1.34P - Introduction to Electrodynamics

Question

Test Stokes' theorem for the function , using the triangular shaded area of Fig. 1.34.

Step-by-Step Solution

Answer

The left and right side gives same result. Hence, strokes theorem is verified. 

Define the Gauss divergence theorem

The integral derivative of a function $f(x,y,z)$ over an open surface area is equal to the volume integral of the function, $\int (\nabla ·v).dl =\oint v.ds.$. The right side of the gauss divergence theorem is the surface integral, that is,$\oint v.da$ The diagram of the triangular path is shown below

Compute the curl of vector v

Let the vector  be defined as$v=xy i+2yz j +3zx k$ and the $\nabla$ operator be 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 as follows:

$\nabla ·v=(\frac{\partial }{\partial x}i +\frac{\partial }{\partial y}j +\frac{\partial }{\partial z}k). (xy i +2yz j + 3xyz k)$

$=y+2z+3x$

Now, compute the left part of gauss divergence theorem as:

$\int (\nabla ·v)dv ={\int }_0^2{\int }_0^2{\int }_0^2 (3x+2z+y) dxdydz$

                      $={\int }_0^{2 }{\int }_0^2(\frac{3}{2}x4 +2z x2 + y x2 ) dydz$

                      $=$ ${\int }_0^2{\int }_0^2(6 + 4z +2y) dydz$

                     $= {\int }_0^2(12+8z +4) dz$

Solve further as: 

$\int (\nabla ·v) dv = (12z +4z$² $+4z)$² ₀ 

                      $= (24+4 x 4 + 4z)-\left(0\right)$

                       $=48$

Compute the left side of strokes theorem

The area vector is given by  $\overrightarrow{da= }$ $dydz i$as the open surface area lies in y-z plane. The left part of the strokes theorem is calculated as:

${\int }_s(\nabla \times v ) . \underset{da}{\to }={\int }_s (-2yi-3zj -xk).(dy dz i )$

                              $= {\int }_s -2ydydz$

The equation of line (i) is $y+z=2.$  . Along this line, z varies 0 to 2 and y varies from 0 to $2-z$. Thus the integral  ${\int }_s(\nabla \times v). \underset{da}{\to }$ $={\int }_{s } -2y dydz$ can be written as:

${\int }_{s } (\nabla \times v) .\underset{da}{\to } ={\int }_0^{2}dz{\int }_0^{2-z} -2y dy$

                                

                                $={\int }_0^2 dz (-2y$²  $/y)$^2-z ₀

                                            _{$={\int }_0^{2}dz (-y$}² $)$^2-z ₀

                                         _{$={\int }_0^{2}dz (-(2 -z)$}²  $- \left(0\right))$

                     Solve further as, 

${\int }_s (\nabla \times v).\underset{da}{\to } ={\int }_0^2 -(4+z$² $-4z)dz$

                               $=-(4z + z$³ $/3 -2z$² $)$²₀

                                 $=-\left(\right(8+ \frac{8}{3}-8 )-(0)$

                                    $=-\frac{8}{3}$

    The surface integral of vector $\underset{v}{\to }$, in the xy plane  is computed as:       

$\int v.da = \int \int (2xzi + (x+2)j + y (z$³  $-3 \left)k\right)(dx dy k)$

               $={\int }_0^2{\int }_0^2 y (z$³ $-3) dxdy$

                $={\int }_0^2{\int }_0^{2}y (0-3 )dx dy$

                   $={\int }_0^2{\int }_0^2-3y dx dy$

                     

Compute the right side of strokes theorem

The differential length vector is given by $\overrightarrow{dl}$ $dx i + dy j + dz k$. The right part of the strokes theorem is calculated as:

$\int v.dl = \int (2xzi +(x+2)j +y(z$³  $-3 \left)k\right)(dxi + dy j + dz k )$

              $= \int \left(xy\right) dx +\left(2yz\right) dy + \left(3zx\right) dz$

              

Along the path (i),$z=x=0$ , thus  $dx=dz=0$and $y=0$. Hence the above integral becomes,$\int v.dl=\int \left(2yz\right)dy.$

Along the path (ii),$x=0$ , $z=2-y$ thus$dx=0$  and $dz=-dy$. Hence the above integral becomes,

$\int v.dl ={\int }_2^{0}2y (2-y)dy$

              $= -{\int }_0^{2}4y-2y$² $dy$

                $=(-2y$² $-\frac{2}{3 }y$³  )² ₀

                $=-(8-\frac{16}{3})$

                 $=-\frac{8}{3}$

Along the path (ii)i, $x=y=0$, thus $dx=dy=0$ and z varies deom 2 to 0.. Hence the integral  $\int v.dl$becomes 0 along path (iii).

Draw a conclusion

The integral of all the three parts are added to give: 

$\int v.dl= 0-\frac{8}{3 } +0$ $=\frac{8}{3}$

Thus the left and right side gives same result. Hence strokes theorem is verified.