Verify this distribution convolution: $E(t,x)\ast (g(x)\delta(t)) = t\int_{\omega\in S^2}{\frac{g(x-t\omega)}{4\pi}dS(\omega)}$ In our class notes we are asked to verify the following equality:

$$E(t,x)\ast (g(x)\delta(t)) = t\int_{\omega\in S^2}{\frac{g(x-t\omega)}{4\pi}dS(\omega)}$$
where $E(t,x)=\frac{1}{4\pi|x|}\delta(t-|x|)$ is the fundamental solution to the wave equation in $\mathbb{R}^3$.

My problem with this is that I am not sure how to deal with the convolution of these two distributions.
First, seeing as they both have compact support, then their convolution (as distributions) is well defined. Moreover, since $f(x)\delta_{a}(x)\ast g(x)\delta_{b}(x) = f(a)g(b)\delta_{a+b}(x)$, where $\delta_{a}(x)=\delta(x-a)$, then I would conclude that $$E(t,x)\ast(g(x)\delta(t)) = \frac{g(x)}{4\pi|x|}\delta(t-|x|).$$ However this is still only a distribution, and I cannot figure out how this would be equivalent to the given solution.
Second, if we appeal to the integral definition of the convolution, then we have $$E(t,x)\ast(g(x)\delta(t)) = \frac{1}{4\pi}\int_{\mathbb{R}^3}\int_{-\infty}^{\infty}{\frac{1}{|y|}\delta(s-|y|)g(x-y)\delta(t-s)}dsdy$$ which gives us an integral involving two delta distributions, again leaving me at a loss for how to proceed.
Any help in clarifying this equality would be greatly appreciated.
 A: So I finally figured this out, and figured I'd post the solution for future reference:
Going with the integral definition of convolution, we have
\begin{align}
E(t,x)\ast(g(x)\delta(t)) &= \frac{1}{4\pi}\int_{\mathbb{R}^3}\int_{-\infty}^{\infty}{\frac{1}{|y|}g(x-y)\delta(s-|y|)}\delta(t-s)dsdy\\
&= \frac{1}{4\pi}\int_{\mathbb{R}^3}|y|^{-1}g(x-y)\left(\int_{-\infty}^{\infty}{\delta(s-|y|)\delta(t-s)}ds\right)dy\\
&= \frac{1}{4\pi}\int_{\mathbb{R}^3}|y|^{-1}g(x-y)\left(\delta_{|y|}(t)\ast\delta_{0}(t)\right)dy\\
&= \frac{1}{4\pi}\int_{\mathbb{R}^3}|y|^{-1}g(x-y)\delta(t-|y|)dy\\
&= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^{\infty}r^{-1}g(x-\tilde{y})\delta(t-r)r^{2}\sin\phi dr d\theta d\phi\\
&= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}\sin\phi\left(\int_{0}^{\infty}rg(x-\tilde{y})\delta(t-r)dr\right) d\theta d\phi\\
&= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}\sin\phi\left\langle\delta(t-r),rg(x-\tilde{y})\right\rangle d\theta d\phi\\
&= \frac{1}{4\pi}\int_{0}^{\pi}\int_{0}^{2\pi}tg(x-t\omega)\sin\phi d\theta d\phi\\
\end{align} where $\omega\in S^{2}$, and since $\delta(t-r)=\delta_{t}(r)$. But $\sin\phi d\theta d\phi = dS(\omega)$, so we get the final equality $$\frac{t}{4\pi}\int_{\omega\in S^{2}}{g(x-t\omega)dS(\omega)}.$$
