= Solution
On the unit upper hemisphere, use <spherical coordinates> with $0\leq\theta\leq\pi/2$ and $0\leq\phi<2\pi$. The <surface mass density> is $\sigma=z=\cos\theta$, while the <surface area element> is $dS=\sin\theta\,d\theta\,d\phi$. Thus the total <mass> is
$$
M=\int_0^{2\pi}\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta\,d\phi
=2\pi\left[\frac12\sin^2\theta\right]_0^{\pi/2}
=\boxed{\pi.}
$$
The answer is in the mass units implicit in the specified unit radius and density law. A useful geometric check is that $z\,dS$ on the unit hemisphere equals the <area element> of its vertical projection onto the horizontal unit disc, whose <area> is $\pi$.
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