Q33P - Introduction to Electrodynamics

Question

Test the divergence theorem for the function $\mathbf{v}\mathbf{=}\mathbf{\left(}\mathbf{xy}\mathbf{\right)}\mathbf{i}\mathbf{+}\mathbf{\left(}\mathbf{2}\mathbf{yz}\mathbf{\right)}\mathbf{j}\mathbf{+}\mathbf{\left(}\mathbf{3}\mathbf{zx}\mathbf{\right)}\mathbf{k}\mathbf{.}$ .Take as your volume the cube shown in Fig. 1.30, with sides of length 2.

                                           

Step-by-Step Solution

Answer

The left and right side of the gauss divergence theorem is equal. Thus the gauss divergence theorem is proved.

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, $\mathbf{\int }\mathbf{(}\mathbf{\nabla }\mathbf{.}\mathit{v}\mathbf{)}\mathit{d}\mathit{l}\mathbf{=}\mathbf{\int }\mathit{v}\mathbf{.}\mathit{d}\mathit{s}\mathbf{.}$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 gauss divergence theorem

Let the vector v  be defined as $v=xy i+2yz j+3zx k$ and the $\nabla$ operator is defined as $\nabla =\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k$ . The divergence of vector v is computed a s follows:

$$\begin{gathered} \nabla .v=\left(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k\right).(xy i+2yz j+3zx k) \\ =y+2z+3x \end{gathered}$$

Now compute the left part of gauss divergence theorem as,

$$\begin{gathered} \int (\nabla .v)dv={\int }_0^2{\int }_0^2{\int }_0^2(3x+2z+y)dxdydz \\ ={\int }_0^2{\int }_0^2\left(\frac{3}{2}\times 4+2z\times 2+y\times 2\right)dydz \\ ={\int }_0^2{\int }_0^2(6+4z+2y)dydz \\ ={\int }_0^2(12+8z+4)dz \end{gathered}$$

Solve further as,

$$\begin{gathered} \int (\nabla .v)dv={(12z+4z^2+4z)}_0^2 \\ =(24+4\times 4+4z)-\left(0\right) \\ =48 .....................\left(1\right) \end{gathered}$$

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 2 and $x=2$ . The area vector becomes $\overrightarrow{da}=dydz.$ .

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

$$\begin{gathered} \int v da={\int }_0^2{\int }_0^{2}xy dydz \\ ={\int }_0^{2}2xy dy \\ =2x{\int }_0^{2}y dy \\ =2x{\left(\frac{y^2}{2}\right)}_0^2 \end{gathered}$$

Solve further as,

 $$\begin{gathered} \int v da=4x \\ =4\left(2\right) \\ =8 \end{gathered}$$

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

The path (iv) goes along the plane y z, where y and z from 0 to 2 and $x=0$ . The area vector becomes $\overrightarrow{da}=-dydz.$ .

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

 

$$\begin{gathered} \int v.da=-{\int }_0^2{\int }_0^{2}xy dydz \\ =\frac{x{y}^2}{2}{\int }_0^{2}dz \\ =0 \end{gathered}$$

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 2 and $y=2$ . The area vector becomes $\overrightarrow{da}=-dxdz$ .

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

$$\begin{gathered} \int v.da=-{\int }_0^2{\int }_0^{2}2yz dxdz \\ ={\int }_0^{2}2yz dz{\int }_0^{2}dx \\ ={\int }_0^{2}4yz dz \\ =y{\left(2z^2\right)}_0^2 \end{gathered}$$

Solve further as,

$$\begin{gathered} \int v.da=4\left(2^2\right) \\ =16 \end{gathered}$$

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

 The path (iv) goes along the plane xz, where  x and z  from 0 to 2 and $y=0$ . The area vector becomes $\overrightarrow{da}=-dxdz$ .

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

$$\begin{gathered} \int v.da=4\left(2^2\right) \\ =0 \end{gathered}$$

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

The path (v) goes along the plane xy, where  x and y  from 0 to 2 and $z=2$ . The area vector becomes $\overrightarrow{da}=dxdy$ .

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

$$\begin{gathered} \int v.da={\int }_0^2{\int }_0^{2}3zx dxdy \\ ={\int }_0^{2}6zx dx \\ =z{\left(3x^2\right)}_0^2 \\ =12z \\ =24 \end{gathered}$$

Step 8: Compute the surface integral of the vector v along path (vi)

The path (vi) goes along the plane xy, where x  and y  from 0 to 2 and $z=0$ . The area vector becomes $\overrightarrow{da}=dxdy$ .

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

$$\begin{gathered} \int v.da=-{\int }_0^2{\int }_0^{2}3zx dxdy \\ =0 \end{gathered}$$

Thus the net value of surface integral through all the paths is computed as follows:

$$\begin{gathered} \int v da= {\int }_{\left(1\right)}v da+{\int }_{\left(2\right)}v da+....+{\int }_{\left(6\right)}v da \\ = 8+0+16+0+24+0 \\ =48 .............\left(2\right) \end{gathered}$$

From equations (1) and (2), the left and right side of the gauss divergence theorem is equal. 

 

Thus the gauss divergence theorem is proved.