What is the integral of three dimensional dirac delta function over a sphere?

Question

Assum $x\in\mathbb R^3$, $t>0$, what is $$\cfrac{1}{4\pi t}\int_{\partial B(x,t)}\delta(y)\,dS(y)$$ where $\delta$ is three dimensional delta function and $\partial B(x,t)$ is the sphere centered at $x$ with radius $t$? Is it some how lower dimension delta function? Thanks

Answer 1

You have to check how this quantity behaves when integrating against test functions, which is how one defines the Dirac delta function and other distributions. Let us set

$$D(x) = \frac{1}{4\pi t} \int_{\partial B(x,t)} \delta(y) \, dS(y).$$

Then $D$ acts on a test function $\phi\in C^\infty_c(\mathbb{R}^n)$ by

$$\langle D,\phi\rangle := \int_{\mathbb{R}^n} D(x) \phi(x)\, dx = \int_{\mathbb{R}^n} \frac{1}{4\pi t} \int_{\partial B(x,t)} \delta(y)\, dS(y) \,\phi(x) \,dx.$$

To see what we get, we change variables $z=y-x$ in the inner integral to obtain

$$\langle D,\phi\rangle = \int_{\mathbb{R}^n} \frac{1}{4\pi t} \int_{\partial B(0,t)} \delta(x+z)\, dS(z) \,\phi(x) \,dx.$$

Now we can swap the order of integration:

$$\langle D,\phi\rangle = \frac{1}{4\pi t} \int_{\partial B(0,t)}\int_{\mathbb{R}^n} \delta(x+z)\phi(x) \, dx\, dS(z) = \frac{1}{4\pi t}\int_{\partial B(0,t)}\phi(-z)\, dS(z),$$

where the last equality is by the definition of the Dirac delta function $\delta$. We can of course replace $-z$ by $z$ by symmetry to obtain

$$\langle D,\phi\rangle = \frac{1}{4\pi t}\int_{\partial B(0,t)}\phi(z)\, dS(z).$$

So $D$ is the distribution that takes a test function $\phi$ and integrates it over the sphere $\partial B(0,t)$ and then divides by $4\pi t$. You can think of this as a delta function supported on $\partial B(0,t)$, i.e.,

$$D(x) = \frac{1}{4\pi t}\delta_{\partial B(0,t)}(x).$$

EDIT: The notation $\delta_A$ for $A\subset \mathbb{R}^n$ in this case refers to the distribution that acts on test functions by

$$\langle \delta_A,\phi\rangle = \int_A \phi \, dx.$$