= Solution
Differentiate the position vector with respect to each of the <spherical polar coordinates>. The derivatives are
$$
\begin{aligned}
\partial_r\mathbf x&=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\\
\partial_\theta\mathbf x&=r(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\\
\partial_\phi\mathbf x&=r\sin\theta(-\sin\phi,\cos\phi,0).
\end{aligned}
$$
Their lengths give the <scale factors of orthogonal coordinates>:
$$
\boxed{h_r=1,\qquad h_\theta=r,\qquad h_\phi=r\sin\theta.}
$$
Dividing the derivatives by these lengths yields
$$
\begin{aligned}
\mathbf e_r&=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta),\\
\mathbf e_\theta&=(\cos\theta\cos\phi,\cos\theta\sin\phi,-\sin\theta),\\
\mathbf e_\phi&=(-\sin\phi,\cos\phi,0).
\end{aligned}
$$
Each has unit length. For example, $\mathbf e_r\cdot\mathbf e_\theta=\sin\theta\cos\theta(\cos^2\phi+\sin^2\phi)-\cos\theta\sin\theta=0$; the other two <dot products> vanish because their first two terms cancel. Thus these are mutually <orthogonal vectors>, and the total differential has the required form. The coordinates degenerate at $r=0$ and on the polar axis, but this does not affect an area integral.
On the <parametrized surface> $\theta=\alpha$, use $r,\phi$ as parameters. The <surface integral> area element is
$$
dS=|\partial_r\mathbf x\times\partial_\phi\mathbf x|\,dr\,d\phi=r\sin\alpha\,dr\,d\phi.
$$
For $0\leq\alpha\leq\pi$, its lateral area, with no closing disk, is therefore
$$
\boxed{A=\int_0^{2\pi}\!\int_0^R r\sin\alpha\,dr\,d\phi=\pi R^2\sin\alpha.}
$$
At $\alpha=0,\pi$ the surface collapses to a line and its area is zero.
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