Compute area of a sphere through a Dirac delta

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): $$\int_{V}{\delta(f(x)) \,dx} = \int_{S = \{x|f(x) =0\}}{\,dA}$$

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.

• @Bernard I can't understand what was edited in the integrals. Can you elaborate for future reference? – lightxbulb Jan 23 at 0:22
• Dirac is a surname, so it has an initial capital. Also I added unbreakable thin spaces ( code: \,) in front of the differentials. – Bernard Jan 23 at 0:28
• – leonbloy Jan 23 at 1:38
• $8 \pi \rho^2/3$ is the correct value for the given integral, it's just a consequence of $\int_{\mathbb R} \delta(x - a) \phi(x) dx = \phi(a)$. However, the correct relation between the volume and the surface integrals is $$\int_{\mathbb R^n} \delta(f(\mathbf x)) \,|\nabla f(\mathbf x)| \,d\mathbf x = \int_{f(\mathbf x) = 0} dA.$$ – Maxim Jan 23 at 13:26
• $$\int_{\mathbb R^3} \delta(x^2 + y^2 + (z - \rho)^2 - \rho^2) \,|\nabla(x^2 + y^2 + (z - \rho)^2 - \rho^2)| \,dx dy dz = \\ 2 \pi \int_0^{\pi/2} \int_0^\infty \delta(r (r - 2 \rho \cos \theta)) \,2 \sqrt{r^2 + \rho^2 - 2 \rho r \cos \theta} \;r^2 \sin \theta \,dr d\theta.$$ Note that the factor $r$ inside the delta function cannot be ignored and the gradient in spherical coordinates is not $(\partial_r f, \partial_\theta f, \partial_\phi f)$ (which is why your Edit 3 cannot be correct). Now if you evaluate this as an iterated integral, you'll get $4 \pi \rho^2$. – Maxim Jan 24 at 18:27

when you write integrating yields... you may be making an unwarranted assumption about what is the "element of area". a simple geometric approach suggests:

$$dA = \rho d(2\theta) \rho \sin 2\theta d\phi = 2 \rho^2 \sin 2\theta d\theta d\phi$$

note that with $$dA$$ thus defined:

$$\int_0^{2\pi}\int_0^{\frac{\pi}2} dA = 4\pi\rho^2$$

• I am not changing the parametrisation nor am I integrating over area though. I know very well how to compute the area as a surface integral. My point is that I want the factors to pop out from the Dirac delta's change. Basically I want this explained in terms of the Dirac delta and what am I doing wrong with it (in the general case not only for sphere). I've seen a property in 1D where the absolute value of derivative pops out of the Dirac delta when applying a transformation $\delta(f(x))$. I would like the multidimensional case, preferably with an example applied to my problem. – lightxbulb Jan 23 at 2:09

I finally figured out why I am getting a 'wrong' result. As expected I cannot substitute with the delta directly since it's a composition with a submersion. However the following equality holds from the coarea formula: $$\int_{R^n}{f(x)\delta(g(x))\,dx} = \int_{g^{-1}(0)}{\frac{f(x)}{|\nabla g(x)|}\,d\sigma(x)}$$ Where $$g:R^n\rightarrow R$$, $$|\nabla g(x)|\ne 0$$, and $$d\sigma$$ is the measure on the surface $$g^{-1}(0)$$. Let us consider the non-normalized uniform probability density function on the sphere with center $$(0,0,0)$$ and radius $$\rho$$ in spherical coordinates: $$p_A(x,y,z) = \delta(r-\rho)r^2\sin\theta$$. Unsurprisingly integrating it yields $$4\pi\rho^2$$: $$\int_{0}^{2\pi}{\int_{0}^{\pi}{\int_{0}^{\infty}{\delta(r-\rho)r^2\sin\theta\,dr}\,d\theta}\,d\phi} = \int_{0}^{2\pi}{\int_{0}^{\pi}{\frac{\rho^2\sin\theta}{1}\,d\theta}\,d\phi} = 4\pi\rho^2$$

