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$$A_z = \mu{\frac{e^{-jBr}}{4\pi r}}∫I(z')e^{jBz'\cos\theta}dz'$$

Midway into my question, I want to compute:

$$-j\left( \frac{\nabla(\nabla\cdot A) }{w\mu\varepsilon} \right).$$

Symbols like $ w, \mu, \varepsilon, B $ are basically constants for things like frequency, permittivity, permeability and $r$ is just radius.

The result would be something like $ A_z \cos\theta \hat r$. I don't really know how to go about doing a divergence on this, and then a gradient. I did calc3 about 4 years ago, and between I have forgotten a lot. I need help at this step, if anyone needed the full problem, just comment.

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closed as off-topic by Kyle Kanos, ACuriousMind, John Rennie, JamalS, Martin Feb 26 '15 at 13:07

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The function $A_z$ is only a function of $r=\sqrt{x^2+y^2+z^2}$ and $\theta$ (the integral over $z'$ is not a function of $z$, but does depend on $\theta$). I.e., $$ A_z=\frac{\mu e^{-iBr}}{4\pi r}{\tt C(\theta)} $$

Since only (apparently) $A_z$ is non-zero we only need to be able to take the derivative w.r.t. $z$ to get $\nabla \cdot A_z$. $$ \nabla \cdot \vec A = \frac{\partial A_z}{\partial z} $$ But since $A_z=A_z(r)$ $$ \frac{\partial A_z}{\partial z}=\frac{\partial A_z}{\partial r}\frac{\partial r}{\partial z}+\frac{\partial A_z}{\partial \theta}\frac{\partial \theta}{\partial z} $$ $$ =\frac{\partial A_z}{\partial r}\cos(\theta)-\frac{\partial A_z}{\partial \theta}\sin(\theta) $$ since, e.g., $$ \frac{dr}{dz}=\frac{z}{r}=\cos(\theta)\;, $$ and so on... Also, there's still another application of the gradient to get your final answer... give it a shot.

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