Find the scalar potential for $\overline{F}$ and the work done in moving an object in this field from $(0, 1, -1)$ to $(\frac{\pi}{2}, -1, 2)$

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

Marks: 8M

Year: Dec 2014

$$\overline{F} = (y^2\ cosx+\ z^3\ )\overline{i} + (2y\ sin\ x-4)\overline{j}+(3x\ z^2+2)\ \overline{k}$$ And $$\overline{r} = x\overline{i\ }+y\ \overline{j} + z\overline{k}$$ $$\boldsymbol{\therefore }\boldsymbol{\ }\overline{dr} = dx\overline{i\ }+dy\ \overline{j} +dz\overline{k}$$

Curl $\overline{F}=\ \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ F_x & F_y & F_z \end{array} \right|$

$\boldsymbol{\therefore }$ Curl $\overline{F}=\ \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ y^2\ cosx+\ z^3 & 2y\ sin\ x-4 & 3x\ z^2+2 \end{array} \right|$

$\boldsymbol{\therefore }$ Curl $\overline{F}=i\ \left[\frac{\partial }{\partial y}\left(3x\ z^2+2\right)-\ \frac{\partial }{\partial z}\left(2y\ sin\ x-4\right)\right]-j\left[\frac{\partial }{\partial x}\left(3x\ z^2+2\right)-\ \frac{\partial }{\partial z}\left(y^2\ cosx+\ z^3\right)\right] +k\left[\frac{\partial }{\partial x}\left(2y\ sin\ x-4\right)-\ \frac{\partial }{\partial y}\left(y^2\ cosx+\ z^3\right)\right]$

$\boldsymbol{\therefore }$ Curl $\overline{F}=i\left(0-0\right)-j\left(3\ z^2-\ 3\ z^2\right)+k\left(2ycosx-2ycosx\right) = 0$

$\boldsymbol{\therefore }\overline{F}$ is a conservative field.

There exists a scalar potential $\mathrm{\oint}$ of $\overline{F}$ such that $\overline{F}=\ \mathrm{\nabla }\oint $

$\boldsymbol{\therefore }$ $\left(y^2\ cosx+\ z^3\right)\ i+\ \left(2y\ sin\ x-4\right)\ j+\ \left(3x\ z^2+2\right)\ k=i\frac{\partial \oint}{\partial x}$ + j$\frac{\partial \oint \ }{\partial y}+k\frac{\partial \oint \ }{\partial z}$

Comparing both the sides , we have

$\frac{\partial \oint }{\partial x}=\ \left(y^2\ cosx+\ z^3\right)\ ,\ \frac{\partial \oint \ }{\partial y}=\ \left(2y\ sin\ x-4\right)\ ,\ \frac{\partial \oint \ }{\partial z}=\ \left(3x\ z^2+2\right)$

Now , $d\oint{=\ }\frac{\partial \oint }{\partial x}\ dx+\ \frac{\partial \oint }{\partial y}\ dy+\ \frac{\partial \oint }{\partial z}\ dz$

$\boldsymbol{\therefore } d\oint{=\ }\ \left(y^2\ cosx+\ z^3\right)\ dx+\left(2y\ sin\ x-4\right)\ dy+\ \left(3x\ z^2+2\right)dz$

Now integrating w.r.t x,y,z variables with treating y and z as constant for x and similarly for y and z.

$\boldsymbol{\therefore }\oint{=(}y^2sinx+\ z^3x\ )+\left(\frac{2y^2}{2}sinx-4y\right)+\left(\frac{3xz^3}{3}+2z\right)+c \\ = y^2sinx+\ z^3x+y^2sinx-4y+\ xz^3+2z$

As the same term is twice it is written only once

$ = y^2sinx+\ z^3x+y^2sinx-4y+\ xz^3+2z+c$

$\boldsymbol{\therefore }\boldsymbol{Scalar}\boldsymbol{\ }\boldsymbol{potential}\boldsymbol{\ }\boldsymbol{of}\boldsymbol{\ }\overline{F}=\ \int{=(}y^2sinx+\ z^3x+y^2sinx-4y+\ xz^3+2z+c)$

Work done = $\int_C{\overline{F}.\overline{dr}}$

$= \int^{\frac{\mathit{\Pi}}{2},\ -1,2}_{0,1,-1}{\left(y^2\ cosx+\ z^3\right)\ dx+\left(2y\ sin\ x-4\right)\ dy+\ \left(3x\ z^2+2\right)dz}$

$\boldsymbol{\therefore }\boldsymbol{\ }$Work done = ${\boldsymbol{[}y^2sinx+z^3x-4y+2z]\ }^{\boldsymbol{(}\frac{\mathit{\Pi}}{2},\ -1,2)}_{\boldsymbol{(}\boldsymbol{0},\boldsymbol{1}\boldsymbol{,-}\boldsymbol{1}\boldsymbol{)}}$

Work done = [(-1) $sin\frac{\mathit{\Pi}}{2}$ + $(2)^3*\ \frac{\mathit{\Pi}}{2}-4\left(-1\right)+2*2]-- [0+0-4(1) +2(-1)] \\ = 15 +4\ \mathit{\Pi}$

$\therefore $ Scalar potential of $\overline{F}=\ \oint{=\left(y^2sinx+\ z^3x+y^2sinx-4y+\ xz^3+2z+c\right)=\ }15 +4 \boldsymbol{\ }\boldsymbol{\mathit{\Pi}}$