# Integration of DiracDelta

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 $4\pi$?

$$\int_{-2}^2 \delta(x^2+y^2+z^2-1)~ dx dy dz = \int_0^\infty dr \int_0^\pi d\theta \int_0^{2\pi}d\phi ~r^2\sin\theta~ \delta(r^2-1)$$ $$= \frac{1}{2}\int_0^\infty d(r^2) \int_0^\pi d\theta \int_0^{2\pi}d\phi ~r\sin\theta~ \delta(r^2-1) = \frac{1}{2}\int_0^\pi d\theta \int_0^{2\pi}d\phi ~ \sin\theta= 2\pi$$
• I think it would be nice to replace the integration region in the first integral by R^3 though. – sebhofer Feb 6 '14 at 9:25