VC3.23

From Exampleproblems

a).Here the curl f = $\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \sin y +z & x\cos y -z & x-y \end{vmatrix}\,$

= $[\frac{\partial}{\partial y}(x-y)-\frac{\partial}{\partial z}(x\cos y - z)]i-[\frac{\partial}{\partial x}(x-y)-\frac{\partial}{\partial z}(\sin y+z)]j+[\frac{\partial}{\partial x}(x\cos y-z)-\frac{\partial}{\partial z}(\sin y+z)]k\,$

= $(-1+1)i-(1-1)j+(\cos y-\cos y)k=0\,$

Since curl f is equal to zero,it follows that f is irrotational.

b).f is irrotational vector implies curl f=0.

$\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (x+2y+az) & (bx-3y-z) & (4x+cy+2z)\end{vmatrix}=0\,$

= $[\frac{\partial}{\partial y}(4x+cy+2z)-\frac{\partial}{\partial z}(bx-3y-z)]i-[\frac{\partial}{\partial x}(4x+cy+2z)-\frac{\partial}{\partial z}(x+2y+az)]j+[\frac{\partial}{\partial x}(bx-3y-z)-\frac{\partial}{\partial y}(x+2y+az)]k=0\,$