Vector potential due to a spinning spherical shell with a non-uniform surface charge distribution In Griffith's Introduction to Electrodynamics 4th edition example 5.11 the solution for the vector potential of a uniformly charged spinning spherical shell is given.

Now let's assume that the surface charge density is given by $\sigma(\theta)=\sigma_0sin(\theta)$ and the sphere is rotating with a constant angular velocity $\omega$ about the $\hat{z}$ axis. So, unlike the figure above we have to put the axis of rotation on $\hat{z}$. The velocity vector will be:
$\vec{v}=\vec{\omega}\times\vec{r'}=\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z}\\
0 & 0 & \omega\\
Rsin\theta'cos\phi' & Rsin\theta'sin\phi' & Rcos\theta'
\end{vmatrix}$
After expanding we get
$\vec{v}=-R\omega sin\theta' sin\phi'\hat{x}+R\omega sin\theta' cos\phi'\hat{y}$
The surface current becomes
$\vec{K}=-R\omega\sigma (sin\theta')^2 sin\phi'\hat{x}+R\omega\sigma (sin\theta')^2 cos\phi'\hat{y}$
We can write it as $\vec{K}=-R\omega\sigma sin\theta' \hat{\phi}$
The vector potential can be written as
$\hat{A}=\frac{-\mu_0R^3\omega\sigma}{4\pi}\hat{\phi}\int_0^{\pi}\int_0^{2\pi}\frac{sin^2\theta'}{\sqrt{R^2+r^2-2Rrcos\alpha}}d\phi'd\theta'$
in which $\alpha$ is taken to be the angle between $\vec{r}$ and $\vec{r'}$ and we can express $cos\alpha$ using the polar and azimuthal angles in the problem as
$cos\alpha=cos\theta cos\theta' + sin\theta sin\theta' cos(\phi-\phi')$
Although I think the formulation of the solution is correct the integral for the vector potential is very difficult for me to solve. I also have tried to integrate using MATHEMATICA but gave up after half an hour since it didn't give me an answer. Is there a better way to solve this problem?
 A: You are missing the $\hat\phi \cdot \hat \phi'$ factor in the integrand.
That is, the current is along $\hat \phi'$, and the resulting vector
potential is along $\hat \phi$, but these are two different directions
in general. So to calculate the $\hat \phi$ component of the vector
potential you need to dot the integral over $\hat \phi'$ with $\hat \phi$.
In addition, I believe there is a trivial sign error:
\begin{equation}
\hat z \times (x\hat x + y \hat y + z\hat z) =
x \hat z \times \hat x+y\hat z \times \hat y = x\hat y-y\hat x =
r\sin\theta \hat \phi \,.
\end{equation}
You can continue and evaluate your integral with this change.
A faster way is to realize $x'$ and $y'$ can be written as linear combinations
of $r' Y_{1m}(\theta',\phi')$, writing
\begin{equation}
\vec A(\vec r) = \frac{\mu_0}{4\pi} \int d^3r' \frac{\vec J(\vec r')}{|\vec r-\vec r'|}
\end{equation}
and then expanding
\begin{equation}
\frac{1}{|\vec r-\vec r'|} = \sum_{\ell m} \frac{4\pi}{2\ell+1} 
Y_{\ell m}(\theta',\phi')^*Y_{\ell m}(\theta,\phi)
 \frac{r_<^\ell}{r_>^{\ell+1}}
\end{equation}
we see that after integrating over the solid angle, using orthogonality
of the spherical harmonics, we get only $\ell=1$
terms with the same
angular function of the $\theta$, $\phi$ coordinates that we had for the
$\theta',\phi'$ coordinates. Therefore the integration is trivial and
\begin{equation}
\vec A (r,\theta,\phi) = \hat \phi
\frac{\mu_0 R^3\omega\sigma_0\sin\theta}{3}\left \{
\begin{array}{cc}
\frac{r}{R^2} & r< R\\
\\
\frac{R}{r^2} & r> R\\
\end{array}
\right .
\end{equation}
A: It seems very difficult to obtain analytically integrated formulas.
By the way, in Landau and Lifshitz book (See PROBLEM 2 of page 125 in "Electrodynamics of Contnuous Media"), the analytical formulas for "magnetic field of a linear current flowing in a circle" is given. In the case of axisymmetry, it would be easy to obtain a practical method of calculation if one allows for numerical integration rather than a complete analytical expression.
