I've been having issues with integrating with a Dirac delta. To compute the area of a sphere centered at $(0,0,0)$ it seems to work just fine:
$$\int_{0}^{2\pi}{\int_{0}^{\pi}{\int_{0}^{\infty}{\delta(r-\rho)r^2\sin\theta\, dr}\,d\theta}\,d\phi} = 4\pi\rho^2$$
Now I will take the same sphere but offset by $(0,0,\rho)$, that is: $x^2 + y^2 + (z-\rho)^2 = \rho^2$. Going to spherical coordinates yields: $r^2\cos^2\phi\sin^2\theta + r^2\sin^2\phi\sin^2\theta + (r\cos\theta-\rho)^2 = \rho^2$, which yields: $r(r-2\rho\cos\theta)=0$, and we can express the sphere in spherical coordinates as: $r(\theta) = 2\rho\cos\theta, \theta \in [0,\pi/2], \phi\in[0,2\pi]$. Integrating yields: $$\int_{0}^{2\pi}{\int_{0}^{\frac{\pi}{2}}{\int_{0}^{\infty}{\delta(r-2\rho\cos\theta)r^2\sin\theta\, dr}\,d\theta}\,d\phi} = \frac{8\pi\rho^2}{3}$$
Now this is not right clearly. The only reason I can think of has to do something with properties of the Dirac delta I am unaware of. Note that I have not studied measure theory. I need the Dirac delta and not a surface integral because I will be using this to compute transformations of probability density functions which I will need to write through a Dirac delta.
Edit: References covering this for engineers/computer science students are welcome.
Edit 2: Taking into account David Holden's answer I came up with the following fact which must hold (I hope it's correct): $$\int_{V}{\delta(f(x)) \,dx} = \int_{S = \{x|f(x) =0\}}{\,dA}$$
Edit 3: I found some more information on the subject: Impulse functions over curves and surfaces Properties_in_n_dimensions Surface area from indicator function Property of Dirac delta function in $\mathbb{R}^n$ Does the coarea formula hold for delta-function?
I believe the issue was that whenever I offset the sphere the Dirac delta changed such that $\delta(f(r)) \rightarrow \delta(g(r,\theta))$ and $g$ was then a non-trivial mapping (so it's not the one-dimensional dirac delta I am used to anymore). Based on the first article I believe that I can rewrite it as a surface Dirac delta $\delta(g(r,\theta)) = \delta_S(r,\theta)$ which yields the surface integral giving a correct result. The other threads and wikipedia state that I should have a normalization by the magnitude of the gradient. I think I am missing an important piece since for the result to be correct this normalization factor should cancel out with something. More precisely: $$\int_V{\delta(r-2\rho\cos\theta)r^2\sin\theta \,dr\,d\theta\,d\phi} = \int_S{\frac{\,d\sigma}{\sqrt{r^2+\rho^2-2r\rho\cos\theta}}}$$ The only idea I have is that somehow the normalization factor will pop out of the $\,d\sigma$. No idea though since it's supposed to be a 'Minkowski content measure' which is way over my head as a computer science student.
To add to this I would also like to be able to solve the same problem with a heaviside function (for integrating the volume of the offset ball). I am unsure whether similar considerations apply there, however if I integrate it, the result seems correct. I still want to make sure this is valid for other volumes also (maybe it's just a coincidence like the sphere centered at $(0,0,0)$). So I would be grateful if somebody with more knowledge on geometric measure theory could clarify all of the points.
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) in front of the differentials. – Bernard Jan 23 '19 at 00:28