Question

$\footnotesize \color{blue} \boxed {\boxed{ \begin{matrix} \rm{If \: f (x,y,z) = {x}^{2}y + {y}^{2} z + {z}^{2}x \: \forall \: (x,y,z) \in\R^3} \\ \rm and \: \nabla = \frac{ \partial}{ \partial x} i + \frac{ \partial}{ \partial y}j + \frac{ \partial}{ \partial z}k \\ \rm then \: the \: of \: \\ \rm \nabla \cdot( \nabla \times \nabla f) + \nabla \cdot(\nabla f) \: at \:(1,1,1) \: is \end{matrix}}}$​

Answer 1

Apologize for not being clean :(

Hope it will help u :)

• for more information I'm adding a few Formulas of Gradient, Divergence and Curl

Answer 2

First, we need to find the curl of the gradient of f:

$\begin{aligned} \nabla f &= \frac{\partial f}{\partial x}i + \frac{\partial f}{\partial y}j + \frac{\partial f}{\partial z}k \\ &= 2xyi + (x^2+2yz)j + 2zk \end{aligned}$

Next, we can find the curl of this vector field:

$\begin{aligned} \nabla \times \nabla f &= \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 2xy & x^2+2yz & 2z \end{vmatrix} \\ &= (2z-2z)i - (0-2y)j + (x^2+2yz-2x)k \\ &= (x^2-2y)i - 2yj + (2yz-2x)k \end{aligned}$

Finally, we can find the divergence of this vector field:

$\begin{aligned} \nabla \cdot (\nabla \times \nabla f) &= \frac{\partial}{\partial x}(x^2-2y) + \frac{\partial}{\partial y}(-2y) + \frac{\partial}{\partial z}(2yz-2x) \\ &= 2x+2z-2 \end{aligned}$

Now, let's find the gradient of f:

$\begin{aligned} \nabla f &= 2xyi + (x^2+2yz)j + 2zk \\ \Rightarrow \nabla \cdot (\nabla f) &= \frac{\partial}{\partial x}(2xy) + \frac{\partial}{\partial y}(x^2+2yz) + \frac{\partial}{\partial z}(2z) \\ &= 2y + 2z \end{aligned}$

Substituting these values into the given expression, we get:

$\begin{aligned} \nabla \cdot(\nabla \times \nabla f) + \nabla \cdot(\nabla f) &= (2x+2z-2) + (2y+2z) \\ &= 2(x+y+2z-1) \end{aligned}$

When we substitute the point (1,1,1) into the expression, we get:

$\begin{aligned} \nabla \cdot(\nabla \times \nabla f) + \nabla \cdot(\nabla f) &= 2(1+1-2) \\ &= 0 \end{aligned}$

So the answer is indeed 0.