= Solution
At height $z>0$, the <elliptic paraboloid> has a circular cross-section $0\le r\le\sqrt z$, of <area> $\pi z$. The <cylindrical coordinates> <volume element> therefore reduces the <volume integral> to
$$
K=\int_0^\infty\int_0^{2\pi}\int_0^{\sqrt z}f'(z)r\,dr\,d\theta\,dz
=\pi\int_0^\infty zf'(z)\,dz.
$$
For the specified $f$, the <derivative> is $e^{-z}(z^{-1/2}+\tfrac12z^{-3/2})$. Thus $zf'(z)$ is integrable at zero and decays exponentially at infinity. Apply <integration by parts> first on a truncated interval and then take the endpoint <limits>:
$$
K=\pi[zf(z)]_0^\infty-\pi\int_0^\infty f(z)\,dz
=\pi\int_0^\infty\frac{e^{-z}}{\sqrt z}\,dz.
$$
Both boundary terms vanish because $zf(z)=-\sqrt z\,e^{-z}$. With $z=u^2$, <integration by substitution> gives the <Gaussian integral>:
$$
\boxed{K=2\pi\int_0^\infty e^{-u^2}\,du=\pi^{3/2}.}
$$
The integrand is singular at the apex but the resulting <improper integral> is convergent.
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