Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the three-dimensional integral $$ \int_{\mathbb R^3} d^3x\,f(x)\delta(g(x)) $$ where $\delta$ is the dirac delta, $f,b:\mathbb R^3\to\mathbb R$ and $g(x) = 0$ on some surface $S$. Is there a way of rewriting the above integral as a surface integral over the level set $S$? Related to this, is there some distributional identity like $$ \delta(g(x)) = \int_S ds dt \frac{\delta^{(3)}(x - X(s,t))}{|g'(X(s,t))|} $$ where $X(s,t)$ is a parameterization of $S$ that would allow one to to this, analogous to the formula $$ \delta(h(x)) = \sum_{x_0\in h^{-1}(0)}\frac{\delta^{(3)}(x-x_0)}{|h'(x_0)|} $$ when the zero set of $h$ is finite?

share|improve this question

1 Answer 1

I refer you to section 4.5 in the WP page (and references therein): simple layer integral. It turns out you can write $$\int_{\mathbb{R}^3} f(\vec{r})\delta(g(\vec{r}))d\vec{r}=\int_{S}\frac{f(\vec{r})}{|\vec{\nabla}g(\vec{r})|}d\sigma$$

A very similar question to this one is here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.