I have an integral over a two-dimensional plane like
$$\int d^2 \mathbf{x} \, \delta(g(\mathbf{x})) f(\mathbf{x}).$$
The delta function restricts the integral to a curve implicitly given by $g(\mathbf{x}) = 0$. I can parametrize this curve in polar coordinates $(r,\theta)$; that is, the delta function is supported on some $r=R(\theta)$.
I need to get rid of the delta function, i.e. change variables to get an integral over $\theta$. This should give me something like
$$ \int \sigma (\theta) d\theta \frac{f(R \cos \theta, R \sin \theta)}{|\nabla g|}.$$
But what measure do I need — what is $\sigma(\theta)$? Is it correct to take the arc length measure
$$ \sigma(\theta)^2 = R'^2 + R^2?$$
Best Answer
Suppose that the curve $C$ given by $g(\mathbf{x})=0$ is smooth and can be parameterized by arc length as $\mathbf{x} = \mathbf{p}(s) = (p_1(s), p_2(s)).$ Let $\mathbf{n}(s) = (n_1(s), n_2(s))$ be the unit normal field to $C,$ pointing to the left when going in the positive direction of $C$. Then a neighborhood of the curve can be parameterized by $\mathbf{x}=\mathbf{p}(s)+t\mathbf{n}(s)$ and we have $$ |\mathbf{p}'| = 1, \quad |\mathbf{n}| = 1, \quad \mathbf{n}' = -\kappa \mathbf{p} $$ where $\kappa$ is the curvature at the actual position on the curve.
The local metrics is given by $$ d\mathbf{x}\cdot d\mathbf{x} = (1-\kappa t)^2 \, ds^2 + dt^2 $$ and the area measure by $$ d^2\mathbf{x} = \left| \frac{\partial\mathbf{x}}{\partial s} \times \frac{\partial\mathbf{x}}{\partial t} \right| \,ds\,dt = (1-\kappa t) \, ds \, dt . $$
It should also be noted that $\nabla g = \pm|\nabla g| \mathbf{n},$ where the sign is $+$ if $g$ grows to the left, $-$ if $g$ grows to the right. Then, to first order, $$ g = t\,\nabla g\cdot \mathbf{n} = \pm t |\nabla g| $$ so $$ \int d^2 \mathbf{x} \, \delta(g(\mathbf{x})) f(\mathbf{x}) = \iint ds\,dt\,(1-\kappa t) \, \delta(\pm t |\nabla g|) f(\mathbf{x}) = \iint ds\,dt\,\frac{1-\kappa t}{|\nabla g|} \, \delta(t) f(\mathbf{p}(s)+t\mathbf{n}(s)) \\= \int ds\,\frac{1}{|\nabla g|} \, f(\mathbf{p}(s)) $$ This result is basically of the same form as you have, so $\sigma(\theta)\,d\theta$ should be the arc length measure.