I have the equation of the lower semicircle of radius $r$ centred at a distance $a+r$ above the x-axis
$$ f(x)=r+a-\sqrt{r^{2}-x^{2}} $$
which I can approximate (for small $x$) as
$$ f(x)\approx a+\frac{1}{2r}x^{2} $$
using standard methods.
I'm doing this to simplify an integral - up to this point the integral becomes
$$ \int_{0}^{c}\int_{0}^{a+\frac{1}{2r}x^{2}}e^{-a\left(x^{2}+y^{2}\right)}\,dy\,dx $$
Then, by converting to a similar integral in polars, the main part of the integral over the region involves
$$ \int_{0}^{\arcsin\left(k\right)}\int_{0}^{P(\theta)}e^{-\beta r^{2}}drd\theta=\frac{1}{2\beta}\int_{0}^{\arcsin\left(k\right)}\left(1-e^{-\beta P^{2}(\theta)}\right)d\theta $$
where $P(\theta)$ is the polar equation of the line
$$ f(x)\approx a+\frac{1}{2r}x^{2} $$
so it would be good if I could find a curve $h(\theta)$ of the form
$$ h(\theta)=\sqrt{\ln\left(g(\theta)\right)} $$
which has the same curvature/second derivative as $f(x)$, and where $h(\theta)$ $=a$ at $\theta=0$ (take theta to be a displacement from the vertical - perhaps plot the original line to a get an idea of what this looks like), resulting in an approximation to $f(x)$, in polars, which is accurate for small $x$ or small $\theta$ (i.e. the final polar approximation to $f(x)$ touches $f(x)$ at $x=0$). This 'root ln' form will get rid of the exponential in the integral.
A first guess might be
$$ h(\theta)=\sqrt{\ln\left(e^{a^{2}}+\theta^{2}\right)} $$
Has anyone seen this sort of technique before? Can anyone think of a good $h(\theta)$?