I recently rediscovered generalized functions and distributions as generalized functions.
I found out many integrals "over the charge distribution" can instead be recast as integrals over all space and replace a density function with a density distribution. For example: The surface area of a sphere. Normally we begin computing the area of the sphere by choosing our coordinates. If we choose the $r\theta\phi$ coordinates, then we may write (for a sphere of radius R centered on the origin):
$$ S = \iint\ dA = \int_{\theta=0}^\pi \int_{\phi=0}^{2\pi}R^2\sin\theta\ d\theta d\phi = 4\pi R^2 $$
In terms of a distribution, we may write that the distribution of the sphere is $\rho=\delta(r-R)$. Now we may write:
$$ S=\iiint \rho\ dV =\int_{r=0}^{\infty} \int_{\theta=0}^\pi \int_{\phi=0}^{2\pi} \delta(r-R) r^2\sin\theta\ d\theta d\phi $$
Which also give the correct result of $S=4\pi R^2$ since $\int_{r=0}^\infty r^2\delta(r-R)dr=R^2$.
I also noted that if we move back to $xyz$ coordinates, we have $\rho=\delta(\sqrt{x^2+y^2+z^2}-R)$. We can employ the expansion $\delta(g(\alpha))=\sum_i \frac{\delta(\alpha-\beta_i)}{\left|g'(\beta_i)\right|}$ where each of the $\beta_i$ are roots of $g(\alpha)$. With respect to $x$, the solutions to $g(x)=0$, where $g(x)=\sqrt{x^2+y^2+z^2}-R$, are given by $\beta_{\pm}=\pm\sqrt{R^2-y^2-z^2}$. The derivative can be evaluated as follows: $g'(x)=\frac{x}{\sqrt{x^2+y^2+z^2}}$. So $g'_+(\beta_+)=\frac{\sqrt{R^2-y^2-z^2}}{R}$ and $g'_-(\beta_-)=\frac{-\sqrt{R^2-y^2-z^2}}{R}$. This gives us:
$$ S=\iiint\ \rho\ dV=\iiint \delta\left(\sqrt{x^2+y^2+z^2}-R\right)\ dxdydz\\ =\iiint\ \frac{\delta(x-\beta_+)}{|g'(\beta_+)|}+\frac{\delta(x-\beta_-)}{|g'(\beta_-)|}\ dxdydz\\ =\iiint\ \frac{R\delta(x-\sqrt{R^2-y^2-z^2})}{\sqrt{R^2-y^2-z^2}} + \frac{R\delta(x+\sqrt{R^2-y^2-z^2})}{\sqrt{R^2-y^2-z^2}}\ dxdydz\\ =\iiint\ \frac{R}{\sqrt{R^2-y^2-z^2}} \left[ \delta(x-\sqrt{R^2-y^2-z^2})+\delta(x+\sqrt{R^2-y^2-z^2}) \right] dxdydz\\ $$
It would seem to be the case that $\int_{x=-\infty}^{\infty}\delta(x-\sqrt{R^2-y^2-z^2})dx=1$ and $\int_{x=-\infty}^{\infty}\delta(x+\sqrt{R^2-y^2-z^2})dx=1$ so that the above integral becomes:
$$ S=\iint\ \frac{2R}{\sqrt{R^2-y^2-z^2}}dydz $$
This does not converge according to Wolfram Alpha as long as the bounds are unaffected. Thus clearly there is something wrong with this derivation. I hypothesize there is some other step involved with solving for the roots or the domain of integration on the yz coordinates is otherwise affected. Perhaps the integrals in $x$ on the delta functions do not sum to 2.
Begining from
$$ S= \iiint\ \frac{\delta(x-\beta_+)}{|{\bf \nabla}g(\beta_+)|}+\frac{\delta(x-\beta_-)}{|{\bf \nabla} g(\beta_-)|}\ dxdydz\\ $$
Here ${\bf \nabla} g = {\bf \hat{r}}$, so $|\nabla g|=1$, and thus we have for S:
$$ S= \iiint \delta(x-\sqrt{R^2-y^2-z^2})+\delta(x+\sqrt{R^2-y^2-z^2}) \ dxdydz $$
Additional question:
Is there a general method for transforming between integrating over some specified region, such as a curve or surface, to integrating over all space and expressing the domain of integration in terms of appropriate distribution functions?
