Gradient, Divergence and Curl | IIT JAM 2014 | Problem 5

If $f(x,y,z)=x^2y+y^2z+z^2x,\quad \forall (x,y,z) \in \mathbb R$ and $\nabla=(\frac{\partial}{\partial x}\hat{i}+ \frac{\partial}{\partial y}\hat{j}+ \frac{\partial}{\partial z}\hat{k} )$ then the value of $\nabla.(\nabla \times \nabla f)+\nabla.(\nabla f)$ at $(1,1,1)$

Scalar Point Function : is a function which assigns a point$(x,y,z) \in \mathbb R^3$ to a scalar. Here $f$ is a scalar point function.

$\nabla=(\frac{\partial}{\partial x}\hat{i}+ \frac{\partial}{\partial y}\hat{j}+ \frac{\partial}{\partial z}\hat{k} ) $

Gradient of a function :($\nabla f) = (\frac{\partial f}{\partial x}\hat{i}+ \frac{\partial f}{\partial y}\hat{j}+ \frac{\partial f}{\partial z}\hat{k} ) $

Divergence of a function ($\nabla.\vec F$) =$ (\frac{\partial}{\partial x}\hat{i}+ \frac{\partial}{\partial y}\hat{j}+ \frac{\partial}{\partial z}\hat{k} ).\vec F $

Curl of a function ($\nabla\times \vec F$) =$ (\frac{\partial}{\partial x}\hat{i}+ \frac{\partial}{\partial y}\hat{j}+ \frac{\partial}{\partial z}\hat{k} )\times \vec F $

Now, $\nabla f = (2xy+z^2) \hat{i}+(2yz+x^2)\hat{j}+(2zx+y^2)\hat{k}$

Therefore $\nabla . (\nabla f)=\frac{\partial}{\partial x}(2xy+z^2)+ \frac{\partial}{\partial y}(2yz+z^2) + \frac{\partial}{\partial z}(2zx+y^2)$

$\quad= 2x+2y+2z $

Again,

$\nabla \times \nabla f = \begin{vmatrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xy+z^2 & 2yz+x^2 & 2zx+y^2\end{vmatrix}= \vec 0$

Therefore $\nabla. (\nabla \times \nabla f)= 0$

then, $\nabla.(\nabla \times \nabla f)+\nabla.(\nabla f) = 2(x+y+z) \bigg|_{(1,1,1)}=6$