Diffraction at the black sphere (or disk) of radius $R$ at large energies: Idealize the S-matrix by
$$
S_l=\left\{\begin{array}{lll}
1 & \text { for } & l>L \equiv k R \gg 1 \\
0 & \text { for } & l \leq L \equiv k R \gg 1
\end{array}\right.
$$
such that all partial waves with an "impact parameter" $b=l / k \leq R$ are supposed to be totally absorbed. Convince yourself that the integrated elastic cross section from $(18.43)$ is identical to the reaction cross section (or inelastic one) defined in (18.45),
$$
\sigma_{\mathrm{el}}=\sigma_{\mathrm{rea}}=\frac{\pi}{k^2} \sum_{l=0}^L(2 l+1) \approx \frac{\pi}{k^2} l^2=\pi R^2 .
$$
Why is the total cross section larger than the area of the disk seen by the beam particles, $\sigma_{\text {tot }}=2 \pi R^2$ ?