Finding the Volume and the Surface Area of a 3-Dimensional Teacup Model

Categories: MathScience


When I am introduced to calculus, I started to believe that calculus has more usage areas than any other subject I learned in math. Amongst the disciplines that use calculus include physics, engineering, economics, statistics, and medicine. Calculus having so many usage areas made me think what can I do with my current knowledge. Solid of revolution took my attention especially since it’s useful when creating everyday objects and furniture. I decided that I might use solid of revolution to create an object myself, like a water glass.

Since I like both designing and mathematics, I believe I can use both in this research to model a glass and find its volume and surface area. When trying to decide for the glass I will be using, I did not want to choose a simple one, so I ended up choosing the Turkish tea cup which is thin in the middle. The thin part of the teacup makes the lower part of the cup to fit in the palm.

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The mouth of the teacup is wide; because it keeps the tea warm in the lower part while cooling the tea on the surface so that it does not burn the mouth.

In this research, I’m aiming to find the volume and the surface area of the teacup I modelled. I wanted to use the subject we learned in class to a real-life situation. For this, I have to draw two graphs to add thickness which I believe will be challenging.

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I am going to use the solid of revolution since I have the revolve it around the axis to form a solid. In lessons, we learned how to find the volume of a solid, but I have to investigate how to calculate the surface area myself.

Solid of Revolution

Imagine a curve, y=f(x) and we draw it like this: The graph of f(x) We learned to find an integral of a graph which is actually finding the area under the graph. In addition to this, we learned that the area surrounded by y = f(x), the x-axis, x = a and x = b, is given by: dx∫ b a y So we take the section of the curve enclosed by ordinates x = a and x = b, then rotate the section through 2 about the x-axis to generate a solid. After this operation, a 3-dimensional π solid is formed, which we call such solid as a solid of revolution.

The solid formed when the graph of f(x) revolved around the x-axis 1 Martin, David, et al. Mathematics For The International Student Mathematics HL. 3rd ed., Haese Mathematics, 2012, pp. 690-691. The area encompassed by y = f(x), the x-axis, x = a and x = b is revolved through 2π about the x-axis to generate a solid, the volume of the solid can be defined by: dxπ ∫ b a y2 A.i.

Graphs of R(x) and r(x), the area they form when they revolved around the axis of revolution and cross-sections which are washers If the area we spin to produce a solid does not border on or cross the axis of revolution, that means that the solid has a hole inside. The cross-sections perpendicular to the axis of revolution is called washers instead of disks. The dimensions of a typical washer are: Outer radius = R(x) Inner radius = r(x) The washers area can be calculated with the formula: (x) R(x) r(x) ( R(x) )A = π[ ]2 − π[ ]2 = π [ ]2 − r(x)[ ]2 Consequently, the definition of volume gives us the formula: (x)dx ( R(x) )dxV = ∫ b a A = ∫ b a π [ ]2 − r(x)[ ]2 2 Weir, M. D., Hass, J., & Giordano, F. R. (2008). Thomas's Calculus Early Transcendentals. (11th ed.), 432. United States of America: Pearson Education.

We are not introduced to finding the surface area of a 3-dimensional object. So, I made some research about this part and explained my findings as follows: Let us take the same curve which has been revolved around the x-axis to form a 3-dimensional solid and split into parts:

  • The solid formed when the graph of f(x) revolved around the x-axis and split into frustums
  • The approximation on every interim gives a distinct portion of the solid, which we call them frustums.

Since we know how to find the surface area of frustums, we can say that the surface area of a frustum is given by: A=2πrl Where: (r )r = 2 1 1 + r2 = radius of the right endr1 = radius of the left endr2 For the frustum on the interim, we can name the boundaries as follows: (x )r1 = f i (x )r2 = f i−1 3. Then we express the length of each of these line segments by , the length of the P P|| i−1 i | | curve will then be defined by the formula: L ≈ ∑ n i=1 P P|| i−1 i | | We can get a definite length by increasing the value of n, using limit.

