I am reading Example 5.11 of Griffiths' EM book and stuck at some step.
Example 5.11 A spherical shell, of radius $R$, carrying a uniform surface charge $\sigma$, is set spinning at angular velocity $\omega$. Find the vector potential it produces at point $\mathbf{r}$ ( Fig. 5.45 )
Solution : It might seem natural to align the polar axis along $\mathbf{\omega}$, but in fact the integration is easier if we let $\mathbf{r}$ lie on the $z$ axis, so that $\mathbf{\omega}$ is tilted at an angle $\psi$. We may as well orient the $x$ axis so that $\omega$ lies in the $xz$ plane, as shown in Figure 5.46. According to Eq.5.66,
$$ \mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4 \pi} \int \frac{\mathbf{K}(\mathbf{r}')}{\mathfrak{r}}da',$$
Then, through some argument, he deduced that
$$\mathbf{A}(\mathbf{r})= \begin{cases} \frac{\mu_0 R \sigma}{3} (\omega \times \mathbf{r} ), & \operatorname{for points inside the sphere}, \\ \frac{\mu_0 R^4 \sigma}{3r^3} (\omega \times \mathbf{r} ), & \operatorname{for points outside the sphere}. \end{cases} \tag{5.68}$$
Then he wrote that, " Having evaluated the integral, I revert to the “natural” coordinates of Fig.5.45, in which $\omega$ coincides with the $z$ axis and the point $\mathbf{r}$ is at $(r, \theta, \phi)$ " :
$$\mathbf{A}(r, \theta, \phi)= \begin{cases} \frac{\mu_0 R \omega \sigma}{3} r \sin \theta \hat{\mathbf{\phi}}, & ( r \le R), \\ \frac{\mu_0 R^4 \omega \sigma}{3} \frac{\sin \theta}{r^2} \hat{\mathbf{\phi}}, & ( r \ge R). \end{cases} \tag{5.69}$$
Q. Why does $(5.69)$ hold? What relationship exists between spherical coordinates of Fig.5.45 and that of Fig. 5.46 ? Can anyone help?
