Show that $\overline{\boldsymbol{F}}$ can be expressed as the gradient of a scalar function. Find the work done in moving a particle in this field from (1,2,-4) to (3,3,2) along the straight line joining the points.

$$\overline{F}$ = $\left(x+2y+az\right)i+\left(bx-3y-z\right)j+\left(4x+cy+2z\right)k$$ and $$\overline{r} = x \overline{i}+y\overline{j}+z\overline{k}$$ $$\therefore dr = dx \overline{i}+dy\overline{j}+dz\overline{k}$$

Since $\overline{F}$ is irrotational

Curl $\overline{F}=0$

$\boldsymbol{\therefore }\boldsymbol{\ }\left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ F_x & F_y & F_z \end{array} \right|=0$

$\therefore \ \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ x+2y+az & {bx-3y-z} & 4x+cy+2z \end{array} \right|=0$

$\therefore i[\frac{\partial }{\partial y}\left(4x+cy+2z\right)-\ \frac{\partial }{\partial z}({bx-3y-z})]-j\left[\frac{\partial }{\partial x}\left(4x+cy+2z\right)+\ \frac{\partial }{\partial z}\left(x+2y+az\right)\right]+k[\frac{\partial }{\partial x}\left(bx-3y-z\right)-\ \frac{\partial }{\partial y}(x+2y+az)]$

$\therefore i\left[c-1\right]-j\left[4-a\right]+k\left[b-2\right]=0i+0j+0k$

Comparing co-efficient of i,j,k

$$\therefore \left[c-1\right]=0$$ $$\therefore \left[4-a\right]=0$$ $$\therefore \left[b-2\right]=0$$ $$\therefore c=1\ ,\ a=4\ ,\ b=2$$ $$\therefore \ \overline{F} = \left(x+2y+az\right)i+\left(bx-3y-z\right)j+\left(4x+cy+2z\right)k$$ $$\because \overline{F}\ is \ irrotational \ , there \ exists \ a \ scalar \ potential \ of \ \overline{F} \ such \ that \ \overline{F}=\ \mathrm{\nabla }\oint $$ $$\therefore \left(x+2y+az\right)i+\left(bx-3y-z\right)j+\left(4x+cy+2z\right)k = i\frac{\partial \oint{}}{\partial x} + j\frac{\partial \oint{}}{\partial y}+k\frac{\partial \oint{}}{\partial z}$$ Comparing co-eeficient of i,j,k we get, $$\frac{\partial \int{}}{\partial x}=\ \left(x+2y+az\right)$$ $$\frac{\partial \int{}}{\partial y}=\ \left(bx-3y-z\right)$$ $$\frac{\partial \int{}}{\partial z}=\ \left(4x+cy+2z\right)$$

Integrating we get ,

$\oint{=(\frac{x^2}{2}}+2yx+4zx)+\left(2xy-\frac{3y^2}{2}-zy\right)+(4zx-2y+\frac{{2z}^2}{2}$) + c

As the same term is twice, it is written only once in this type of integration.

$\therefore \ \int{=\frac{x^2}{2}}+2yx+4zx+2xy-\frac{3y^2}{2}-zy+4zx-2y+z^2+$c

$\therefore $ Scalar potential of $\overline{F}=\ \int{=\ 1/2[}x^2+4yx+8zx-3y^2-2zy+{2z}^2+$c

Now work done in moving a particle in this field = $\int_C{\overline{F}.\overline{dr}}$

$=\int_C{(\left(x+2y+4z\right)}\ dx+\left(2x-3y-z\right)dy+\left(4x+y+2z\right)dz$

Integrating we get ,

$=[x^2+4yx+8zx-3y^2-2zy+{2z}^2]^{(3,3,2)}_{(1,2,-4)}$

$=[9/2 +2*3*3 +4*2*3 -3 * 9/2 -3*2*-4] -- [1/2+2*2*1 + 4*(-4)*1-3* 2 - 2*(-4) +16] =24.5$

$\boldsymbol{\therefore }\boldsymbol{Work}\boldsymbol{\ }\boldsymbol{done}\boldsymbol{\ }\boldsymbol{in}\boldsymbol{\ }\boldsymbol{moving}\boldsymbol{\ }\boldsymbol{a}\boldsymbol{\ }\boldsymbol{particle}\boldsymbol{=}\boldsymbol{24}.\boldsymbol{5}$