= Solution
For the outward orientation of a <spherical cap>, its <vector area element> is
$$
d\mathbf S=\hat{\mathbf r}\,a^2\sin\theta\,d\theta\,d\phi,\qquad
\hat{\mathbf r}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta).
$$
In the <surface integral> defining the <vector area>, the $x$ and $y$ components vanish by integration over a complete azimuth. The remaining component is
$$
S_z=2\pi a^2\int_0^{\pi/4}\cos\theta\sin\theta\,d\theta
=\frac{\pi a^2}{2}.
$$
Consequently
$$
\boxed{\mathbf S=\left(0,0,\frac{\pi a^2}{2}\right)}.
$$
This also equals the area of the cap's projection onto the horizontal plane, directed upwards.
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