To give you a better picture, what I mean with my comment, consider
Stokes Theorem
$$\int_D db^{n-1}=\int_{\partial D}b^{n-1}$$
be $b^{n-1}$ an arbitrary 1-form
$$b^{1}= a_1 dx + a_2 dy$$
leads to
$$\int_D f(x,y)dxdy = \int_D\left(\frac{\partial a_2}{\partial x}-\frac{\partial a_1}{\partial y}\right)dxdy$$
Without loss of generality, $a_1=0$ one needs to solve $f(x,y)=\frac{\partial a_2}{\partial x}$ for $a_2$.
Now you need the parametrization for the boundary functions. For the first three bounding functions $$(x=0,y\in[-h,\eta(0)]),(x\in[0,L],y=-h),(x=L,y\in[-h,\eta(L)])$$ this is straight forward. For the bounding function I assume you know or can evaluate the function in $(x(t),y(t)) \wedge t\in[0,1]$. So I suggest using two Gaussian quadratures. First from $[lb,peak]$ and second from $[peak, rb]$.
Given you can obtain information on the peak and integrate $f$, this method will give you very high accuracy and very good convergence.
Supplementing Edit
As bigge pointed out, knowing $f$ does not guarantee to find an parametrization easily. Of course this is true. But even if one is not able to solve $f(x,y)=\frac{\partial a_2}{\partial x}$ analytically, one can still use this approach. Assuming Gaussian quadrature techniques one can read it as
$$a_2(x,y)=\int_{x_0}^x f(t,y)dt =\int_{-1}^1 f(r,y) \frac{dt}{dr} dr\approx \sum_k w_k f(r_k,y) \frac{dt}{dr}$$ With e.g. $x_0=0$ chosen arbitrarily. Now the scaling factor $\frac{dt}{dr}$ depends on the integration range $x$. Since we still integrate line integrals along the boundary, $x$ is not arbitrary but specific set of quadrature points for evaluating $\int_{\partial D} a_2(x,y) dy$.
So the final term is a double sum for each boundary integral as expected for an integral over a 2-D domain.
In that form it is simply a very sophisticated 2D Simpson's rule.
In my opinion, it is usually more challenging to find a good parametrization of the bounding curves than evaluation of $f$.
x=x(t), y=y(t)
the integral of the bounding curve can be solved by quadrature techniques. $\endgroup$ – Bort Apr 7 '15 at 8:55