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 cm^{2}
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.
Spot | Coordinate in x | Coordinate in y |
---|---|---|
C | 0.03 | 3.37 |
D | 3.50 | 5.35 |
E | 9.83 | 7.73 |
F | 10.80 | 7.89 |
G | 12.60 | 7.80 |
H | 13.87 | 7.34 |
I | 14.70 | 6.77 |
J | 15.07 | 6.31 |
K | 15.47 | 5.61 |
L | 15.85 | 4.90 |
M | 17.52 | 4.54 |
N | 17.97 | 4.84 |
O | 18.23 | 5.79 |
P | 18.54 | 6.38 |
Q | 16.30 | 4.57 |
R | 18.77 | 6.57 |
S | 19.20 | 6.71 |
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.
Points | Command used | Function found |
---|---|---|
C, D, E | Polynomial({C, D, E}] | f(x) = −0.02x^{2} + 0.64x + 3.35 |
E, F, G | Polynomial({E, F, G}] | g(x) = −0.08x^{2} + 1.17x − 2.13 |
G, H, I | Polynomial({G, H,I}] | h(x) = −0.15x^{2} + 3.73x − 14.64 |
I, J, K, L | Polynomial({I, J, K, L}] | p(x) = 0.47x^{3} − 21.85x^{2} + 337.83x − 1725.98 |
L, Q, M, N | Polynomial({L,Q, M, N}] | q(x) = −0.01x^{3} + 0.91x^{2} − 22.57x + 172.27 |
N, O, P | Polynomial({N,O, P}] | r(x) = −3.07x^{2} + 114.83x − 1066.94 |
P, R, S | Polynomial({P, R, S}] | s(x) = −0.76x^{2} + 29.12x − 272.84 |
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.
Function | Start x - value | End x - value |
---|---|---|
f(x) | 0.00 | 9.83 |
g(x) | 9.83 | 12.60 |
h(x) | 12.60 | 14.70 |
p(x) | 14.70 | 15.85 |
q(x) | 15.85 | 17.97 |
r(x) | 17.97 | 18.54 |
s(x) | 18.54 | 19.20 |
Then, once I had entered the data, I obtained the functions restricted to their respective domains, as shown in Figure 12 in the algebraic view and in Figure 13, in the graphical view.
In this process of finding the respective upper and lower limits of each of the functions, I was also able to recognize the potential inaccuracies of this method. I was able to realize that the silhouette delimited by the functions that I had found did not fit exactly with the real measurements of the pot. Specifically, I noticed this with the height of the pot, as in the GeoGebra graphical view it was 19.2 cm, while its actual measurement was 19 cm. Despite that, this method would be much more accurate than the previous one, so it was extremely valuable to continue.
Even though at this point I already had all the data necessary to be able to use the formula to find the surface area of the pot, being a highly visual person, it was difficult for me to understand all of my progress if I could not see it in a visual manner.
This is why, taking advantage of the fact that I was using the GeoGebra program, I decided to take the opportunity to model the pot in the 3D graphic view. For this, I also had to research some videos on YouTube until I finally figured out how to model the pot in 3D successfully.
After watching Jara's video (2021), I discovered that in order to rotate the functions in 360° and see the true shape of the pot, which was a surface of revolution, I had to use the “Slider” option in GeoGebra . In total, I would have to do this seven times in order to rotate the seven functions I had previously found. Figure 14 shows how I did this procedure for the first slider.
In the “Nombre” [Name] section, in this case I placed the letter “α” but for the following sliders, I placed other letters of the Greek alphabet to be able to distinguish them. Also, in the “Min” section, I put “0”, in “Max”, “360”, and in “Incremento” [Increase], “1”, for all the sliders. This way, you could then effectively rotate them 360°.
This is how I was able to create the seven sliders that I would need to be able to rotate all the functions that made up my flower pot figure. These can be seen in Figure 15.
After having created these, I would now be able to rotate each of the functions in 3D with their respective domains. For this, still guided by Jara's video (2021), I opened the GeoGebra graphic view in 3D and only made visible the functions that already had their respective upper and lower limits.
Then, I used the “Axial Rotation” tool in GeoGebra, first selecting one of the functions, then the x axis and finally, after asking me for the angle by which it would rotate, choosing the name of its corresponding slider created previously.
For example, for the first rotation, I selected the function “t(x)”, then the x axis as I would with the rest of the rotations and finally, after being asked for the angle of rotation, I selected the angle “α” that I had previously created. I would then repeat this process with the other six remaining functions.
Finally, when I had finished carrying out this process, I was able to form my pot in 3D. Figure 16 demonstrates how all functions rotated by 360° looked in the GeoGebra algebraic view and Figure 17 in the 3D graphical view.
