to give extremely good estimates of stretching free energy even for layers with curvature radii comparable to their thickness.[@ballbrush91]

The work to add an A-block is independent of the location of the free end, and so may be conveniently taken to be the work to add a chain with its conformation very near the surface, simply $\Omega_{\scriptscriptstyle A}
\mu(z_d)$. Hence the total free energy of a chain is obtained by integrating up to the desired coverage $\sigma$, $$f_{\mit str} = {3\pi^2\over 8\sigma} 
\int_0^{\sigma}d{\sigma}' \left[
{n_{\scriptscriptstyle A}\over  R_{\scriptscriptstyle A}^2}h_{\mit
\scriptscriptstyle A}^2\left({\sigma}'\right)
+ {n_{\scriptscriptstyle B}\over  R_{\scriptscriptstyle B}^2}h_{\mit
\scriptscriptstyle B}^2\left({\sigma}'\right)\right]. 
\label{eq:fstr0}$$ Using eqs \[eq:htcov0\]-\[eq:htcov\] we can write $$\begin{aligned}
d\sigma &=& {dh_{\scriptscriptstyle A}\over\Omega \phi}\,
{a(\beta + h_{\scriptscriptstyle A}/R)\over a(\beta)}, \\
d\sigma &=& {dh_{\scriptscriptstyle B}\over\Omega (1-\phi)}\,
{a(\beta - h_{\scriptscriptstyle B}/R)\over a(\beta)}.\end{aligned}$$ Changing variables from $\sigma$ to $h_{\scriptscriptstyle A}$ and $h_{\scriptscriptstyle B}$ and using eqs \[eq:htcov0\]-\[eq:htcov\], we rewrite eq \[eq:fstr0\] as $$f_{\mit str}={\pi^2 R^2 \over 8 v(1)} \left[
{n_{\scriptscriptstyle A} \over \phi R_{\scriptscriptstyle A}^2}
I_{\scriptscriptstyle A}
 +
{n_{\scriptscriptstyle B} \over (1-\phi) R_{\scriptscriptstyle B}^2}
I_{\scriptscriptstyle B}\right] 
\label{eq:fstr}$$ where $$\begin{aligned}
I_{\scriptscriptstyle A} &=& 3 \int_0^\beta\!\!\!dy\, a(\beta-y) y^2 
\label{eq:IA}\\
I_{\scriptscriptstyle B} &=& 3 \int_0^{1-\beta}\!\!\!dy\, a(y+\beta) y^2\;.
\label{eq:IB}\end{aligned}$$ At this point we make contact with previous calculations of asymmetric block 