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How do you find the divergence of a field $f(\vec{r})=\vec{r}\exp(r^2)$ where $\vec{r}$ is the position vector and $r$ is its magnitude? In other words, how does one evaluate $\nabla\cdot[\vec{r}\exp(r^2)]$? I think we could do it by writing $f(\vec{r})$ as a column vector and differentiating each component wrt their variable and I get $\exp(r^2)(3+2r^2)$ is that right? Is there a quicker way to do this?

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up vote 2 down vote accepted

The divergence of a field is symbolically written as $$ \nabla\cdot f\tag{1} $$ since, in $\mathbb{R}^3$, $\nabla=\mathbf{i}\frac{\partial}{\partial x}+\mathbf{j}\frac{\partial}{\partial y}+\mathbf{k}\frac{\partial}{\partial z}$, and taking the symbolic dot product with $f=\mathbf{i}f_1+\mathbf{j}f_2+\mathbf{k}f_3$ yields $$ \frac{\partial}{\partial x}f_1+\frac{\partial}{\partial y}f_2+\frac{\partial}{\partial z}f_3\tag{2} $$ In your function, $$ \begin{array}{}f_1=xe^{x^2+y^2+z^2}&f_2=ye^{x^2+y^2+z^2}&f_3=ze^{x^2+y^2+z^2}\end{array}\tag{3} $$ So $$ \begin{align} \nabla\cdot f &=(1+2x^2)e^{x^2+y^2+z^2}+(1+2y^2)e^{x^2+y^2+z^2}+(1+2z^2)e^{x^2+y^2+z^2}\\ &=(3+2r^2)e^{r^2}\tag{4} \end{align} $$ which is exactly what you got.

As for a "quicker way," you could have precomputed that for $f(\vec{r)}=\vec{r}g(r)$, $$ \begin{align} \nabla\cdot f &=3g(\vec{r})+\vec{r}\cdot\nabla g(\vec{r})\tag{5} \end{align} $$ and that might make the computation of $(4)$ easier.

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Thank you, robjohn! – vec Feb 11 '12 at 10:43

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