Polishing the Surface Area of a Solid of Revolution

The vase depicted in Figure 1 is a very precious asset to my family as it remains a memory of my dead grandfather. However, over time, this sentimental family heirloom has grown water spots, stains, and other flaws that take away from its aesthetic appeal. Hence, polishing it is beneficial in getting rid of these flaws and returning the vase to its previous state. Therefore, me and my family would like to polish it from the outside, in order to maintain the vase in good condition while preserving its value and longevity.

This exploration aims to find the surface area of this vase by integration in order to determine how much polishing paint is needed to cover it. Additionally, I purposefully chose to do my mathematical exploration on this topic since I'm passionate about calculus, and I want to gain a deeper understanding of this subject by attempting more complex applications, outside of the scope of the Analysis and Approaches Standard Level Mathematics course. Hence, I am hoping that after this investigation I will gain a greater comprehension of differentiation and integration.

I have learned from my AA SL mathematics course that the area underneath a curve can be found using a definite integral. However, the syllabus does not cover how to find the surface area of a figure that revolves of curves around the axis of revolution, such as my vase. Therefore, I did some research and I found the formula used to find the area of a surface of revolution. Hence the formula: \(S=\int_{a}^{b} 2 \pi f(x) \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x\)

This formula looks quite new to me, so I will decipher it to be sure I understand. The formula can be divided to simplify the explanation of its origin, such as the arc length formula which will then be incorporated to figure out the whole formula.

To find the distance between points B and C from Figure 2, the distance formula derived from the Pythagorean theorem should be applied:

\[d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\]

If we label the change in x with dx and label the change in y with dy, they can be comprised into the distance formula and integrated from points A to D which would give the length of the entire curve. Hence, it would give an infinite number of lengths of small line segments that would make up for the integrity of the curve. Therefore, the formula:

\[L=\int_{a}^{b} \sqrt{(d x)^{2}+(d y)^{2}}\]

However, it can simplified by isolating dx from each term:

\[L=\int_{a}^{b} \sqrt{(d x)^{2}\left(1+\frac{(d y)^{2}}{(d x)^{2}}\right)}\]

Taking out dx of the square root:

\[L=\int_{a}^{b} d x \sqrt{1+\left(\frac{(d y)}{(d x)}\right)^{2}}\]

Move the dx to the back:

\[L=\int_{a}^{b} \sqrt{1+\left(\frac{(d y)}{(d x)}\right)^{2}} d x\]

\(\frac{(d y)}{(d x)}\) is the derivative of y with respect to x, thus the arc length formula:

\[L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x\]

A solid of revolution is a figure revolving around an axis. However, in my case, I am focussing on revolutions around the x-axis. If the points A to D revolved around the x-axis, it would form a figure of revolution consisting of small circles such as the one depicted in Figure 3. Therefore, in order to calculate its area, multiplying the circumference of the circles created by the curve's rotations by its length is needed. The circumference of a circle is defined by the following formula:

\[C=2 \pi r\]

Where the radius is equal to the distance from the axis of revolution to the curve, hence the function's value.

\[C=2 \pi f(x)\]

Therefore the surface area formula is defined as follows:

\[S=2 \pi f(x) \cdot \text{ Length of the arc }\]

As it is now known as the arc length formula due to the previous section, it can be incorporated into the formula as so:

\[S=\int_{a}^{b} 2 \pi f(x) \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x\]

Now that I have successfully derived the formula and the way it should be applied, I can proceed to find the function of my vase in order to find the surface area.

In order to obtain the best fit of a function I have to first model my vase. I initially decided to divide the vase into three functions as I saw each one would have different segments of increase and decrease (figure 4).

I, therefore, inserted an image of my vase on Geogebra. However, to make it as accurate as possible I had to first measure the length of my vase, which I discovered to be 46.5 cm. I then had to find the midpoint of the width of my vase using a ruler so that I could correctly input the length in the middle of the vase in order to make both sides as symmetrical as possible (figure 5). Additionally, I made sure that both points A and B lie on the same X coordinates. For simplification, I rounded the x-coordinates to two significant numbers to the right of the decimal point. Additionally, the terms in each function were rounded to three significant numbers to the right of the decimal point for more precision. Except for the first term of f₃(x) as its value was a significantly small number.