Define the k:
$$
k^2=\frac{4(R\text{sin}\theta)r}{(R\text{sin}\theta+r)^2+(z-R\text{cos}\theta)^2},
$$
where the observation point is $(r,z)$ (Note, the definition of $r$ differs from the figure). The vector potential from the ribbon region $(\theta,\theta+d\theta)$ is expressed as
$$
dA_{\phi}(r,z,\theta)=\frac{\mu_0dI}{\pi k}\sqrt{\frac{R\text{sin}\theta}{r}}\left[(1-\frac{1}{2}k^2)K(k)-E(k)\right]\;\;\;\cdots(1)
$$
where $K(k)$ and $E(k)$ are complete eliptic integral of the first and second kind. Since the  electric current of the ribbon is
$$
dI=\sigma(\theta) R d\theta\:v
=\sigma(\theta) R d\theta(\omega R\text{sin}\theta)
=R^2\omega\sigma(\theta)\text{sin}\theta d\theta, \;\;\;\cdots(2)
$$
substituting (2) into (1), and make integration, we get,
$$
  A_\phi(r,z)=\int_0^{\pi}dA_{\phi}\;\;\;\cdots(3)
$$
The (3) is easy to calculate numerically provided the observation position $(r,z)$ is not on the spherical surface.
A: Since you ask "Is there a better way to solve this problem?", here's one:
Instead of calculating the vector potential, you can instead calculate a magnetic scalar potential $\psi$ defined such that $\vec{B} = - \vec{\nabla} \psi$.  This is permissible at all points with $r \neq R$, since at these points we have $\mu_0 \vec{J} = \vec{\nabla} \times \vec{B} = 0$.  Moreover, since $\vec{\nabla} \cdot \vec{B} = 0$, we have $\nabla^2 \psi = 0 $, allowing us to use the machinery (familiar to us from electrostatics) of axially symmetric solutions to Laplace's equation.  I will give an outline of the procedure for this problem below, but I will skip a lot of steps, which I encourage you to fill in yourself.  (Also, I did this quickly, and it's entirely possible that I've dropped a sign or a factor of 2 somewhere in here.)
The general solution to Laplace's equation for $\psi$ (assuming good behavior at the origin and at infinity) will be
$$
\psi = \begin{cases} \displaystyle\sum_{\ell=0}^\infty A_\ell \left( \frac{r}{R}\right)^\ell P_\ell(\cos \theta) & r < R \\
\displaystyle \sum_{\ell=0}^\infty B_\ell \left( \frac{R}{r}\right)^{\ell+1}  P_\ell(\cos \theta) & r > R
\end{cases}  \tag{1}
$$
The boundary conditions are different from the electrostatic case, though; we must have
$$
\hat{r} \cdot (\vec{\nabla} \psi_\text{int} - \vec{\nabla} \psi_\text{ext} ) = 0 \qquad \hat{r} \times (\vec{\nabla} \psi_\text{int} - \vec{\nabla} \psi_\text{ext} ) = -\mu_0 \vec{K}. \tag{2}
$$
The first boundary condition (it can be shown) implies that
$$
B_\ell = - \frac{\ell}{\ell+1} A_\ell. \tag{3}
$$
The second one (combined with Eq. (3)) yields
$$
\sum_{\ell = 0}^\infty \frac{A_\ell}{\ell + 1} P_\ell^1(\cos \theta) = R \mu_0 K_\phi(\theta) \tag{4}
$$
where $P_\ell^1(\cos \theta)$ is an associated Legendre function.  The orthogonality relations for associated Legendre functions then imply that
$$
A_\ell = \frac{2\ell + 1}{2 \ell} R \mu_0 \int_0^\pi P_\ell^1(\cos \theta) K_\phi(\theta) \sin \theta \, d \theta.
$$
You can then calculate as many $A_\ell$ coefficients as you like.  Unfortunately, for the case you've described, with $K_\phi(\theta) \propto \sin^2 (\theta)$, there are infinitely many non-zero coefficients, so an exact closed-form solution is (probably) not possible using this technique; but you can get as close as you like.
