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Mathematics AA SL
Mathematics AA SL
Sample Internal Assessment
Sample Internal Assessment

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Rationale
Aim
Research question
Introduction
Conclusion
Bibliography

What will be the cost of painting the Eiffel Tower?

What will be the cost of painting the Eiffel Tower? Reading Time
11 mins Read
What will be the cost of painting the Eiffel Tower? Word Count
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Table of content

Rationale

My dad is the owner of a paint shop. That doesn't only include selling the definite colour of paint, one needs to have an adequate knowledge of the quantity required for painting a certain area, so that cost can be estimated.

 

It is very much essential because everyone asks for a budget before getting anything painted.

 

I love the smell of paints. I have been visiting the shop for a long time now. I love the environment at the shop. There are times when customers ask for suggestions to choose colors, I very willing participate in the discussion.

 

At school, I am taught to calculate area. This really helped me to guess the quantity of paint a customer would require when he came with measurements. Once the quantity was guessed, I could predict the cost. It was correct. I felt extremely happy and proud.

 

That was not all. One day, dad asked me to calculate how much paint will be required to paint the exterior of a customer's house. I could not solve. It wasn't completely symmetric. My mathematics formulae could not be applied.

 

This urged me to study more. I devoted lot of time into this. I was so much into this, that every time I came across a structure that could be painted, I would try estimating the quantity of paint so that I could calculate the cost.

 

Once I visited the Eiffel tower. Such a beautiful architecture it is! I was completely mesmerized.

 

I started to gather information about it. I read about its architecture, its height and other features. Owing to my habit, the thought of estimating the cost of painting the Eiffel tower did not leave me. I tried finding about it but in vain. This is the reason I am doing this IA.

Aim

The main motive of this IA is to study the surface area of the Eiffel Tower. Moreover, as Eiffel Tower is one of the most visited tourist destinations in the world, it needs to be maintained in a regular basis. As it is made up of iron, it needs to be painted very often to avoid rusting. So, calculation of approximate cost of painting the Eiffel Tower is also another motive of this IA.

Research question

What is the surface area of the Eiffel Tower and the cost of painting it?

Introduction

  • What is Eiffel Tower?

Eiffel Tower has designed by French civil engineer and architect, Gustave Eiffel. It took almost 2 years to complete the construction of one of the tallest monuments in the World and also one of the most visited tourist destinations.

Figure 1 - Measurement Of Eiffel Tower
Figure 1 - Measurement Of Eiffel Tower
  • Measurement of Eiffel Tower:

It is a cone shaped monument with a height of 324 m and 100 m in width.

  • Rate of Painting in Paris

In reference to the cost of painting, usually 1 liter of paint will usually cover 11square meters of area (theoretically). The cost of paint varies with a high range. The cost of paint ranges between $30 to $120 dollars per gallon of paint depending upon the colour and quality of the paint.

  • Methodology of Calculation:

In the process of calculation, as the structure of Eiffel Tower is symmetrical, I will find the surface area of half of any face of the Eiffel Tower and then I will multiply it by 8 to get the entire surface area of the Eiffel Tower.

Process of calculation

Process of Calculation: In this investigation, I will calculate the total surface area of three above mentioned types of Eiffel Tower. As the Eiffel Tower is not a regularly shaped structure, i.e., it has empty spaces in between the structure, thus, first of all, total surface area including the empty spaces will be calculated. Then, the area of the spaces will be calculated and subtracted from the total surface area. Desmos is used to find the equation of the curve.

Section 1

Figure 2 - Place The Image Of Wine Glass In Desmos Software.
Figure 2 - Place The Image Of Wine Glass In Desmos Software.

So, we can observe that the equation of curve for the above - mentioned limits, i.e., from x = 0 to x = 1.5.

