Examining the Use of Limits in 3D Applications
Introduction
As a student in my school’s (Concordia International School Shanghai) Honors Precalculus course, we were assigned a large end-of-year project. This year’s project had two main parts: Apply the knowledge we learned in the course this year in any sort of practical application, and find the surface area of some 3D object.
The limits we learned in class were mostly one-dimensional, concerning lines and points. Even the most complicated concept we learned, the Riemann sum, only applied to two dimensions, leaving the third dimension largely unexplored. I thought that I could try to extrapolate what I learned and expand one dimension further.
I felt that I could implement my idea and combine the two parts of the project if I examined a cup. As volume and surface area are both important metrics for any cup, I could try to apply Riemann sums to volume in the first part (Part A) and then find the surface area of the cup in the second part (Part C). To make this project more challenging, I decided to model the cup as a section of a paraboloid, which is essentially a 3D parabola.
As I use the same cup in both parts, I will start by defining this cup. The sides of the cup will be modeled by the 3D function f(x,y)=x²+y², bounded by the planes f(x,y)=15 and f(x,y)=30. The cup would then look something like this:
Remember that in 3D space, the z-axis is the vertical (up-down) axis while the x and y axes are horizontal.
Part A: Volume
To apply Riemann sums to volume, I can take advantage of the fact that the cup is circular. This means that I can use cylinders in place of rectangles. For example, if I estimate the volume of the cup with 3 cylinders (shown below):
I would get a volume estimation of 375π cubic units.
This is an overestimation, and I will slightly deviate from the topic to discuss different methods of estimation.
- Overestimation: This is the method I demonstrate above. Overestimations are done when the largest value is set as the width of the rectangle or in this case the radius of the cylinder. This means that the cylinders will always be larger than the actual volume, resulting in an overestimation.
- Underestimation: Underestimations are done when instead the smallest value is set as the radius of the cylinder. This means that the cylinders will always be smaller than the actual volume, resulting in an underestimation.
- Average: This method takes the average of the overestimate and underestimate. This is generally the most accurate method, as it somewhat cancels out the deviation of the other methods by adding them together.
As I will be using an infinite number of infinitely thin cylinders, all methods would approach the same volume. Here I will use the top of the cylinder and (initially) overestimate.
To find the radii of the cylinders, I can work with the cross-section of the paraboloid, as the paraboloid is circular when viewed from above. For the parabola, I will use the z-axis as the y-axis when graphing in the x-y plane while I keep the x-axis the same. The cross-section of the paraboloid can be modeled by the function f(x)=x².
The cup would be bounded by the lines h and g. When the parabola is graphed, it is evident that the radius is the x value at the given z value. If I use the inverse function of the parabola, I can flip this relationship and the radius would instead be the z value at the given x value. This would be a lot easier to work with.
To find the inverse of the cross-section function, I simply switched x with z and f(x) with x. The inverse function would then be x=z² or z=±sqrt(x), better rewritten as R(x)=±sqrt(x) or just R(x)=sqrt(x) as radius must be a positive value. I will refer to this later as R(x).
The cup would remain bounded by h and g.
With radius out of the way, I can move on to finding the volume. The general logic is to find the volume up to h, then subtract the volume up to g from that.
I will construct and solve an equation below to find the volume up to h. It may look complicated, but if I put it into words it should sound something like this:
The volume up to line h is equal to the limit of the sum of the volume of n cylinders, each cylinder’s height being 1/n the total distance to line h, as n approaches infinity.
Something similar can be done for the volume up to g.
Subtracting one from the other, I get a volume of 675π/2 cubic units, or around 1060.288 cubic units.
This part was relatively simple, though the same cannot be said for the next part.
Part C: Surface Area
At first, I went about this part much the same as the first. I would still use cylinders, though now I will use the lateral surface area of the cylinders instead of the volume of the cylinders.
I can use an infinite number of infinitely thin cylinders to find the lateral surface area of the cup, then use the area of a circle to find the area of the base of the cup. Adding these values together should yield the total surface area of the cup.
The base has a radius of sqrt(15) units and applying the area of a circle formula (A=πr²) I find that the area of the base is 15π square units.
The lateral surface area of a cylinder can be found with the formula A=2πrh. I think of this as the side of a cylinder being essentially a rolled-up rectangle, with its height being the height of the cylinder and its width being the perimeter, or circumference, of the base. I will also use an overestimation here, much as I have in the previous part. Once again, I will use R(x), as well as the lines h and g.
I intend to use the same method as the previous part as well, subtracting the area up to g from the area up to h. I will start with h.
I run into a problem. The formula for the sum of consecutive square roots is not known. I can try to write it out and derive it, and I find that:
For some unknown constants a.
This sum quickly spirals out of hand, as I now need to deal with multiple different simplified radicals, all with different coefficients. These radicals are also not just square roots of primes but can also be square roots of composite numbers such as 6 or 35. Considering that this formula needs to generalize to all real numbers, this method hits a wall (For example, if n=99999, I would have countless different radicals to account for). In short, I cannot use the traditional method and must find another method.
