Why divergence of radius vector equals 3?

Question

Ok, I know $div \vec r=3$.

$\vec r = r_x\vec i+r_y\vec j + r_z\vec k$.

$(\frac{\partial}{\partial x}i+\frac{\partial}{\partial y}j+\frac{\partial}{\partial z}k, r_xi+r_yj+r_zk)=\frac{\partial r_x}{\partial x} + \frac{\partial r_y}{\partial y} + \frac{\partial r_z}{\partial z}$

Now this supposedly gives $1+1+1$. How do I start seeing that?

Can you add the full question to your post? As written, it is not clear what you're asking. — DreamConspiracy, Sep 06 '18 at 16:17
I don't understand it. What is the definition of $\mathbf{r}?$ — saulspatz, Sep 06 '18 at 16:20
@user3600124 I still don't understand. Can you post the full exercise that you are working on, and explain all parts that you have attempted to solve? — DreamConspiracy, Sep 06 '18 at 16:21

score 3 · Answer 1 · edited Sep 06 '18 at 16:47

3

Note that the radial vector has its component as $$ \vec{r} =\langle x,y,z \rangle $$ Thus the divergence is $$ \operatorname{div} \vec{r} = \frac {\partial x}{\partial x} +\frac {\partial y}{\partial y}+\frac {\partial z}{\partial z} =1+1+1 =3$$

edited Sep 06 '18 at 16:47

md2perpe

26,770

answered Sep 06 '18 at 16:33

Mohammad Riazi-Kermani

68,728

I will memorize it – user3600124 Sep 06 '18 at 16:52

Why divergence of radius vector equals 3?

1 Answers1