= Solution
In <cylindrical coordinates> the <spherical cap> has $1<z<\sqrt2$, $0<r<\sqrt{2-z^2}$ and $0<\theta<2\pi$. Its <mass density> is $kz$ and the <volume element> is $r\,dr\,d\theta\,dz$. Therefore
$$
M=\int_1^{\sqrt2}\int_0^{2\pi}\int_0^{\sqrt{2-z^2}}kzr\,dr\,d\theta\,dz=\pi k\int_1^{\sqrt2}z(2-z^2)\,dz=\boxed{\frac{\pi k}{4}}.
$$
This is the <mass of a spherical cap with height-proportional density> with $R=\sqrt2$ and $c=1$.
For its curved <surface area>, write $z=\sqrt{2-r^2}$ with $0\leq r\leq1$. The <surface integral> of one becomes
$$
A=\int_0^{2\pi}\int_0^1\sqrt{1+\left(\frac{dz}{dr}\right)^2}\,r\,dr\,d\theta=2\pi\sqrt2\int_0^1\frac{r}{\sqrt{2-r^2}}\,dr=\boxed{2\pi(2-\sqrt2)}.
$$
It agrees with the <curved area of a spherical cap>, $2\pi Rh$, for $h=\sqrt2-1$. The planar closing disk is not part of the requested curved surface.
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