# Another mathamatician neede to check this! CURL-GRAD-DIV

## Homework Statement

I have repeated this problem over and over again so either the question is wrong or my likely its my method thats at fault. Please can someone check this for me.

Prove this is true:
$$\nabla\times(\nabla \times F)=-\nabla^{2}F+\nabla(\nabla.F)$$

for $$F=x^{2}z^{3}i$$

## The Attempt at a Solution

Im getting for the LHS:

$$(\nabla \times F)= 3x^{2}z^{3} j$$

$$\nabla\times(\nabla \times F)= -6x^{2}z i + 6xz^{2}k$$

and for the right:

$$-\nabla^{2}F=-2z^{3} i$$

$$(\nabla.F)=2xz^{3}$$

$$\nabla(\nabla.F)=2z^{3}i + 6xz^2k$$

$$-\nabla^{2}F+\nabla(\nabla.F)=6xz^{2} k$$

$$\Rightarrow \nabla\times(\nabla \times F)\neq-\nabla^{2}F+\nabla(\nabla.F)$$
for $$F=x^{2}z^{3}i$$

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andyc10;2201561 and for the right: [tex said:
-\nabla^{2}F=-2z^{3} i[/tex]

$$(\nabla.F)=2xz^{3}$$

$$\nabla(\nabla.F)=2z^{3}i + 6xz^2k$$

$$-\nabla^{2}F+\nabla(\nabla.F)=6xz^{2} k$$

$$\Rightarrow \nabla\times(\nabla \times F)\neq-\nabla^{2}F+\nabla(\nabla.F)$$
for $$F=x^{2}z^{3}i$$
isn't $$\nabla$$.$$\nabla$$ = $$\nabla$$^2= d^2/dx^2+d^2/dy^2+etc.

ugh the text never comes out right why is line 1 not equal to negative line 3?

Last edited:
and for the right:

$$-\nabla^{2}F=-2z^{3} i$$
This is not correct.

Cyosis
Homework Helper
It seems you've taken the scalar Laplacian and then made a vector out of it. Make sure you use the vector Laplacian and you will get the extra term you need to proof the equality.

Vid
This question doesn't make any sense. Using the vector laplacian is trivial since it's defined by using the other two terms in your equations.

cristo
Staff Emeritus
This question doesn't make any sense. Using the vector laplacian is trivial since it's defined by using the other two terms in your equations.
Presumably, given that we are in Cartesian coordinates, you use the fact that $$\nabla^2\vec{F}=\nabla^2F_x\mathbf{i}+\nabla^2F_y\mathbf{j}+\nabla^2F_z\mathbf{k}$$

This is not correct.
http://en.wikipedia.org/wiki/Vector_Laplacian

$$A_{x}=x^{2}z^{3}$$

$$A_{y}=0$$

$$A_{z}=0$$

$$\nabla^{2}A_{x}=2z^{3}$$

$$\Rightarrow\nabla^{2}A=2z^{3}i$$

$$\Rightarrow-\nabla^{2}F=-2z^{3}i$$

is this not right?

Last edited by a moderator:
cristo
Staff Emeritus
The normal Laplacian is given by $$\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}$$. You already stated that $A_x=x^2 z^2$. Now all you have to do is apply the normal Laplacian to A_x.