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Double Integrals Over Regions in the Plane

How to think about integrals over regions in the plane
When thinking about integrals over regions in the plane, the first thing I think of is a roller coaster. The xy plane is the base of the roller coaster, and as you walk around under it, the roller coaster changes height. This is how functions of the form f (x, y) generally behave; you can pick any "base" or domain for the roller coaster and find the volume of the resulting three-dimensional shape using double integration with respect to x and y. You can also approximate the volume by taking a rectangular region R around your domain and splitting it into a number of boxes, as discussed below.
(Image from https://www.google.com/url?sa=i&url=https%3A%2F%2Fwww.architecturaldigest.com%2Fgallery%2Froller-coaster-designs&psig=AOvVaw2XBThbDuZkz6JaLWpHsX9p&ust=1632786586027000&source=images&cd=vfe&ved=0CAsQjRxqFwoTCICUnZ_qnfMCFQAAAAAdAAAAABAD)

How to compute the volume of the approximation
The double integral of f (x, y) over R is given by Rf(x,y)dA=limm,nmi=1nj=1f(xij,yij)ΔxΔy. Here, the region R is given by [a, b] x [c, d]: [a, b] is divided into [xi,xi+1] which has length bam. Similarly, [c, d] is divided into [yj,yj+1] which has length dcn. Pick a sample point (xij,yij) in Rij. Then, f(xij,yij)(bamdcn) approximates the volume using a double Riemann sum.

(Image from https://www.usna.edu/Users/oceano/raylee/SM223/Ch15_1_Stewart(2016).pdf#:~:text=Volumes%20and%20Double%20Integrals%20The%20sum%20in%20Definition,for%20a%20function%20of%20a%20single%20variable.%5D%2016)

For example, consider the function f(x,y)=13(x2)(2y) on the plane bounded by y=ex5 and y=ln(x)+5 from [0, 7] x [0, 7].
This region has been split into boxes such that m and n are both equal to 7. Using upper-right endpoints (where if the point lies within R but outside of the domain D, the height is zero), we compute the following: V7i=17j=1f(xij,yij)ΔxΔy =f(1,1)ΔA+f(1,2)ΔA+f(1,3)ΔA... =0.671+2.671+61+... =1182.02units3
How to compute the actual volume using integration
There are two scenarios to think about when finding the actual volume with integration. They are pictured below (source: https://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx ):
For our function, we are working with scenario 1. For this region "D" given by D=(x,y)|axb,g1(x)yg2(x) [a,b] x [c,d], the following integration applies: Df(x,y)dA=RF(x,y)dA=badcF(x,y)dydx=bag2(x)g1(x)F(x,y)dydx=bag2(x)g1(x)f(x,y)dydx. "R" is the rectangle [0, 7] x [0, 7] defined in our example that we used for the approximation. Also, the above comes from Fubini's theorem (source: https://sites.math.washington.edu/~morrow/335_12/fubini.pdf), which states that if f is continuous (like our function f (x, y) is), then Rf(x,y)dA=badcf(x,y)dydx=dcbaf(x,y)dxdy. Now on to the computation! 70ln(x)+5ex513(x2)(2y)dydx=70e5(lnxex5+5)(e5x2lnx+x2ex+5e5x2)3dx =e10(12348e10ln2(7)+115248e10ln(7)2295e14+270248e10+273241283.95units3.
Why I chose this function
My research is in developing cancer treatments against a colorectal cancer cell line LS174T CRC. Colorectal cancer is the third most common diagnosed malignancy and the fourth-leading cause of death worldwide. Within the next 10 years, this cancer's burden on society is projected to increase by 60% with a survival rate of only 14.2% today.1 Taking what I've learned working with this cancer, I decided to make a function that (very roughly) demonstrates its tumor growth.
In order for a tumor to grow, it needs energy. We get energy from glucose and the production of ATP. Glucose is taken into cells by a transporter called GLUT1 and ATP is generated through oxidative phosphorylation. Treatments exist that limit GLUT1 activity (called BAY-876), so I developed the synthetic route for a new drug (DBI-1) that can disrupt oxidative phosphorylation. Together, this drug cocktail should (hopefully) treat colorectal cancer more effectively. In terms of the function I picked, if x represents the untreated oxidative phosphorylation and y represents the treated GLUT1 transporter, the value of f (x, y) should increase quickly in the x-direction and more slowly in the y-direction. I would model the combined treatment data on tumor growth, but I don't have these results back yet! Here, my DBI-1 is pictured for my fellow chemistry nerds:
I messed with different coefficients and exponents for a while before deciding on the ones to use. The value of the function f (x, y) represents relative tumor size at certain values of ATP production (x) and glucose uptake (y) and best fits the desired behavior for growth with respect to each variable. The tricky thing about finding a good function was that we don't have data for "how much glucose" and "how much ATP" -- these are hard things to measure, but it's easy to measure size! That's why I worked backwards and found a function with the correct shape first to demonstrate the relationship between these variables instead of the exact values; we know our tumor size and can then see how much glucose we may have compared to ATP at that point to determine which drug (BAY-876 or DBI-1) is most effective there. I chose the given domain 1) for the fun shape and 2) because there is a limit in biology on how big a tumor gets before an organism dies, so this was a reasonable bound.
Here is the 3D model. In the y-direction, we can see the steps are more gradual than the x-direction, as expected. From the top, we can see the leaf-like shape of the domain. I used centimeters to make a 7x7 cm region, but the print will be scaled down to accomodate for its height being too tall for the printers.


Reference List
1. Julianne E. Jenkins. Colorectal Cancer: Risk, Diagnosis and Treatments. Nova; 2011. Accessed September 27, 2021. https://search-ebscohost-com.ezproxy.uky.edu/login.aspx?direct=true&db=nlebk&AN=440798&site=ehost-live&scope=site

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