Volume Under an Egg Carton

 

Fubini's Theorem asserts that for a function $f(x,y)$ the volume under the surface which is bounded by a region $[a,b] \times [c,d]$ can be found either by first integrating with respect to y, or with respect to x. In either case, the other variable is treated like a constant. 

This means that if we make the bounds of y between c and d, and the bounds of x between a and b, then the two following integrals are equivalent!

$$\int_{c}^{d}\int_{a}^{b}{f(x,y)dxdy}$$

$$\int_{a}^{b}\int_{c}^{d}{f(x,y)dydx}$$

This really only works easily for continuous functions. The fact that we can integrate x and y (or any collection of independent variables) in any order is surprisingly fundamental to quantum chemistry. For the three-dimensional wavefunction for electronic orbitals for a molecule, to determine the electronic surface, each electron is given a volume integral over the whole space, for each spin. If we couldn't integrate freely in any direction, matter as we know it wouldn't exist!

Here, $\Psi_1$ represents wavefunctions for electrons with spin 1 and $\Psi_2$ represents electrons with spin 2, s represents energetic states, and r represents a basis set of functions for each particle in an N particle system. 

$$(\Psi_1, \Psi_2)  = \sum_{s_N}^{} \dots \sum_{s_2}^{} \sum_{s_1}^{} \iiint_V{r_1 dxdydz } \dots \iiint_V{r_N dxdydz}$$

Additionally, a region can also be bounded by functions of a certain variable. For instance, we could force the bounds of y to be $g(x)$ and $h(x)$ to yield the integral below: 

$$\int_{a}^{b}\int_{g(x)}^{h(x)}{f(x,y)dydx}$$

However, for the purpose of brevity, I choose to enclose my function in a rectangular region. I was thinking about how much I liked egg cartons, but how I hated eggs. Likewise, I didn't want to do anything as high flying and complicated as a wavefunction, though it would have been fun. So, I decided to make a surface which looked like an egg carton. Below I give my surface, which I created with Geogebra. 

I chose $\displaystyle f(x,y) = 3\cos(x) - 4\cos(y)$ and bound x and y between $[0,4]$ and $[0,4]$ respectively since it roughly captured the geometry of an egg carton. 

I can find the volume under this region with the following integral:

$$V = \int_{0}^{4}\int_{0}^{4}{3\cos(x) - 4\cos(y) dxdy}$$

$$V = \int_{0}^{4}{3\sin(x) -4x\cos(y)\Big|_{0}^{4}dy}$$

$$V = \int_{0}^{4}{3\sin(4)-16\cos(y)dy}$$

$$V = (3y\sin(4) - 16\sin(y)\Big|_{0}^{4}$$

$$V=(3*4\sin(4) -16\sin(4)$$

$$V = -4\sin(4) \approx 3.027$$

My approximation had a high error, and the computed volume was $V \approx 4.8 u^3$. I believe this was due to the high step size, and that integrating over a more dense grid would solve this problem.

To make a model of this surface, I divided the region into 20 squares and found the righthand endpoints. My model spanned 4 inches in both the x and y directions, and spanned almost 11 inches from top to bottom due to the height of the wave in the positive and negative z direction. 

Once again, I used Python to calculate the how $f(x,y)$ changed over the region with my chosen subdivision. I give the output of this code below, as written to a CSV file, made from a Pandas DataFrame. 

My code only spans 20 lines, so I supply it as an image, rather than an interactable widget.