# divergence

• Mar 27th 2008, 11:31 AM
heathrowjohnny
divergence
Let $\bold{v} = (r \cos \theta) \bold{\hat{r}} + (r \sin \theta) \bold{\hat{\theta}} + (r \sin \theta \cos \phi) \bold{\hat{\phi}}$.

So we can check that $\int_{V} (\nabla \cdot \bold{v}) d \tau = \int_{A} \bold{v} \cdot d \bold{a}$.

But why do we compute the divergence as follows: $\nabla \cdot \bold{v} = \frac{1}{r^2} \frac{\partial}{\partial r}(r^{2}r \cos \theta) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (\sin \theta \ r \sin \theta) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} (r \sin \theta \cos \phi)$?
• Mar 27th 2008, 12:31 PM
topsquark
Quote:

Originally Posted by heathrowjohnny
Let $\bold{v} = (r \cos \theta) \bold{\hat{r}} + (r \sin \theta) \bold{\hat{\theta}} + (r \sin \theta \cos \phi) \bold{\hat{\phi}}$.

So we can check that $\int_{V} (\nabla \cdot \bold{v}) d \tau = \int_{A} \bold{v} \cdot d \bold{a}$.

But why do we compute the divergence as follows: $\nabla \cdot \bold{v} = \frac{1}{r^2} \frac{\partial}{\partial r}(r^{2}r \cos \theta) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (\sin \theta \ r \sin \theta) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \phi} (r \sin \theta \cos \phi)$?

I"m not sure what you are asking. This is the divergence formula you use when working in spherical-polar coordinates. Are you asking how do we get this form?

-Dan
• Mar 27th 2008, 01:26 PM
heathrowjohnny
how do you get that form?