Subsequently, I started plotting different points on the border of my vase. After plotting the points of the first function (orange colored line in Figure 6), I observed that the best-fit line is a polynomial function, also known as a cubic function, to the third degree. Forwarding to the second function (red colored line in Figure 6), I initially thought that the line of best fit would be a sin wave due to its particular curvature that reminded me of it. Later on, I discovered that the line of best fit is indeed a sin wave. This made me feel more confident in the unit circle chapter as I remember greatly battling over this course. Finally, for the last segment (blue colored line in Figure 6), I was heavily contemplating if a polynomial function or a sin wave would be the most appropriate. However, after many rearrangements of the points attempting to make the sine wave an adequate fit, I discovered that the polynomial function is the best fit as it passed the closest to my plotted points without having to rearrange them.

Hence the function of my vase can be defined as follows:

\[f(x)=\left\{\begin{array}{cc}f_{1}(x)=0.581 \mathrm{x}^{3}-2.193 \mathrm{x}^{2}+1.680 \mathrm{x}+6, \quad 0 \leq \mathrm{x}<2.17 \\f_{2}(x)=\sin (0.271 \mathrm{x}-2.811)+6.725, \quad 2.17 \leq \mathrm{x}<16.87 \\f_{3}(x)=0.0007 \mathrm{x}^{3}-0.059 \mathrm{x}^{2}+1.461 \mathrm{x}-2.862, & 16.87 \leq \mathrm{x}<46.5\end{array}\right.\]

Now that I have found the function defining the vase, I can input them in the surface area of the revolution formula. Firstly, I will be differentiating the equations without the use of the calculator. However, by observing the complexity of the functions, a solution by hand would exceed the scope of this exploration. Thus, I will be using a GDC calculator to evaluate the integrals for each function.

\[f_{1}(x)=0.581 x^{3}-2.193 x^{2}+1.680 x+6, \quad 0 \leq x<2.17\]

First, I have to obtain the derivative of f₁(x):

\[\begin{gathered}f_{1}^{\prime}(x)=0.581 x^{3}-2.193 x^{2}+1.680 x+6 \\f_{1}^{\prime}(x)=3 \cdot 0.581 x^{3-1}-2 \cdot 2.193 x^{2-1}+1 \cdot 1.680 x^{1-1}+0 \cdot 6 \\f_{1}^{\prime}(x)=1.743 x^{2}-4.386 x+1.680\end{gathered}\]

Now that I have found the derivative of f₁(x), I can put it in the surface area formula:

\[A_{1}=\int_{0}^{2.17} 2 \pi \cdot\left(0.581 x^{3}-2.193 x^{2}+1.680 x+6\right) \cdot \sqrt{1+\left(1.743 x^{2}-4.386 x+1.680\right)^{2}} d x\]

Evaluated by inputting the equation in the GDC Calculator:

\[=101.028 \mathrm{~cm}^{2}\]

\[f_{2}(x)=\sin (0.271 x-2.811)+6.725, \quad 2.17 \leq x<16.87\]

First, I have to obtain the derivative of f₂(x):

\[f_{2}^{\prime}(x)=1.785 \sin (0.271 x-2.811)+6.725\]

However, I noticed that the function of f₂(x) is "function of a function". Hence, I will be using the chain rule in order to differentiate the complex function. The chain rule is represented by the following formula:

\[f^{\prime}(x)=u^{\prime}(v(x)) \cdot v^{\prime}(x)\]

Where,

\[\begin{gathered}u(x)=1.785 \sin (0.271 x-2.811)+6.725, v(x)=0.271 x-2.811 \\u^{\prime}(x)=1.785 \cos (0.271 x-2.811)+0 \cdot 6.725, v^{\prime}(x)=1 \cdot 0.271 x^{1-1} \\f_{2}^{\prime}(x)=1.785 \cos (0.271 x-2.811) \cdot 0.271\end{gathered}\]

Now that I have found the derivative of f₂(x), I can put it in the surface area formula:

\[A_{2}=\int_{2.17}^{16.87} 2 \pi \cdot(1.785 \sin (0.271 x-2.811)+6.725) \cdot \sqrt{1+(1.785 \cos (0.271 x-2.811) \cdot 0.271)^{2}} d x\]