Having all the necessary data to use the formula and a visualization in GeoGebra that allowed me to recognize this entire process in a visual way, I was now able to use the formula for the surface area of a revolution to find the areas of each of the segments.
To do this, I first found the derivatives of each of the functions I had found and collected all the other data necessary to be able to apply the formula in Figure 18. For the upper and lower limits, I relied on those I had already found in Figure 11.
Segment’s function | Derivative of the segment’s function | Lower limit | Upper limit |
---|---|---|---|
−0.02x^{2} + 0.64x + 3.35 | −0.04x + 0.64 | 0.00 | 9.83 |
−0.08x^{2} + 1.17x − 2.13 | −0.16x + 1.77 | 9.83 | 12.60 |
−0.15x^{2} + 3.73x − 14.64 | −0.3x + 3.73 | 12.60 | 14.70 |
0.47x^{3} − 21.85x^{2} + 337.83x − 1725.98 | 1.41x^{2} − 43.7x + 337.83 | 14.70 | 15.85 |
−0.01x^{3}+ 0.91x^{2} − 22.57x + 172.27 | −0.03x^{2} + 1.82x − 22.57 | 15.85 | 17.97 |
−3.07x^{2} + 114.83x − 1066.94 | −6.14x + 114.83 | 17.97 | 18.54 |
−0.76x^{2} + 29.12x − 272.84 | −1.52x + 29.12 | 18.54 | 19.20 |
Having all this in order, I now applied the formula for each of these segments, as seen in Figure 19.
Segment | Formula applied and result |
---|---|
C, D, E | \(\displaystyle \int^{9.83}_0\bigg[2π × (−0.02x_2 + 0.64x + 3.35) × \sqrt{1 + (−0.04x + 0.64)^2}\bigg]dx= 393. 53 \,cm^2\) |
E, F, G | \(\displaystyle\int_{9.83}^{12.6}\bigg[2π × (−0.08x^2 + 1.77x − 2.13) × \sqrt{1 + (−0.16x + 1.77) ^2}\bigg] dx = 133. 51 \,cm^2\) |
G, H, I | \(\displaystyle \int_{12.6}^{14.7}\bigg[2π × (−0.15x^2 + 3.73x − 14.64) × \sqrt{1 + (−0.3x + 3.73)^2}\bigg] dx = 117. 48 \,cm^2\) |
I, J, K, L | \(\displaystyle\int^{15.84}_{14.7}\bigg[2π × (0.47x^3 − 21.85x^2 + 337.83x − 1725.98) ×\sqrt{1 + (1.41x^ − 43.7x + 337.83)^2} dx = 94. 53 \,cm^2\) |
L, Q, M, N | \(\displaystyle\int^{17.97}_{15.85}2π × (−0.01x^3 + 0.91x^2 − 22.57x + 172.27) × \sqrt{1 + (−0.03x ଶ + 1.82x − 22.57) ^2}\bigg]dx = 41. 33 \,cm^2\) |
N, O, P | \(\displaystyle \int^{18.54}_{17.94}\bigg[2π × (−3.07x^2+ 114.83x − 1066.94) × \sqrt{1 + (−6.14x + 114.83)^2}\bigg] dx = 63.22 \,cm^2\) |
P, R, S | \(\displaystyle \int^{19.2}_{18.54}\bigg[2π × (−0.76x^2 + 29.12x − 272.84) × \sqrt{1 + (−1.52x + 29.12)^2}\bigg]dx = 35.72 cm^2\) |
Having this data, now it was possible to find the total surface area by adding them as follows -
total surface area = 393.53 + 133.51 + 117.48 + 94.53 + 41.33 + 63.22 + 35.72
total surface area = 879. 32 cm^{2}
Having found the surface area of the pot using two different methods, I now had a good idea of the approximate surface area I would need to cover. Although I recognize that both methods were subject to errors, I do consider that the result I obtained using the integral method was the most accurate between the two.
Despite this, I decided to calculate how much paint I would need based on both methods, since I could use the lower value as the minimum amount of paint I would need and the higher value as the maximum. In this way, I would make sure that I did not lack paint, but that there was not much left over either, thus optimizing not only the material used but the money I would have to spend to carry out the project.
However, the “Pouring” technique did not simply involve using the same amount of paint as the surface area, so I had to look up how much paint was actually needed for this method. According to Voorhies (2018), this is 1 ml per 4 cm2 of surface area. Therefore, to calculate how much paint I would need, I applied the following formula based on this conversion for the results of both methods -
\(acrylic \,paint \,needed \,(ml) = \frac{surface \,area \,of \,the \,pot \,(cm^2)}{4}\)
Starting with the paint I would need based on the result of the first method to find the surface area, I applied the formula as follows -
\(acrylic \,paint \,needed \,(ml) =\frac{ 990.28}{ 4 }= 247. 57 \,ml\)
Therefore, I used this value as the maximum amount of paint I would need to proceed with my project.