In other words, the definite length will be given by the formula: L = lim n→∞ ∑ n i=1 P P|| i−1 i | | Let’s define , with that, we can then estimate the length of the y (x ) (x )Δ i = yi − yi−1 = f i − f i−1 line segments as follows: P P|| i−1 i | | = √(x ) y )i − xi−1 2 + ( i − yi−1 2 = √Δx y2 + Δ 2 By the Mean Value Theorem* we know that on the interim, there is a point called so xi* that: (x ) (x ) (x )(x )f i − f i−1 = f ′ i* i − xi−1 y (x )ΔxΔ i = f ′ i* Therefore, the length can now be expressed like this: P P|| i−1 i | | = √(x ) y )i − xi−1 2 + ( i − yi−1 2 = √Δx Δx2 + f (x )[ ′ i* ]2 2 Δx = √1 + f (x )[ ′ i* ]2 In the end, we can say that the true length of the curve can be defined like this: dx L = ∫ b a√1 + f (x)[ ′ ] 2 So, the surface area of the frustum on the interval is defined as follows: π( )Ai = 2 2 f (x )+f (x )i i−1 P P|| i−1 i | | πf (x ) Δx ≈ 2 i* √1 + f (x )[ ′ i* ]2 Meaning the surface area of the entire solid is then: πf (x ) Δx S ≈ ∑ n i=1 2 i* √1 + f (x )[ ′ i* ]2 And we can calculate the exact surface area by getting the limit as n goes to infinity: πf (x ) Δx S = lim n→∞ ∑ n i=1 2 i* √1 + f (x )[ ′ i* ]2 πf (x) dx = ∫ b a 2 √1 + f (x)[ ′ ]2.

We, therefore, finally represent this integral to be the area of the surface cleared by the graph of f(x) from a to b. πf (x) dx ∫ b a 2 √1 + f (x)[ ′ ]2 *Mean Value Theorem It states that, in a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. This means that if f is a continuous function on the closed interval [a,b], and differentiable on the open interval (a,b), then there exists a point c in (a,b) such that: (c)f ′ = b−a f (b)−f (a).


I started by taking a glass I found and then draw its shape on the graph paper. I derived the function from the graph I drove in the paper and found these results: Figure 5. Graphs of the curves of teacup (red) First equation: (x 5)(x 1.77)(x 7) 0−2071265 − 8 − 5 − 1 + 2 (blue) Second Equation: (x 5)(x 5.34)(x 4.72) 0−1574807.65 − 9 − 4 − 2 + 2 (green) First limit: x=95 (purple) Second limit: x=85 Then I took the shape and revolved around the x-axis to create a 3-dimensional model, to make the shape of my cup more visible.

The solid of revolution, the cup, is now look like: Figure 6. The 3-dimensional model of the solid formed when graphs are revolved through the axis of revolution A. Volume I will be using the Washer Method to find the area enclosed by my two graphs.

In the Washer Method, I stated that the formula to be used is: (x)dx ( R(x) )dxV = ∫ b a A = ∫ b a π [ ]2 − r(x)[ ]2 Where: (x) (x 5)(x 1.77)(x 7) 0R = −2071265 − 8 − 5 − 1 + 2 (x) (x 5)(x 5.34)(x 4.72) 0r = −1574807.65 − 9 − 4 − 2 + 2 So the area enclosed by R(x) and r(x) can be calculated like this: ( (x 5)(x 1.77)(x 7) 0 )dx∫ 85 π [ −2071265 − 8 − 5 − 1 + 2 ] 2 − (x 5)(x 5.34)(x 4.72) 0[ −1574807.65 − 9 − 4 − 2 + 2 ] 2 21650.68= 1 I have to say that, till now I only calculated the upper area.