Evaluated by inputting the equation into the GDC Calculator:

\[=632.571 \mathrm{~cm}^{2}\]

\[f_{3}(x)=0.0007 x^{3}-0.059 x^{2}+1.461 x-2.862, \quad 16.87 \leq x \leq 46.50\]

First, I have to obtain the derivative of f₃(x):

\[\begin{gathered}f_{3}^{\prime}(x)=0.0007 x^{3}-0.059 x^{2}+1.461 x-2.862 \\f_{3}^{\prime}(x)=3 \cdot 0.0007 x^{3-1}-2 \cdot 0.059 x^{2-1}+1 \cdot 1.461 x^{1-1}-0 \cdot 2.862 \\f_{3}^{\prime}(x)=0.0021 x^{2}-0.188 x+1.461\end{gathered}\]

Now that I have found the derivative of f₃(x), I can put it in the surface area formula:

\[A_{3}=\int_{16.87}^{46.50} 2 \pi \cdot\left(0.0007 x^{3}-0.059 x^{2}+1.461 x-2.862\right) \cdot \sqrt{1+\left(0.0021 x^{2}-0.188 x+1.461\right)^{2}} d x\]

Evaluated by inputting the equation in the GDC Calculator:

\[=1329.23 \mathrm{~cm}^{2}\]

\[\begin{gathered}A_{t}=A_{1}+A_{2}+A_{3} \\A_{t}=101.028 \mathrm{~cm}^{2}+632.571 \mathrm{~cm}^{2}+1329.23 \mathrm{~cm}^{2} \\A_{t}=2062.829 \mathrm{~cm}^{2}\end{gathered}\]

In conclusion, now that I have found the total surface of revolution for the vase, I can subsequently figure out how many milliliters of paint are needed to polish the whole area of my vase. It is estimated that 1 milliliter of paint is needed per 1 square centimeter of surface area, thus:

\[\begin{gathered}\text{ ml of paint needed }=1 x \text{ surface area of vase in }\mathrm{cm}^{2} \\\text{ ml of paint needed }=2062.829 \mathrm{ml}\end{gathered}\]

This exploration has brought me many strengths. Firstly, I was greatly content that I was able to find the surface area of the vase as I can now successfully polish it myself which means my grandfather's legacy can continue. Additionally, I became experienced with a whole new scope of calculus. Furthermore while choosing this topic, I developed my skills regarding the GDC calculator. I observed that I am much more familiar with it now that I had to solve multiple different equations. As a result, my paste accelerated, which is a crucial benefit for the paper 2 external assignment. Furthermore, I became more experienced with a whole new program that I had not previously worked with.

While there were advantageous aspects during this mathematical investigation, there were limits in my findings as well. For instance, I had to round my results in order to simplify the calculating process, which limited my findings as I wanted to find the most exact measurements possible. Another factor limiting my final results was the tiny curbs on the neck and foot of the vase that I was not able to compose due to the complexity of the shape. Additionally, the vase was handmade which meant it couldn't be perfectly symmetrical. Consequently, my vase's model is limited to a certain extent by the rounding of the results, small curbs that I neglected to include, and the impossibility of symmetry. Finally, although I learned how to utilize the Geogebra software, it was initially very difficult for me to find out how to adequately operate it. Hence, a lot of time was needed so that I could familiarize myself with the software and effectively utilize it.

How to find the area of a surface of revolution. dummies. (2023, September 8).
https://www.dummies.com/article/academics-the-arts/math/calculus/how-to-find-the-area-of-a-surface-of-revolution-190931/

Paint calculator: Quick and easy interior wall paint calculator. Chameleon decorators and windows restoration. (n.d.).
https://chameleon-decorators.co.uk/paint-calculator/#:~:text=How%20much%20paint%20do%20I%20need%20per%20m2%3F,metre%20of%20wall%20or%20ceiling

Arc length. Math is Fun Advanced. (n.d.). https://www.mathsisfun.com/calculus/arc-length.html

Calculator Suite. GeoGebra. (n.d.). https://www.geogebra.org/calculator