Note that the division by $$1$$ is to emphasize that $$|\nabla g| = 1$$. That is, I have used the coarea formula above even if it may seem unnecessary (but as we'll see later it is actually important for other mappings, and this is simply a special case where we have the standard delta function). Now let us compute the area of the translated sphere. Going to Cartesian coordinates gives us: $$p_B(x,y,z) = \frac{p_A(r,\theta)}{|r^2\sin\theta|} = \delta(\sqrt{x^2+y^2+z^2}-\rho)$$ (we have used the invertible pdf transformation theorem). Translating by $$(0,0,\rho)$$ yields: $$p_C(x,y,z) = p_B(x,y,z-\rho)$$, where the Jacobian of this transformation is $$1$$. Finally going back to spherical coordinates we have: $$p_D(r,\theta) = \delta(\sqrt{r^2\cos^2\phi\sin^2\theta + r^2\sin^2\phi\sin^2\theta + r^2\cos\theta^2 + \rho^2 - 2r\rho\cos\theta}-\rho)r^2\sin\theta = \\ = \delta(\sqrt{r^2+\rho^2-2r\rho\cos\theta}-\rho)r^2\sin\theta$$ We compute the gradient of the mapping as: $$\nabla g(r,\theta) = \frac{1}{2\sqrt{r^2+\rho^2-2\rho\cos\theta}}(2r-2\rho\cos\theta, 2\frac{r}{r}\sin\theta,0)$$. Finally $$|\nabla g(r,\theta)| = 1$$. We may compute $$g^{-1}(0) = \{(2\rho\cos\theta, \theta, \phi),\theta \in [0,\frac{\pi}{2}], \phi \in [0,2\pi]\}$$. The surface area element is $$dA = 2\rho^2\sin2\theta\,d\theta\,d\phi$$. Then finally: $$\int_{0}^{2\pi}{\int_{0}^{\frac{\pi}{2}}{\int_{0}^{\infty}{p_D(r,\theta)\,dr}\,d\theta}\,d\phi} = \int_{0}^{2\pi}{\int_{0}^{\frac{\pi}{2}}{\rho^2\sin2\theta\,d2\theta}\,d\phi} = 4\pi\rho^2$$

Now let us consider a slightly different variant: $$p_A(r) = \delta(r^2 - \rho^2)r^2\sin\theta$$, $$|\nabla g(r)| = 2r$$ $$\int_{0}^{2\pi}{\int_{0}^{\pi}{\int_{0}^{\infty}{\delta(r^2-\rho^2)r^2\sin\theta\,dr}\,d\theta}\,d\phi} = \int_{0}^{2\pi}{\int_{0}^{\pi}{\frac{\rho^2\sin\theta}{2\rho}\,d\theta}\,d\phi} = 2\pi\rho$$

Rather surprisingly (at least for me) we get a different result, which however for the $$\delta$$ defined as is, is supposedly correct (I believe that the result being $$2\pi\rho$$ is just a lucky coincidence). So one has to be careful about the mapping.

After transforming to cartesian coordinates, translating and returning to spherical coordinates we get $$p_D(r, \theta) = \delta(r^2-2\rho\cos\theta)r^2\sin\theta$$, $$|\nabla g(r,\theta)| = 2\sqrt{r^2+\rho^2-2r\rho\cos\theta}$$. Using the coarea formula once again:

$$\int_{0}^{2\pi}{\int_{0}^{\frac{\pi}{2}}{\int_{0}^{\infty}{p_D(r,\theta)\,dr}\,d\theta}\,d\phi} = \int_{0}^{2\pi}{\int_{0}^{\frac{\pi}{2}}{\frac{\rho^2\sin2\theta}{2\sqrt{\rho^2}}\,d2\theta}\,d\phi} = 2\pi\rho$$

In conclusion, it seems that it is not correct to substitute directly when the delta function is composed with a function different than the identity (or $$\pm const$$). In that specific case the coarea formula has to be used. Additionally we seem to have the relationship $$\delta_S(x) = \delta(g(x))|\nabla g(x)|$$, where $$S=g^{-1}(0)$$:

$$\int_{R^n}{f(x)\delta(g(x))|\nabla g(x)|\,dx} = \int_{R^n}{f(x)\delta_S(x)\,dx} = \int_{S}{f(x)\,d\sigma(x)}$$

I very much appreciate the input from Maxim and David Holden, which ultimately helped me figure this out.

