the surface integral of the vector v along path (ii)

The plath (ii) goes along the plane xy, where x and y from 0 to 1 and z = 0 . The area vector becomes $\overrightarrow{da}=dydz$ .

 

The surface integral of vector $\stackrel{\rightharpoonup }{v}$ , along the path (ii) is computed as:

 $$\begin{gathered} \int (\nabla \times v)·da=-{\int }_0^2{\int }_0^2{\int }_0^2\left(\left(4z^2-2x\right)i+\left(2z\right)j\right)dydz i \\ =-{\int }_0^2{\int }_0^2{\int }_0^{2}2\left(0\right) \\ =0 \end{gathered}$$

Step 5: Compute the surface integral of the vector v along path (iii)

The path (iii) goes along the plane xz, where x and z from 0 to 1 and y = 1 . The area vector becomes $\overrightarrow{da}=dydz$ .

 

The surface integral of vector $\stackrel{\rightharpoonup }{v}$, along the path (iii) is computed as:

 $$\begin{gathered} \int (\nabla \times v)·da={\int }_0^1{\int }_0^1\left(\left(4z^2-2x\right)i+\left(2z\right)j\right)dxdz i \\ ={\int }_0^1{\int }_0^1\left(0\right) \\ =0 \end{gathered}$$ 

Step 6: Compute the surface integral of the vector v along path (iv)

The path (iv) goes along the plane xz, where   and   from 0 to 1 and y = 0 . The area vector becomes $\stackrel{\rightharpoonup }{da}=-dxdz$ .

 

The surface integral of vector $\stackrel{\rightharpoonup }{v}$ , along the path (iii) is computed as: 

 $$\begin{gathered} \int (\nabla \times v)·da=-{\int }_0^1{\int }_0^1\left(\left(4z^2-2x\right)i+\left(2z\right)j\right)dxdyz i \\ =-{\int }_0^1{\int }_0^1\left(0\right) \\ =0 \end{gathered}$$

 

Step 7: Compute the surface integral of the vector v along path (v)

The path (iv) goes along the plane xy, where x and y from 0 to 1 and z = 1 . The area vector becomes $\stackrel{\rightharpoonup }{da}=dxdz$ .

 

The surface integral of vector $\stackrel{\rightharpoonup }{v}$ , along the path (v) is computed as: 

 $$\begin{gathered} \int (\nabla \times v)·da=-{\int }_0^1{\int }_0^1\left(\left(4z^2-2x\right)i+\left(2z\right)j\right)dxdyz i \\ =-{\int }_0^1{\int }_0^{1}2{\left(x\right)}_0^1{\left(y\right)}_0^1 =2 \end{gathered}$$

 

Therefore, the total value of surface integral is sum of integral of all surfaces, as

 

 $$\begin{gathered} \int (\nabla \times v)·da=-\frac{2}{3}+0+0+2 \\ =\frac{4}{3} \end{gathered}$$

 

Thus, the corollary 1 says that $\int (\nabla \times v)·da$ depends on the boundary line of the open surface but not on the surface itself