= Solution
Use the same axes as in (a) and choose the outward orientation. On the curved <spherical lune>, $\mathbf n=\mathbf e_r$ and $dA=a^2\sin\theta\,d\theta\,d\varphi$. Its <vector area> is
$$
\mathbf S=a^2\int_{-\beta/2}^{\beta/2}\int_0^\pi
(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)
\sin\theta\,d\theta\,d\varphi.
$$
The $y$ and $z$ components cancel. The $x$ component is $a^2(\pi/2)\,2\sin(\beta/2)$, hence
$$
\boxed{\mathbf S=\pi a^2\sin\frac\beta2\,\mathbf e_x}.
$$
Here $\mathbf e_x$ points along the azimuthal bisector; another coordinate choice rotates this vector accordingly. Its magnitude is the signed projected <area>, not the scalar curved <area> $2\beta a^2$. Reversing the orientation reverses its sign.
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