Edit: A very interesting read I found later: https://www.mathpages.com/home/kmath663/kmath663.htm It certainly helps to understand the problem from an intuitive point of view also.

To summarize the discussion in the comments, the definition of $$\delta(f)$$ is derived from postulating two basic properties: the substitution rule $$\int_{\mathbb R^n} \delta(f(\mathbf x)) \,\phi(\mathbf x) \,d\mathbf x = \int_U \delta(f(\mathbf x(\mathbf u))) \,\phi(\mathbf x(\mathbf u)) \left| \det D \mathbf x(\mathbf u) \right| d\mathbf u$$ and $$\int_{\mathbb R^n} \delta(x_1) \,\phi(\mathbf x) \,d\mathbf x = \int_{\mathbb R^{n - 1}} \phi(\mathbf x) \rvert_{x_1 = 0} \,dx_2 \cdots dx_n.$$ If you try to set $$\small \int \delta(f(\mathbf x)) d\mathbf x = \int_{f(\mathbf x) = 0} dS = \int_{2 f(\mathbf x) = 0} dS = \int \delta(2 f(\mathbf x)) d\mathbf x,$$ you violate the first rule. If you try to set $$\small \iint \delta(r - f(\theta)) \phi(r, \theta) dr d\theta \neq \int \phi(f(\theta), \theta) d\theta,$$ you violate the second rule. If you adopt the definition that the rest of the world is using derived from the two stated properties, you get the identity $$\int_{\mathbb R^n} \delta(f(\mathbf x)) \left| \nabla f(\mathbf x) \right| \phi(\mathbf x) \,d\mathbf x = \int_{f(\mathbf x) = 0} \phi(\mathbf x) \,dS(\mathbf x),$$ which is formally the same as the coarea formula because both are essentially the same change of variables formula. The first two formulas in your question will in fact be correct, while the last two will be incorrect.

• There are some small details, like the coarea formula being more general than the substitution. I think Federer developed it not so long ago too either. But for the problem at hand both work equally well. Thank you for all of the input. After I figured everything out, I found many references of this in physics books, so it certainly helps asking the right question. – lightxbulb Jan 29 at 19:32
• Also what first two formulae are you referring to? The second equality in my question is wrong. All the equalities in my answer should be correct however. @Maxim – lightxbulb Jan 29 at 22:10
• I refer to the formulas $\iiint = 4 \pi \rho^2$ and $\iiint = 8 \pi \rho^2/3$. If you still insist that the second one is wrong, then we haven't made any progress. I specifically tried to explain above why insisting that it should give some other answer is incompatible with the formula you're trying to derive. – Maxim Jan 29 at 22:43
• I am not sure how you managed to integrate that thing. Using the coarea formula the second formula gives you $|\nabla f| = \sqrt{1 + 4\rho^2\sin^2\theta / r^2}$ in the denominator. And $2\rho\sin2\theta\,d\theta\,d\phi$ in the numerator with integration limits $\phi \in [0,2\pi]$ and $\theta \in [0,\pi/2]$. My issue is really that $r^2$ in the denominator. I am assuming it's supposed to be $r = 2\rho\cos\theta$. But even then I don't think you'll get $8\pi\rho^2/3$. – lightxbulb Jan 30 at 0:33
• First, $dS = 2 \rho^2 \sin 2 \theta \,d\theta d\phi$. Then $$2 \pi \int_0^{\pi/2} \frac {2 \rho^2 \sin 2 \theta} {\sqrt{1 + \frac {4 \rho^2 \sin^2 \theta} {r^2}}} \bigg\rvert_{r = 2 \rho \cos \theta} \,d\theta = \frac {8 \pi \rho^2} 3.$$ But this exercise is completely pointless if you ignore the reason why $$2 \pi \int_0^{\pi/2} \int_0^\infty \delta(r - 2 \rho \cos \theta) \,r^2 \sin \theta \,dr d\theta = 2 \pi \int_0^{\pi/2} r^2 \sin \theta \rvert_{r = 2 \rho \cos \theta} \,d\theta$$ gives the same result. – Maxim Jan 30 at 23:59