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

Marks: 5M

Year: May 2014

$\because$ $\overline{\ F}$ is irroational

$\nabla \ X\ \overline{\ F}=0$

$\therefore \left| \begin{array}{ccc} \overline{i} & \overline{j} & \overline{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ axy+bz^3 & 3x^2-cz & 3xz^2-y \end{array} \right| = 0$

$\therefore \overline{i}\ [\frac{\partial }{\partial y}(3xz^2-y)-\ \frac{\partial }{\partial z}(3x^2-cz)]-\overline{j}(\ \frac{\partial }{\partial x}(3xz^2-y)-\ \frac{\partial }{\partial z}(axy+bz^3)] +\ \overline{k}\left[\frac{\partial }{\partial x}\left(3x^2-cz\right)-\frac{\partial }{\partial y}\left(axy+bz^3\right)\right]=0$

$\therefore \overline{i}\ \left[-1-c\right)-\ \overline{j}\left[3z^2-3bz^2\right]+\ \overline{k}\left[6x-ax\right]$

$\therefore \overline{i}\left[c-1\right]-\ \overline{j}\left[3z^2-3bz^2\right]+\overline{k}\left[6x-ax\right]=0 \overline{i}+0\overline{j}+0\overline{k}$

Comparing like terms on both the sides,

$\therefore c-1=0 \\ \therefore c=1 \\ \therefore 3z^2-3bz^2=0 \\ \therefore 3z^2 \left(1-b\right)=0 \\ \therefore b = 1 \\ \therefore 6x-ax=0 \\ \therefore x\left(6-a\right)=0 \\ \therefore a=6 ,b= 1, c = 1$