Question

Radial field

Answer 1

\[\large\rm F=<x,y,z>(x^2+y^2+z^2)=r \left| r \right|^2 ~~Calculate~ the~
divergence \]\[\rm \color{ green}{F=\frac{ <x,y,z> }{ x^2+y^2+z^2} } \]\[\rm\color{green}{\nabla \cdot F = \frac{d}{dx}(\frac{ x }{ x^2+y^2+z^2 }) +\frac{ d }{ dy }(\frac{ y }{ x^2+y^2+z^2 })+\frac{ d }{ dz }(\frac{ z }{ x^2+y^2+z^2})}\]
\[\rm \color{green}{=\frac{ x^2+y^2+z^2 }{ (x^2+y^2+z^2)^2 }=\frac{ 1 }{ x^2+y^2+z^2 }= \frac{1}{r}}\]

Answer 2

double-n, you say that:

\(\large F= <x,y,z> (x^2+y^2+z^2) \)

then you do the Div on:

\(\large F=\dfrac{ <x,y,z> }{ x^2+y^2+z^2} \)

so it's a total fook-up :(

you should in the end go spherical-Div on this kinda thing cos it's just: \(\vec F (\vec r) = r^3 \, \vec {e}_{r}\).

but get yr *** over to Socratic and one can go from there.

the owners here have decided that zero-IQ, caveman-trolly-banter rules the day :(

not my cuppa(!), maybe yours!!!

Answer 3

ok what is it ???
help or an insult ??

Answer 4

\(\vec F= <x,y,z> (x^2+y^2+z^2) \)

\(\nabla \cdot \vec F = <\partial_x,\partial_y, \partial_z > \cdot <x,y,z> (x^2+y^2+z^2)\)

\(= \partial_x ( x (x^2+y^2+z^2)) +\partial_y (y (x^2+y^2+z^2)) + \partial_z( z (x^2+y^2+z^2)) \)

Product Rule, if you like:

\(= ( (x^2+y^2+z^2) + x \cdot 2x) +(x^2+y^2+z^2) + y \cdot 2y) + ((x^2+y^2+z^2) + z \cdot 2z)) \)

\(= 3(x^2+y^2+z^2) + 2x^2 + 2y^2 + 2z^2 \)

\(= 5(x^2+y^2+z^2) = 5r^2\)


That looks dreadful, I know, but with a soupçon of practise, you should spot the pattern and do it in way fewer lines.

Even a largely cosmetic change like this makes it look less mind-numbingly boring:

\(\vec F= <x,y,z> \rho \), where \(\rho(x,y,z) = (x^2+y^2+z^2)\)

So:

\(\nabla \cdot \vec F = <\partial_x,\partial_y, \partial_z > \cdot <x,y,z> \rho \)

\(= <\partial_x,\partial_y, \partial_z > \cdot <x \rho,y \rho,z \rho> \)

\(= \partial_x ( x \rho) +\partial_y (y \rho) + \partial_z( z \rho) \)

And the symmetry becomes more obvious?!

In spherical you have:

\(\vec F= r^3 \vec {e_r} \)

Spherical div is:

\(\nabla \cdot \vec F = {1 \over r^2}{\partial \left( r^2 F_r \right) \over \partial r}
+ {1 \over r\sin\theta}{\partial \over \partial \theta} \left( F_\theta\sin\theta \right)
+ {1 \over r\sin\theta}{\partial F_\varphi \over \partial \varphi}\)

[Really *** latex but i ripped it from Wiki]

But you can drop most of that as \(r\) is the only indie variable so it is:

\(\nabla \cdot \vec F = {1 \over r^2}{\partial \left( r^2 F_r \right) \over \partial r}\)

\( = {1 \over r^2}{\partial \left( r^2 \color{red}{r^3} \right) \over \partial r}\)

\( = {1 \over r^2}{\partial \left( r^5 \right) \over \partial r}\)

\( = {1 \over r^2} 5 r^4 = 5r^2\)

And, to be sure, clearly it was meant as an insult of the most severe type

Answer 5

Thanks.

Answer 6

4 both