 

y = - 0.5 x + 1.85

 

From the equation, I can state that y = f (x). Thus, we can use the formula of Surface area as:

 

\(Surface\ Area=\displaystyle\int\limits ^b_a2\pi ydB\)

 

\(dB=\sqrt{1+(\frac{dy}{dx})^2dx}\)

 

Therefore, by differentiating y with respect to x, we get,

 

\(\frac{dy}{dx}=\frac{d(-0.5x+1.85)}{dx}\)

 

\(=>\frac{dy}{dx}=\frac{d(-0.5x)}{dx}+\frac{d(1.85)}{dx}\)

 

\(=>\frac{dy}{dx}=-0.5\)

 

\(∴dB=\sqrt{1+(-0.5)^2dx}\)

 

\(∴surface\ Area=\displaystyle\int\limits^{1.5}_02\pi(-0.5x+1.85)\sqrt{1+(-0.5)^2}dx\)

 

\(=\displaystyle\int\limits^{1.5}_0\sqrt5\bigg(\frac{37}{20}-\frac{x}{2}\bigg)dx\)

 

\(=\frac{37\pi}{4\sqrt{5}}\displaystyle\int\limits^{1.5}_01dx-\frac{\sqrt{5\pi}}{2}\displaystyle\int\limits^{1.5}_{0}xdx\)

 

\(=\frac{37\pi}{4\sqrt{5}}\times[x]^{1.5}_0-\frac{5\pi}{2}\times[\frac{x^2}{2}]^{1.5}_{0}\)

 

\(=\frac{37\pi}{4\sqrt{5}}\times1.5-\frac{\sqrt{5}\pi}{4}\times2.25\)

 

= 15.54

Figure 3
Figure 3

So, we can observe that the equation of curve for the above-mentioned limits, i.e., from x = 0 to x = 1.2.

 

y = - 1. 14 x+ 0.49 x - 1. 14

 

From the equation, I can state that y = f(x). Thus, we can use the formula of Surface area as:

 

\(Surface\ Area=\displaystyle\int\limits^b_a2\pi ydB\)

 

\(dB=\sqrt{1+\big(\frac{dy}{dx})^2}dx\)

 

Therefore, by differentiating y with respect to x, we get,

 

\(\frac{dy}{dx}=\frac{d(-1.14x^2+0.49x+1.14)}{dx}\)

 

\(=>\frac{dy}{dx}=\frac{d(-1.14x^2)}{dx}+\frac{d(0.49x)}{dx}+\frac{d(1.14)}{dx}\)

 

\(=>\frac{dy}{dx}\) = - 2.28 x + 0.49

 

\(∴dB=\sqrt{1+(-2.28x+0.4)^2dx}\)

 

\(∴Surface\ Area\)

 

\(=\displaystyle\int\limits^{12}_02\pi (-1.14x^2+0.49x-1.14)\sqrt{1+(-2.28x+0.4)^2}dx\)

 

\(\displaystyle\int2\pi\sqrt{\bigg(\frac{2}{5}-\frac{57x}{25}\bigg)^2+1}\ \ \bigg(-\frac{57x^2}{50}+\frac{49x}{100}+\frac{57}{50}\bigg)dx\)

 

\(=\displaystyle\int\bigg(-\frac{57\pi x^2\sqrt{57x-10)^2+625}}{625}\bigg)+\)

 

\(=\frac{49\pi x\sqrt{57x-10)^2+625}}{1250}+\frac{57\pi \sqrt{(57x-10)^2+625}}{625}\bigg)dx\)

 

Applying linearity:

 

\(=\frac{-57\pi}{625}\displaystyle\int x^2\sqrt{(57x-10)^2+625}\ dx\)

 

\(+\frac{49\pi}{1250}\displaystyle\int x\sqrt{(57x-10)^2+625}\ dx\)

 

\(+\frac{57\pi}{625}\displaystyle\int\sqrt{57x-100)^2+625}\ dx\)

 

Know solving

 

\(\displaystyle\int x^2\sqrt{(57x-10)^2+625}\ dx\)

 

Substitude, u = 57x - 10

 

\(\frac{du}{dx}=57\)

 

\(​​​​​​​dx=\frac{1}{57}du\)

 

\(x^2=\frac{(u+10)^2}{3249}\)

 

\(=\frac{1}{185793}\displaystyle\int(u+10)^2\sqrt{u^2+625}\ du\)

 

Now solving: 

 

