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I just came across the following curiosity, and I'm not sure whether it's a bug or just me missing something. Here's a minimal example: I defined the following "function"

In[1]:= S[x_, y_, z_] := DiracDelta[x^2 + y^2 + z^2 - 1];

which is supposed to parametrize the surface of a unit sphere. However, when I integrate over the entire space ($x,y,z \in [-2,2]$ should do) I get $2\pi$ as a result

In[2]:= Integrate[S[x, y, z], {x, -2, 2}, {y, -2, 2}, {z, -2, 2}]

Out[2]:= 2 Pi

Shouldn't this be rather $4\pi$?

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I think Mathematica is correct here:

$$\int_{-2}^2 \delta(x^2+y^2+z^2-1)~ dx dy dz = \int_0^\infty dr \int_0^\pi d\theta \int_0^{2\pi}d\phi ~r^2\sin\theta~ \delta(r^2-1)$$ $$= \frac{1}{2}\int_0^\infty d(r^2) \int_0^\pi d\theta \int_0^{2\pi}d\phi ~r\sin\theta~ \delta(r^2-1) = \frac{1}{2}\int_0^\pi d\theta \int_0^{2\pi}d\phi ~ \sin\theta= 2\pi$$

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  • $\begingroup$ Hmpf... you beat me to it ;) +1 $\endgroup$ – sebhofer Feb 6 '14 at 9:23
  • $\begingroup$ I think it would be nice to replace the integration region in the first integral by R^3 though. $\endgroup$ – sebhofer Feb 6 '14 at 9:25
  • $\begingroup$ Damn, you are absolutely right! In order to get 4*Pi I would have to replace the argument of the delta"function" by (Sqrt[x^2+y^2+z^2]-1). Cheers! $\endgroup$ – André Feb 6 '14 at 9:39
  • $\begingroup$ Unexpected but true. Here's a Generalisation: In[175]:= Integrate[ DiracDelta[(x^2 + y^2 + z^2)^a - 1], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}, {z, -Infinity, Infinity}] Out[175]= (2*Pi)/Abs[a] $\endgroup$ – Dr. Wolfgang Hintze Aug 17 '14 at 20:55

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