Question

\(F=(\dfrac{x}{r}, \dfrac{y}{r},\dfrac{z}{r})\) where \(r=\sqrt{x^2+y^2+z^2}\)
Find div F, please, help

Answer 1

@misty1212

Answer 2

if F=<P,Q,R>
\[Div(F)=P_x +Q_y +R_z\]

Answer 3

you might simplify calculations using the chain rule for partial derivatives
P=P(x,r)
\[\frac{\partial P}{\partial x} =(\frac{\partial P}{\partial x})_r+\frac{\partial P}{\partial r}\frac{\partial r}{\partial x}\]

Answer 4

you have \(<\partial_x, \partial_y, \partial_z> \bullet <\dfrac{x}{r}, \dfrac{y}{r}, \dfrac{z}{r}>\)

looking purely at the \(\partial_x \left( \dfrac{x}{r}\right)\) bit, as the symmetry does the rest, you can go at it from quotient rule

so \(\left. \partial_x \left( \dfrac{x}{r}\right) \right|_{y,z = const}= \dfrac{\partial_x(x)r - x \partial_x(r)}{r^2} = \dfrac{r - x. \dfrac{x}{r}}{r^2}\)


\(=\dfrac{r^2 - x^2}{r^3}\)

and where \(\partial_x (r) = \left. \partial_x (\sqrt{x^2 + y^2 + z^2}) \right|_{y,z = const} = \dfrac{x}{r}\)


repeat for y and z, and add it all up....

i am sure you could also switch to spherical div, and maybe do this more efficiently.

Answer 5

Thanks for that. since F is symmetric so we have the same for y, z hence
the result is \(\dfrac{1}{r^3} \) but it is not, the result is 2/r. I don't know how

Answer 6

from
\[\left. \partial_x \left( \dfrac{x}{r}\right) \right|_{y,z = const}= \dfrac{\partial_x(x)r - x \partial_x(r)}{r^2} = \dfrac{r - x. \dfrac{x}{r}}{r^2}\]


you get
\[ \partial_x \left( \dfrac{x}{r}\right) = \dfrac{r^2 - x^2}{r^3}\]

so applying the symmetry: \( \nabla \bullet <\dfrac{x}{r}, \dfrac{y}{r}, \dfrac{z}{r}> = \dfrac{r^2 - x^2}{r^3} + \dfrac{r^2 - y^2}{r^3} + \dfrac{r^2 - z^2}{r^3} = \dfrac{3r^2 - r^2}{r^3}\)

\(= \dfrac{2r^2 }{r^3} = \dfrac{2 }{r}\)

Answer 7

Thank you so much.