\(\displaystyle\int(u+10)^2\sqrt{u^2+625}\ du\)

 

\(=\displaystyle\int\limits\big(u^2\sqrt{u^2+625}+20u\sqrt{u^2+625}+100\sqrt{u^2+625}\big)du\)

 

Now solving: 

 

\(\displaystyle\int u^2\sqrt{u^2+625}\ du\)

 

\(u=25\ tan \theta\rightarrow\theta=arctan\bigg(\frac{u}{25}\bigg)\)

 

\(du=25sec^2\theta \ d\theta\)

 

\(=\displaystyle\int15625\ sec^2\theta\ tan^2\theta\sqrt{625\ tan^2\theta+625}\ du\)

 

\(=390625\displaystyle\int sec^3\theta\ tan^2\theta\ d\theta\)

 

Now solving:

 

\(\displaystyle\int sec^3 \theta\ tan^2\theta \ du\)

 

\(=\displaystyle\int sec^3\theta(sec^2\theta-1)du\)

 

\(=\displaystyle\int sec^5\theta\ - sec^2\theta-1)du\)

 

\(=\displaystyle\int sec^5\theta\ d\theta- \displaystyle\int sec^3\theta\ d \theta\)

 

Now solving:

 

\(\displaystyle\int sec^5\theta\ d\theta\)

 

\(=\frac{sec^3\theta\ tan\theta}{4}+\frac{3}{4}\displaystyle\int \theta\ d\theta\)

 

Now solving:

 

\(\displaystyle\int sec^3 \theta\ d\theta\)

 

\(=\frac{sec\theta\ tan\theta}{2}+\frac{1}{2}\displaystyle\int sec\theta\ d\theta\)

 

Now solving:

 

\(\displaystyle\int sec\theta\ d\theta\)

 

\(=In(tan \theta+sec \theta)\)

 

plug in solved integral:

 

\(\frac{sec\theta\ tan \theta}{2}+\frac{1}{2}\displaystyle\int \sec\theta\ d\theta\)

 

\(=\frac{In\ (tan\theta+sec\theta)}{2}+\frac{sec\theta \ tan\theta}{2}\)

 

plug in solved integral:

 

\(\frac{sec^3\theta\ tan \theta}{4}+\frac{3}{4}\displaystyle\int sec^3\theta\ d\theta\)

 

\(=\frac{3\ In\ (tan \theta+sec \theta)}{8}+\frac{sec^3 \theta \tan\theta}{4}+\frac{3\ sec \theta\ \tan\theta}{8}\)

 

Now solving:

 

\(\displaystyle\int sec^3\theta\ \ d\theta\)

 

\(=\frac{In\ (tan \theta+sec\theta)}{2}+\frac{secθ\ \tanθ}{2}\)

 

plug in solved integral:

 

\(\displaystyle\int sec^5θ\ dθ-\displaystyle\int sec^3θ\ dθ\)

 

\(=\frac{In\ (tanθ+secθ)}{8}+\frac{sec^3θ\ tanθ}{4}-\frac{secθ\ tanθ}{8}\)

 

Now: 

 

\(3902625\displaystyle\int sec^3θ\ tanθ^2θ\ dθ\)

 

\(=\frac{390625\ In(tanθ+secθ)}{8}+\frac{390625\ sec^3θ\ dθ}{4}\)

 

\(\frac{-390625secθ\ tanθ}{8}\)

 

\(θ=arctan\bigg(\frac{4}{25}\bigg)\)

 

\(tan=\bigg(arctan\bigg(\frac{4}{25}\bigg)\bigg)=\frac{4}{25}\)

 

\(sec=\bigg(arctan\bigg(\frac{4}{25}\bigg)\bigg)=\sqrt{\frac{4^2}{625}+1}\)

 

\(=\frac{-39065\ In \Big(\sqrt{\frac{u^2}{625}\ +\ 1}\ +\ \frac{u}{25}\Big) }{8} \ +\frac{156254\bigg(\frac{u^2}{625}+1\bigg)^{\frac{3}{2}}}{4}\)

 