For the bottom area (126 to 138), since this part does not have a hole, I will calculate this part with the original function: dxπ ∫ b a r(x)2 Where: (x) (x 5)(x 5.34)(x 4.72) 0r = −1574807.65 − 9 − 4 − 2 + 2 The bottom part can be calculated as follows: [ (x 5)(x 5.34)(x 4.72) 0] dx 6316.64π ∫ 95 85 −15 74807.65 − 9 − 4 − 2 + 2 2 = 1 So in the end, we found the area enclosed by the graphs as 284767.32 which means the volume of the cup is 84.77cm2 3 B.

Surface area To facilitate the calculation, I divided the overall graph into two different parts: Part I (inside) Part II (outside) (x) (x 5)(x 1.77)(x 7) 0f = −2071265 − 8 − 5 − 1 + 2 (x) (x 5)(x 5.34)(x 4.72) 0f = −15 74807.65 − 9 − 4 − 2 + 2 Figure 7. Graphs divided into two parts to facilitate the calculation In this part, the formula to be used is: πf (x) dx ∫ b a 2 √1 + f (x)[ ′ ]2 Part I (for the equation ):(x) (x 5)(x 1.77)(x 7) 0f = −2071265 − 8 − 5 − 1 + 2 π[ (x 5)(x 1.77)(x 7) 0] dx ∫ 85 2 −2071265 − 8 − 5 − 1 + 2 √1 + (x 5)(x 1.77)(x 7) 0 [ −2071265 − 8 − 5 − 1 + 2 ] ′ 2 3924.38= 1 Part II (for the equation ):(x) (x 5)(x 5.34)(x 4.72) 0f = −1574807.65 − 9 − 4 − 2 + 2 π[ (x 5)(x 5.34)(x 4.72) 0] dx ∫ 95 2 −1574807.65 − 9 − 4 − 2 + 2 √1 + (x 5)(x 5.34)(x 4.72) 0 [ −1574807.65 − 9 − 4 − 2 + 2 ] ′ 2 6301.54= 1 Since the bottom surface is formed by circles, we can easily find the area by using the well-known formula . Both inner and outer radiuses are equal and 20, so the additional rπ 2 surface area of the bottom part is: x π20 513.282 2 = 2 Therefore, our final surface area calculated is 32739.2, which can be expressed as 27.39cm3 2.


In the end, I found the volume of my cup as , and the surface area as . 84.77cm2 3 27.39cm3 2 Both of the properties are realistic to form a Turkish teacup. Since teacups do not have a constant ratio and only have the speciality of being thin in the middle, I’m unable to check my results. That’s why I wish I had used an existing teacup, in that way I could be able to compare my findings with an original one.

However, I believe my model was an average one, but I’m unable to justify since I can’t find the volume and surface area values of different cups in the same style. In addıtion, my values have an uncertainty of ±0,01 caused by the digits of the numbers after the dot. I did not state the error one by one for every property since it’s a small error and I was looking for the approximate values. For further investigations like this, I suggest using a real cup to calculate these properties.


Calculating the volume and surface area by drawing the cup graph with two graphs was really helpful for me to minimize the error since it added thickness to my model and created a more possible structure. I’m also glad that I chose the numbers in the graph according to true values in mm, this allowed me to calculate the possible values in and . So, instead of mc 3 mc 2 finding a random value of volume and surface area, I actually found the real ones.

Drawing the shape of the teacup on the graph paper also helped me significantly to create a realistic model. My results are now meaningless since my cup is imaginary. This method of calculating volume and surface area of the cups can be used by glass manufacturers to model a glass with definite volume or surface area.

It’s possible to notice that I did not calculate the volume and the surface area for the mouth part of the cup individually since I believe it will add an insignificant value for both properties. This research caused a research question to rise in my mind; what is the relation between the price of a glass with its volume and its surface area? Since I only calculated these properties for one type of glass, I am unable to answer this question, but further researches can take place to investigate this.

Updated: Feb 06, 2024
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Finding the Volume and the Surface Area of a 3-Dimensional Teacup Model. (2024, Feb 06). Retrieved from

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