Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: Dec 2014

$$\overline{\boldsymbol{F}}\boldsymbol{=\ }{\boldsymbol{x}}^{\boldsymbol{2}}\boldsymbol{i}\boldsymbol{+}\boldsymbol{xy}\boldsymbol{\ }\boldsymbol{j}\boldsymbol{+}\boldsymbol{0}\boldsymbol{k}$$ $$\mathrm{\nabla }\mathrm{.}\overline{F}=\ \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \ {\boldsymbol{x}}^{\boldsymbol{2}} & \boldsymbol{xy} & 0 \end{array} \right|$$ $$\therefore \mathrm{\nabla }\mathrm{.}\overline{F}=i\left[\frac{\partial }{\partial y}\ \left(0\right)-\ \frac{\partial }{\partial z}\ xy\ \right]-j\ \left[\ \frac{\partial }{\partial x}\ \left(0\right)-\ \frac{\partial }{\partial z}x^2\right]+k\ \left[\frac{\partial }{\partial x}\left(xy\right)-\ \frac{\partial }{\partial y}\left(x^2\right)\right]$$ $$ = 0 i - 0j + yk$$

The rectangle lies in xy plane .

In xy plane unit normal is $\overline{N}=\ \overline{k}$ and ds = dx.dy

$$therefore \ \overline{\boldsymbol{N}}.\left(\overline{\boldsymbol{\mathrm{\nabla }}\boldsymbol{\ }}\boldsymbol{\ }\boldsymbol{X}\boldsymbol{\ }\overline{\boldsymbol{F}}\right) = \overline{\boldsymbol{k}}\boldsymbol{.\ }\left(\boldsymbol{y}\overline{\boldsymbol{k}}\right)\boldsymbol{=}\boldsymbol{y}$$

$\therefore $ By stokes theorem. $$\int_{\boldsymbol{C}}{\overline{\boldsymbol{F}}.\overline{\boldsymbol{dr}}} = \int_{\boldsymbol{R}}{\int{\overline{\boldsymbol{N}}\boldsymbol{\ }\left(\overline{\boldsymbol{\mathrm{\nabla }}\boldsymbol{\ }}\boldsymbol{\ }\boldsymbol{X}\boldsymbol{\ }\overline{\boldsymbol{F}}\right)\boldsymbol{ds}}}$$ $$\int_{\boldsymbol{C}}{\overline{\boldsymbol{F}}.\overline{\boldsymbol{dr}}} = \int_{\boldsymbol{R}}{\int{\boldsymbol{\ }\boldsymbol{y}\boldsymbol{\ }\boldsymbol{dxdy}}}$$ ![enter image description here][1] Now consider vertical strip This vertical strip will slide from x = 0 to x=a 1) Outer limit x : x=0 to x=a 2) Inner limit y : y=0 to y = b $$\int_C{\overline{F}.\overline{dr}}=\int^a_0{\int^b_0{y\ dx\ dy}}$$ $$ = \int^a_0{{\left[\frac{y^2}{2}\right]}^b_0}\ dx$$ $$ = b^2\ /\ 2 {[x]}^a_0=\boldsymbol{a}{\boldsymbol{b}}^{\boldsymbol{2}}\boldsymbol{/}\boldsymbol{2}$$