Calculating the surface area of a pot to find the amount of paint needed to cover it using the “Pouring” method

For a while now, I've been thinking about what I could give to my mom for her birthday. Therefore, I was looking for some ideas on the internet until it occurred to me that I could give her something related to gardening, since this is one of her greatest passions. However, I didn't want to just buy her a gift, package it up, and hand it to her; I wanted it to have a part of my essence to make it more special.

Therefore, I decided to go to a flower shop and I bought this 19 cm high and 16 cm wide (Figure 1) pot for S/. 15. This seemed perfect to me, because despite looking a bit simple, it occurred to me that I could customize it in my own way to make it unique. For this reason, I began to investigate the various techniques that existed for painting pots until I found one that seemed extremely interesting to me, known as “Pouring”.

This involved using acrylics and mixing them with a medium, which helps obtain a lighter consistency, in order to create abstract shapes. However, all of these materials can be quite expensive and much of them can be wasted if there is no knowledge of the necessary quantities.

This is why, in this investigation, I will focus on calculating the surface area of the pot with two methods that I had never used before: using integrals and a more traditional method.

This will be done in order to know exactly how much paint I should purchase and to determine the approximate cost of this project. In this way, I can then decide if this is a feasible option or if perhaps I should opt for a method that is less expensive.

Using masking tape to find the approximate surface area of the pot

Before knowing the method to find the surface area using integrals, I decided to use a fairly quick and easy technique to perform that occurred to me when I was just thinking about how to calculate how much paint I would need to paint my pot. For this, I decided to use the materials shown in Figure 2, including a 1.9 cm wide masking tape that was cut into segments and used to cover the entire surface of the pot. In this way, I could then remove each of these and find their total area.

However, as I cut and attached the masking tape to the pot, I realized that the result would be quite inaccurate. This was because the masking tape wrinkled when stuck on curved surfaces and it was difficult for all the segments to cover the entire pot, as seen in Figure 3. Despite this, I decided to continue with this method since the result could serve as a guide as to what would be the maximum amount of paint I would need.

At the end of this process, I measured the lengths of each of the segments that I had used to wrap the pot. These have been represented in Figure 4 and can also be seen in Figure 5.

Then, to find the surface area of the outer part of the pot, I added the lengths of all the segments and multiplied them by the width.

surface area = (38.5 + 32.5 + 28.5 + 38.5 + 46.5 + 48.5 + 48.7 + 48.0 + 45.5 + 42.0 + 38.5 + 34.5 + 31.0) × 1.9

surface area = 521.2 × 1.9 = 990. 28 cm²

With this, I now had an idea of what the surface area of the pot was. However, I knew there had to be a better way to find it with all the technology out there now, so I didn't limit myself to just using this method.

Taking into account that I knew that the area measurement that I had previously found was not very precise, I decided to use a more exact mathematical method, which made use of integrals and technological programs.

During the Mathematics Analysis and Approaches classes at Standard Level, I had already learned about the use of definite integrals. This concept, which is the inverse of derivatives, is used to approximate the area under a function, dividing it into fine vertical slabs that cannot be measured individually.

Additionally, in class, I had also learned how to use integrals to find the volume of solids of revolution, such as this flower pot. The formula for this was the following -

\(Volume \,of \,a \,solid \,of \,revolution: \int_a^b \pi [f(x)]^2 \, dx\)

However, for this project, I was not interested in finding the volume but rather the surface area of it, which was something we had not learned until now. Therefore, I wondered if the formula for this would be something similar, so I began to investigate how the area of a solid of revolution could be found. Finally, on a page created by Svirin (n.d.) I found that the formula, which had some similarity to the previous one, was the following -

\(Surface \,area \,of \,a \,solid \,of \,revolution: \int _a^b 2πf(x)\sqrt{1 + [f ́(x)]^2}dx\)

This would help me find the area of the pot more accurately, which would be helpful in finding the total amount of paint required to paint it and the cost of the project. However, in order to use the formula, I first had to find the following components of it -

f(x) = segment's  function

f '(x) = derivative of the segment's  function

b = upper limit on the x − axis of the segment′s function

a = lower limit on the x − axis of the segment′s function

To start using this method, the first thing I decided to do was model my pot in GeoGebra, adding it as an image and then plotting the points through which the functions that would form its figure would pass. These have been represented in Figure 6 and can also be seen in Figure 7.

However, although it may seem like it, this was not an easy process, since I had to constantly change the coordinates of the points and add others, such as Point Q, so that later the functions could be as similar as possible to the silhouette of the flowerpot.

Once all the points were drawn on the graph, I used the “Polynomial (List of Points)” command to find a variety of functions that would help me form the silhouette of the pot. However, in order to find these, I couldn't simply put all the points in the command, since I would get a function of an extremely high degree that would not resemble the silhouette, as seen in Figure 8.

For this reason, I decided to group the points in groups of three or four in order to find second and third degree functions that could resemble the silhouette of the flower pot in the Cartesian plane. In total, I was able to group the points into seven groups, which were used to find different function, each resembling a certain part of the pot, as shown in Figure 9.

So, after I finished finding all the functions, I could look at them in the GeoGebra graphical view, as shown in Figure 10.

However, it was difficult to visualize the pot's silhouette made by the functions if their upper and lower limits were not set. Thus, I used the following command to restrict the domains of each function to only what I would need: “(Function, Start x − value, End x − value)”

For this, I filled in the data required by the command as shown in Figure 11, using the functions I had found. For the start and end x - values, I relied on the 17 points I had already plotted on the graph, subtly changing some values to fit the figure better. Specifically, this occurred with the start x - value of f(x) since even though the coordinate on the x axis of the first plotted point was “0.03”, it needed to be “0.00” to achieve greater accuracy.