= Solution
Assume $\phi$ has period $2\pi$, $f>0$ in the regular bulk, and consider a unit proper length in $z$. The proper transverse area element is $dA=f\,dr\,d\phi$, while the mass density is $T^{00}/c^2$. The <axial boundary term in the mass of a cylindrical spacetime> is therefore
$$
\boxed{\mu_{\rm bulk}=\int_0^R\frac{T^{00}}{c^2}\,2\pi f\,dr
=-\frac{c^2}{4G}\int_0^Rf''\,dr
=\frac{c^2}{4G}[f'(0^+)-f'(R^-)].}
$$
Choose the usual deficit branch $0\le\lambda<1$, so $f_{\rm out}=(1-\lambda)r$. Continuity of the induced metric requires $f(R^-)=(1-\lambda)R$. If there is no distributional thin shell at $R$, continuity of the <extrinsic curvature> also requires $f'(R^-)=1-\lambda$. A jump in $f'$ would add a surface-layer stress and cannot simply be discarded.
\b[Using the printed inner condition] $f'(0^+)=0$, these assumptions yield
$$
\boxed{\mu_{\rm bulk}=\frac{c^2}{4G}(\lambda-1).}
$$
This is the literal smooth-bulk answer. It is negative for the ordinary deficit range, so it must not be silently replaced by the usual positive cosmic-string mass formula.
For comparison, a regular axis with proper radial coordinate and angular period $2\pi$ requires $f(0)=0$ and $f'(0)=1$, since the circumference-to-radius ratio must approach $2\pi$. Under that different regular-axis condition the same integral gives
$$
\boxed{\mu_{\rm regular}=\frac{\lambda c^2}{4G},\qquad
\delta=2\pi\lambda=\frac{8\pi G\mu_{\rm regular}}{c^2}.}
$$
The original PDF really prints the derivative limit zero. It describes either a nonregular axis or an inner geometry different from an ordinary smooth string core; any additional singular-axis matter is not included in the bulk integral above and would need its own stress specification. Standard regular-core matching can also require an offset in the exterior radial origin. Thus the regular-axis formula is a qualified physical interpretation, not a derivation from the printed zero. For a general exterior branch replace $1-\lambda$ by $|1-\lambda|$ in the boundary derivative.
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