\(-\frac{15625\sqrt{\frac{u^2}{625}+1}}{8}\)

 

Now solving:

 

\(\displaystyle\int u\sqrt{u^2+625}\ du\)

 

θ = u2 + 625

 

\(\frac{dθ}{du}=2u\bigg|du=\frac{1}{2u}dθ\)

 

\(=\frac{1}{2}\displaystyle\int\sqrt{θ}\ dθ\)

 

\(=\frac{2θ\frac{3}{2}}{3}\times\frac{1}{2}=\frac{θ\frac{3}{2}}{3}\)

 

\(=\frac{(u^2+625)\frac{3}{2}}{3}\)

 

Now solving:

 

\(\displaystyle\int\sqrt{u^2+625}du\)

 

\(u=tan(θ)\rightarrowθ=arctan\bigg(\frac{u}{25}\bigg)\)

 

du = 25 sec2 θd θ

 

\(=\displaystyle\int25sec^2 θ\sqrt{625tan^2θ+625}d θ\)

 

\(=625\displaystyle\int sec^3 θd θ \)

 

\(\frac{625\ In\bigg(\sqrt{\frac{u^2}{625}+1} \ +\frac{4}{25}\bigg)}{2}+\frac{25u\sqrt{\frac{u^2}{625}+1}}{2}\)

 

plug in solved:

 

\(=\frac{-140625\ In\bigg(\sqrt{\frac{u^2}{625}+1}+\frac{u}{25}}{8}+\frac{20\big(u^2+625\big)^{\frac{3}{2}}}{3}\)

 

\(+\frac{15625u\big(\frac{u^2}{625}+1\big)^{\frac{3}{2}}}{4}-\frac{5625u\sqrt{\frac{u^2}{625}+1}}{8}\)

 

\(=\frac{-15625\ In\bigg(\sqrt{\frac{u^2}{625}+1}+\frac{u}{25}}{164616}+\frac{20\big(u^2+625\big)^{\frac{3}{2}}}{555579}\)

 

\(+\frac{15625u\bigg(\frac{u^2}{625}\bigg)^{\frac{3}{2}}}{740772}-\frac{625u\sqrt{\frac{u^2}{625}+1}}{164616}\)

 

Now solving:

 

\(\displaystyle\int x\sqrt{(57x-10)^2+625}\ dx\)

 

\(=\displaystyle\int \frac{(57x-10)\sqrt{(57x-10)^2+625}}{57}\)

 

\(+\frac{10\sqrt{(57x-10)^2+625}}{57}dx\)

 

\(=\frac{1}{57}\displaystyle\int(57x-10)\sqrt{(57x-10)^2+625}\ dx\)

 

\(+\frac{10}{57}\displaystyle\int\sqrt{(57x-10)^2+625}dx\)

 

Now solving:

 

\(\displaystyle\int(57x-10)(\sqrt{57x-10)^2+625\ dx)}\)

 

substitute:

 

\(u=(57x-10)^2+625\)

 

\(\frac{du}{dx}=114(57x-10)\)

 

\(dx=\frac{1}{114(57x-10)}du\)

 

\(=\frac{1}{144}\displaystyle\int\sqrt{u}du\)

 

\(=\frac{1}{144}\times\frac{2}{3}u^{\frac{3}{2}}\)

 

\(=\frac{u^\frac{2}{2}}{171}\)

 

\(=\frac{\bigg((57x-10)^2+625\bigg)^{\frac{3}{2}}}{171}\)

 

Now:

 

\(\displaystyle\int\sqrt{(57x-10)^2+625}\ dx\)

 

\(=\frac{1}{57}\displaystyle\int\sqrt{u^2+625}\ du\)

 

\(=\frac{625\ In\bigg(\frac{u^2}{625}+1+\frac{u}{25}\bigg)}{2}+\frac{25u\frac{u^2}{625}+1}{2}\)

 

plug in solved integral:

 

\(\frac{1}{57}\displaystyle\int\sqrt{u^2+625}\ du\)

 

\(=\frac{625\ In\bigg(\sqrt{\frac{u^2}{625}}+1+\frac{u}{25}\bigg)}{114}+\frac{25u\sqrt{\frac{u^2}{625}+1}}{114}\)

