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Math Help - divergence

  1. #1
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    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) ?
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    Quote Originally Posted by heathrowjohnny View Post
    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
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  3. #3
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    how do you get that form?
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