"The question at hand is on the very bottom of this post."
The double integral here is an example I came up with.
$$V=\int _0^2\int _0^12x^2+3y^2-5\kern10.0ptdxdy\\ \int _0^2\left[ \int _0^12x^2+3y^2-5\kern10.0ptdx\right] dy \\\left[ \frac{2}{3}x^3+3xy^2-5x\right] _{x=0}^{x=1}\\ \left[ \frac{2}{3}+3y^2-5\right] \\ \int _0^2\frac{2}{3}+3y^2-5\kern10.0ptdy\\ \left[ \frac{2}{3}y+y^3-5y\right] _{y=0}^{y=2}\\ \left[ \frac{4}{3}-2\right] \\ \left[ -\frac{2}{3}\right] $$
Now if you want to in essence prove that the answer$$-\frac{2}{3}$$ is correct and show that by way of riemann sums, then there is a formula you can use. However I have only seen it used once. So here it is. $$V={\lim_{n\to \infty}} \sum _{i=1}^n {\lim_{m\to \infty}}\sum _{j=1}^mf\left( x_i^{\times },y_j^{\times }\right) \Delta y\Delta x$$ This is the double riemann sum formula. It is not mentioned anywhere on the internet except once and never mentioned in an advanced calc course. But anyway, the steps are pretty much the same for any multivariable problems when using this formula.
It is not that much different when computing a single definite integral using the regular limit of riemann sum formula. Let's look how it can be used to show that the double integral is correct.
$$f=2x^2+3y^2-5\kern10.0pt;\kern10.0pt\left[ 0,1\right] \times \left[ 0,2\right]$$ I'm using right endpoints here. $$x_i^{\times }=\frac{i}{n};\kern10.0pty_j^{\times }=\frac{2j}{m}$$
- just represents the binary operation of the two riemann sums.
Delta x is. $$\Delta x=\frac{b-a}{n}\\ \Delta x=\frac{1}{n}$$ Delta y is. $$\Delta y=\frac{d-c}{m}\\ \Delta y=\frac{2}{m}$$
We can compute the double riemann sum now.
$$\left[{\lim_{n\to \infty}} \sum _{i=1}^n\left[{\lim_{m\to \infty}}\sum _{j=1}^mf\left( x_i^{\times },y_j^{\times }\right) \Delta y\right] \Delta x\right]$$
You just focus on the most inner riemann sum first like you would with multiple integrals.
Now.
$${\lim_{m\to \infty}}\sum _{j=1}^mf\left( x_i^{\times },\frac{2j}{m}\right) \frac{2}{m}\\ {\lim_{m\to \infty}}\sum _{j=1}^m\left[ 2x^2+3\left( \frac{2j}{m}\right) ^2-5\right] \frac{2}{m}\\ {\lim_{m\to \infty}}\sum _{j=1}^m\left[ 2x^2+3\left( \frac{2j}{m}\right) \left( \frac{2j}{m}\right) -5\right] \frac{2}{m}\\ {\lim_{m\to \infty}}\sum _{j=1}^m\left[ 2x^2+\left( \frac{6j}{m}\right) \left( \frac{2j}{m}\right) -5\right] \frac{2}{m}\\ {\lim_{m\to \infty}}\sum _{j=1}^m\left[ 2x^2+\frac{12j^2}{m^2}-5\right] \frac{2}{m}\\ {\lim_{m\to \infty}}\sum _{j=1}^m\left[ \frac{4x^2}{m}+\frac{24j^2}{m^3}-\frac{10}{m}\right] \\ {\lim_{m\to \infty}}\left[ \frac{x^2}{m}\sum _{j=1}^m4m+\frac{24}{m^3}\sum _{j=1}^m\frac{m\left( m+1\right) \left( 2m+1\right) }{6}-\frac{1}{m}\sum _{j=1}^m10m\right] \\ {\lim_{m\to \infty}} \left[ \frac{4mx^2}{m}+\frac{24m\left( m+1\right) \left( 2m+1\right) }{6m^3}-\frac{10m}{m}\right] \\ {\lim_{m\to \infty}}\left[ 4x^2+\frac{4\left( m+1\right) \left( 2m+1\right) }{m^2}-10\right] \\ {\lim_{m\to \infty}}\left[ 4x^2+\frac{\left( 4m+4\right) \left( 2m+1\right) }{m^2}-10\right] \\ {\lim_{m\to \infty}}\left[ 4x^2+\frac{8m^2+4m+8m+4}{m^2}-10\right] \\ {\lim_{m\to \infty}}\left[ 4x^2+\frac{8m^2}{m^2}-10\right] \\ \left[ 4x^2+8-10\right] \\ \left[ 4x^2-2\right]$$
We computed the first riemann sum and now have the function $$\left[ 4x^2-2\right] $$ left which is now just a standard riemann sum.
We now use that function to compute for $$\Delta x$$
So this is what we have left.
$$\left[ 4x^2-2\right] \\ {\lim_{n\to \infty}}\sum _{i=1}^nf\left( \frac{i}{n}\right) \frac{1}{n}\\ {\lim_{n\to \infty}}\sum _{i=1}^n\left[ 4\left( \frac{i}{n}\right) ^2-2\right] \frac{1}{n}\\ {\lim_{n\to \infty}}\sum _{i=1}^n\left[ 4\left( \frac{i}{n}\right) \left( \frac{i}{n}\right) -2\right] \frac{1}{n}\\ {\lim_{n\to \infty}}\sum _{i=1}^n\left[ \left( \frac{4i}{n}\right) \left( \frac{i}{n}\right) -2\right] \frac{1}{n}\\ {\lim_{n\to \infty}}\sum _{i=1}^n\left[ \frac{4i^2}{n^2}-2\right] \frac{1}{n}\\ {\lim_{n\to \infty}}\sum _{i=1}^n\left[ \frac{4i^2}{n^3}-\frac{2}{n}\right] \\ {\lim_{n\to \infty}}\left[ \frac{4}{n^3}\sum _{i=1}^n\frac{n\left( n+1\right) \left( 2n+1\right) }{6}-\frac{1}{n}\sum _{i=1}^n2n\right] \\ {\lim_{n\to \infty}}\left[ \frac{4n\left( n+1\right) \left( 2n+1\right) }{6n^3}-\frac{2n}{n}\right] \\ {\lim_{n\to \infty}}\left[ \frac{2\left( n+1\right) \left( 2n+1\right) }{3n^2}-2\right] \\ {\lim_{n\to \infty}}\left[ \frac{\left( 2n+2\right) \left( 2n+1\right) }{3n^2}-2\right] \\ {\lim_{n\to \infty}}\left[ \frac{4n^2+2n+4n+2}{3n^2}-2\right] \\ {\lim_{n\to \infty}}\left[ \frac{4n^2}{3n^2}-2\right] \\ \left[ \frac{4}{3}-\frac{6}{3}\right] \\ \left[ -\frac{2}{3}\right]$$
This is the answer using the limit of the double riemann sum formula. If you want to do the limit of a triple riemann sum, then add another riemann sum to this formula. It will be the same process with an extra step.
As for an inquiry as I know this was mainly an educational post, why is this not really taught? I could see it being useful with plotting rectangles in 2 dimensional and 3 dimensional objects. You can draw out a 2 dimensional riemann sum with the formula I used if you really want to see riemann sums in action on 2d planes. Suffice to say, maybe just maybe will someone here take interest in this formula I brought to light.