Riemann sums are the only way I know of to find the exact surface area of the cup. If I cannot find the exact value, I can at least try to estimate.
While definitely not as desirable as exact values, estimations are commonly used in a range of practical applications. For instance, the wing area of a plane is usually accepted as the top-down area (planform area), even though the wings are curved in a shape called an airfoil. Much like how the hypotenuse of a right triangle is not equal to its legs, the actual surface area of a wing is not equal to the planform area. In this case, the planform area only serves as an estimate, yet it is widely used and accepted.
I will then move on to my estimations, and I examine 3 different methods of estimation below.
1. Cross-sectional method
This new* method understands the idea of a cylinder in a different way. Instead of understanding 2πrh as 2πr*h or circumference*height, I can instead understand it as 2rh*π or cross-sectional area*π.
This makes intuitive sense as π is the ratio between the diameter and circumference, and the cross-section acts as a sort of 2D diameter to the surface area of the cylinder.
However, this is only perfectly accurate if the object in question is a cylinder with a rectangular cross-section. Here, where the cross-section is a section of parabola, I would expect some error.
Again, the cross-section of a paraboloid is a parabola. In this specific question where the paraboloid is described by f(x,y)=x²+y², the cross-section can be described by f(x)=x². To find the cross-section, I can find the area under s(x)=-x²+15 (Area up to g) and subtract it from the area under t(x)=-x²+30, within the domains [-sqrt(15), sqrt(15)] and [-sqrt(30), sqrt(30)], respectively.
The formula for the area under a curve (Riemann sum) is as follows:
The area of the cross-section is then 40sqrt(30)-20sqrt(15) square units.
Multiplying by π, the estimated surface area of the sides is 40πsqrt(30)-20πsqrt(15) square units, which is around 444.942 square units.
Adding in the base of the cup, the estimated total surface area is 40πsqrt(30)-20πsqrt(15)+15π square units, which is around 492.066 square units.
2. Frustum method
A frustum is essentially a cone with its top chopped off. This is useful in this case as the cup heavily resembles a frustum.
In this diagram, the cross-section of the frustum is BCED, where the cross-section of the cone the frustum is cut out of is ABC. The cup would look like an upside-down version of this frustum.
The formula of the lateral area of the frustum is:
Plugging in the numbers:
the estimated surface area of the sides is π(15sqrt(2)+30)sqrt(9-sqrt(2)) square units, which is around 443.131 square units.
Adding in the base of the cup, the estimated total surface area is π(15sqrt(2)+30)sqrt(9-sqrt(2))+15π square units, which is around 490.255 square units.
3. Average cylinder method
This method is the simplest and involves finding the average radius of the cup (average of top and bottom) and then using this as a radius for a cylinder. This is different from the frustum method. While in 2D triangles may be chopped and reconnected to look like a rectangle, this is not the case for the frustum and the cylinder.
Increasing the number of cylinders would increase the accuracy of this method, but here I will only examine the efficacy of using one cylinder. The same can be said for the frustum method.
Using this method, the estimated surface area of the sides is 15πsqrt(30)+15πsqrt(15) square units, or around 440.618 square units.
Adding in the base of the cup, the estimated total surface area is 15πsqrt(30)+15πsqrt(15)+15π square units, which is around 487.742 square units.
Using this paraboloid surface area calculator I found online, I get a lateral surface area of around 447.454 square units (The math used in the calculator is beyond me but is likely linked to some way to find the sum of consecutive square roots, which, again, I don’t have enough time for). Adding in the base of the cup (15π square units), the surface area is around 494.578 square units.
In summary, the surface areas found by the different methods are:
- Cross-sectional method: 492.066 square units
- Frustum method: 490.255 square units
- Average cylinder method: 487.742 square units
- Calculator (cheat): 494.578 square units
Evidently, the cross-sectional method is the best, though only marginally. The percent error of this method is only 0.508%. However, this error does scale up, especially when considering something like mass production. If a manufacturer based their calculations on this estimate and bought just enough material to produce 100000 cups, they would only produce 99492 cups, 508 less than the target.
The greatest downside to the cross-sectional method is that it cannot be increased in detail. This can be done for the frustum and cylinder methods, as I can add more frustums/cylinders, but the cross-sectional method is already as detailed as it gets. This means that the frustum and cylinder methods are arguably better as they can become more accurate, while the cross-sectional method has no potential to do so.
Conclusion
In conclusion, the volume of this cup is around 1060.288 cubic units and the estimated surface area is 492.066 square units, though using a calculator the exact surface area is closer to 494.578 square units.
More importantly, in this article, I have examined:
- The use of Riemann sums in a 3D context
- The failure of Riemann sums in certain 2D contexts
- The use of estimations for surface area
- A novel approach to surface area estimation (cross-sectional method)
With my current knowledge, I can only take this problem this far. However, as my understanding of math broadens, I will likely be able to get over the sum of square root “dead end” and be able to accurately find the surface area of this cup.
