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

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

Let the vector be defined as$v=xyi+2yzj+3zxk$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:

$\nabla ·v=(\frac{\partial }{\partial x}i+\frac{\partial }{\partial y}j+\frac{\partial }{\partial z}k).(xyi+2yzj+3xyzk)$

$=y+2z+3x$

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

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

$={\int }_0^2{\int }_0^2(\frac{3}{2}x4+2zx2+yx2)dydz$

$=$${\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$+4z)$²₀

$=(24+4x4+4z)-\left(0\right)$

role="math" localid="1650075090878" $=48$

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

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

$={\int }_s-2ydydz$

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-2ydydz$can be written as:

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

$={\int }_0^{2}dz(-2y$² $/y)$^2-z₀

_{$={\int }_0^{2}dz(-y$}²$)$^2-z₀

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

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

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

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

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

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

$={\int }_0^2{\int }_0^2-3ydxdy$

The differential length vector is given by $\overrightarrow{dl}$$dxi+dyj+dzk$. The right part of the strokes theorem is calculated as:

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

role="math" localid="1650083307191" $=\int \left(xy\right)dx+\left(2yz\right)dy+\left(3zx\right)dz$

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 )²₀

$=-(8-\frac{16}{3})$

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

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).

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.