Starting with the paint I would need based on the result of the first method to find the surface area, I applied the formula as follows -
\(acrylic \,paint \,needed \,(ml) =\frac{ 990.28}{ 4 }= 247. 57 \,ml\)
Therefore, I used this value as the maximum amount of paint I would need to proceed with my project.
Having these results, I now knew the range of the amount of paint I would need to carry out the project, taking into account that I would have to buy at least 220 ml and that I would definitely not need more than 250 ml. So, my next step was to start searching online for the various stores that offered acrylic paint. For this, I decided to specifically look at stores that were relatively close to my house, since I did not want to pay for an extremely expensive home delivery service in case I ended up opting for this pot painting method.
Therefore, I decided to focus on the stores “Spondylus”, which is known for the high- quality art supplies it offers, and “Tai Loy”, a slightly cheaper option. The best options I found are displayed in Figure 20.
After thinking about it, I decided that option 1 would be the best. This is because, although option 3 had a lower price per milliliter, option 1 offered a greater variety of colors than any of the other options and was from a reputable French brand recommended by several artists. Plus, it still had a lower price per milliliter than options 2 and 4 and had the lowest overall price of all. Additionally, if I had paint left over, I could recover a little bit of the money spent by selling it to my friend whose an artist that I’m sure would be interested.
However, these would not be the only materials I would need to be able to perform this method, since I would also have to get a special medium for the “Pouring” technique. For this reason, I also began to investigate how much of it I would need, the options for sale that were available and their respective costs.
After researching it in depth and after reading Carreiro's (n.d.) blog specialized in the “Pouring” technique, I discovered that it was recommended to use three parts of medium for one part of acrylic paint. Therefore, taking into account that I would not need more than 250 ml of acrylic paint, I multiplied this value by three, thus obtaining that I would need approximately 750 ml of medium.
Having this information, I searched the two stores where I had also looked for the acrylic paint. However, I could only find one option in “Spondylus” (Figure 21) for a liter of medium for S/. 232, which seemed quite expensive to me. For this reason, I began to investigate if there was another alternative to store-bought mediums.
In this process, after reading another blog by Carreiro (n.d.), I found that mediums could be done at home using white glue and water. This was a perfect alternative since these were materials that I already had in my house and, therefore, I could save money in that regard.
In conclusion, in this exploration, it has been possible to successfully determine the surface area of the pot using both a traditional method, which involves the use of masking tape, and a more advanced and precise mathematical one that uses integrals and programs such as GeoGebra. Through these methods, I was able to calculate how much paint I would need to purchase and estimate the approximate cost of the proposed project. Also, after this research, I have decided that I will carry out this project, since I discovered that the costs of the proposed materials were not as high as I initially thought and that it would really be a gift that my mother would love.
This research also allowed me to recognize the importance of the concept of integrals, since I was able to apply it to a problem in my daily life. Furthermore, I reflected on the fact that although neither of the two methods used could be completely accurate, I was able to observe the advantages and disadvantages of each in the process. This led me to understand the importance of using different approaches to address the same problem. However, greater attention to precision in the execution of both methods would have decreased the variation in the results.
Therefore, in the future, it would be interesting to continue investigating whether there are other alternatives to calculate the area of a surface of revolution, such as this pot, in order to corroborate the results obtained. Additionally, a next step could be to determine the volume of the pot to know how much soil would fit inside it and what types of plants could be grown there.
Artemiranda. (2020) What is Pouring? Definition and basic techniques. Artemiranda Fine Arts Blog. https://www.artemiranda.es/blog/index.php/que-es-el-pouring-definicion- y-tecnicas-basicas/
BYJU'S. (2018, April 15). Integration in Maths - Definition, Formulas and Types. https://byjus.com/maths/integration/
Carreiro, J. (n.d.). How to prepare pouring paint. Fluid Acrylic. https://acrilicofluido.com/curso/pintura-fluida-casera/
Carreiro, J. How to prepare homemade Pouring medium. Fluid Acrylic. https://acrilicofluido.com/curso/medium-casero/
Jara, FJ [Jaramáticas]. (2021). Solids of Revolution with Geogebra | Disc method. [Video File]. Youtube. https://youtu.be/RIbkpzVl7x0?si=gP_ZUVgh8UDI5opp
Svirin, A. (n.d.). Area of a Surface of Revolution. Math24. https://math24.net/area-surface- revolution.html
Voorhies, D. (2018). How Much Paint to Use for an Acrylic Pour? Left Brained Artist. https://leftbrainedartist.com/how-much-paint-to-use-for-an-acrylic-pour