# Evaluating an Integral with a Cross Product

Problem:

A infinite thin spherical shell with radius $$r_0$$, centered around the origin, is homogeneously charged with surface-charge-density $$\sigma$$ and rotates with the constant angular verlocity $$\mathbf{\omega}$$

Solution:

I'm not going to evaluate the integral fully. I do have the solution to it. It's a lot of work which seems fine. I just don't seem to get the first step in the evaluation. Anyway, here we go:

We know, for the vector potential we have:

$$\mathbf{A}(\mathbf{r})=\frac{1}{c}\int\frac{\mathbf{j}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}dV' \tag{1}$$

We further know that, in general, for a point charge at position $$\mathbf{r}_q$$ we have the charge density

$$\rho_q(\mathbf{r})=q\delta(\mathbf{r}-\mathbf{r}_q) \tag{2}$$

and if that point charge moves with $$\mathbf{v}$$ we have the current density

$$\mathbf{j}_q(\mathbf{r})=q\mathbf{v}\delta(\mathbf{r}-\mathbf{r}_q) \tag{3}$$

We also know from mechanics that

$$\mathbf{v}=\boldsymbol{\omega}\times \mathbf{r}_q \tag{4}$$

So now the homogeneous charged spherical shell has the charge density

$$\rho(\mathbf{r})=\sigma\delta(r-r_0) \tag{5}$$

so for the current density we get

$$\mathbf{j}(\mathbf{r})=\sigma (\boldsymbol{\omega} \times \mathbf{r})\delta(r-r_0) \tag{6}$$

so for the vector potential we get $$\mathbf{A}(\mathbf{r})=\frac{1}{c}\int\frac{\sigma (\boldsymbol{\omega}\times\mathbf{r}')\delta(r'-r_0)}{|\mathbf{r}-\mathbf{r}'|}dV'=\frac{r_0^3\sigma\boldsymbol{\omega}}{c}\times\int d\Omega'\frac{\mathbf{e}_r'}{|\mathbf{r}-r_0\mathbf{e}_{r'}|} \tag{7}$$

Question:

How do we explain the second $$=$$ in (7)? I can't really see how we "evaluated" the cross product and the delta function here.

## 2 Answers

Your problem is in the integral:

$$\mathcal I \equiv\int dv' \frac{\boldsymbol \omega \times \mathbf r'}{|\mathbf r - \mathbf r'|} \delta(r'-r_0)$$

First of all, the cross product is a bilinear operation, meaning that since $$\boldsymbol \omega$$ is independent of the integration variable $$\mathbf r'$$, you can take it out of the integral:

$$\mathcal I = \boldsymbol \omega \times \int dv' \frac{ \mathbf r'}{|\mathbf r - \mathbf r'|} \delta(r'-r_0)$$

Now if we use spherical coordinates, this is going to be:

$$\mathcal I = \boldsymbol \omega \times \int_0^\infty dr' \int_{4 \pi} d\Omega' \ r'^2 \frac{ \mathbf r'}{|\mathbf r - \mathbf r'|} \delta(r'-r_0)$$

Since $$\mathbf r' = r' \hat{\mathbf e}_{r'}$$, this is:

$$\mathcal I = \boldsymbol \omega \times \int_{4 \pi} d\Omega' \int_0^\infty dr' \ r'^3 \frac{ \hat{\mathbf e}_{r'}}{|\mathbf r - r' \hat{\mathbf e}_{r'}|} \delta(r'-r_0)$$

Now by the sifting property of the Dirac delta, $$\int_0^\infty dr' f(r') \delta(r'-r_0) = f(r_0)$$ for $$r_0 > 0$$, resulting in:

$$\mathcal I = r_0^3 \ \boldsymbol\omega \times \int_{4 \pi} d\Omega' \ \frac{ \hat{\mathbf e}_{r'}}{|\mathbf r - r_0 \hat{\mathbf e}_{r'}|}$$

Seems like it switched to spherical coordinates, then evaluated the integral with respect to the radial coordinate, which just leaves the 2D integral with respect to the solid angle or angular coordinates. Is this clear?