 

\(=\frac{625\ In\bigg(\sqrt{\frac{(57x-10)^2}{625}}+1+\frac{57x-10}{25}\bigg)}{114}+\)

 

\(\frac{25(57x-10)\sqrt{\frac{(57x-10)^2}{625}+1}}{114}\)

 

plug in the solved integral:

 

\(=\frac{3125\ In\bigg(\sqrt{\frac{(57x-10)^2}{625}}+1+\frac{57x-10}{25}\bigg)}{3249}+\frac{\bigg((57x-10)^2+625\bigg)^{\frac{3}{2}}}{9747}\)

 

\(+\frac{125(57x-10)\sqrt{\frac{(57x-10)^2}{625}+1}}{3249}\)

 

Now solving:

 

\(\displaystyle\int\sqrt{(57x-10)^2+625}\ dx\)

 

Using previous result:

 

\(=\frac{625\ In\bigg(\sqrt{\frac{(57x-10)^2}{625}}+1+\frac{57x-10}{25}\bigg)}{114}\)

 

\(+\frac{25(57x-10)\sqrt{\frac{(57x-10)^2}{625}+1}}{114}\)

 

plug in the solved integral in first equation:

 

\(=\frac{14201\ \pi\ In\bigg(\sqrt{\frac{(57x-10)^2}{625}}+1+\frac{57x-10}{25}\bigg)}{25992}+\frac{\pi\big((57x-10)^2+625\big)^{\frac{3}{2}}}{1353750}\)

 

\(-\frac{25\ \pi(57x-10)\bigg(\frac{(57x-10)^2}{625}+1\bigg)^{\frac{3}{2}}}{12996}+\frac{14201\ \pi(57x-10)\sqrt{\frac{(57x-10)^2}{625}+1}}{649800}\)

 

\(=\displaystyle\int\pi\bigg(8875625\ In\big(\big|\sqrt{(57x-10)^2+625}+57x-10\big|\big)\)

 

\(-(370386x^3-233928x^2-690327x\)

 

\(+118810)\sqrt{(57x-10)^2+625}\bigg)16245000\)

 

\(=2.000742\)

 

Effective Surface Area = 15.54-9.000 = 6.54 sq. unit

Section 2

Figure 4 - Place The Image Of Wine Glass In Desmos Software
Figure 4 - Place The Image Of Wine Glass In Desmos Software

So, we can observe that the equation of curve for the above-mentioned limits, i.e., from x = 1.2 to x = 1.95.

 

y = 1.15

 

From the equation, I can state that y = f(x). Thus, we can use the formula of Surface area as:

 

Surface Area = \(\displaystyle\int\limits2\pi ydB\)

 

\(dB=\sqrt{1+(\frac{dy}{dx})^2}dx\)

 

Therefore, by differentiating y with respect to x, we get,

 

\(\frac{dy}{dx}=\frac{d(1.15)}{dx}\)

 

\(=>\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+0}dx=dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{1.85}_{1.2}2\pi(1.15)dx\)

 

\(\displaystyle\int\limits^{1.85}_{1.2}2\pi(1.15)dx\)

 

\(=2\pi(1.15)\displaystyle\int\limits^{1.85}_{1.2}dx\)

 

\(=2\pi(1.15)[x]^{1.85}_{1.2}dx\)

 

\(=2\pi(1.15)[1.85-1.2]\)

 

= 4.69

Section 3

Figure 5
Figure 5

So, we can observe that the equation of curve for the above-mentioned limits, i.e., from x = 1.95 to x = 3.1.

 

y = - 0.3x + 1.5

 

From the equation, I can state that y = f(x). Thus, we can use the formula of Surface area as:

 

\(Surface\ Area=\displaystyle\int\limits ^b_a2\pi ydB\)

 

\(dB=\sqrt{1+(\frac{dy}{dx})^2}dx\)

 

Therefore, by differentiating y with respect to x, we get,

 

\(\frac{dy}{dx}=\frac{d(-0.3+1.5)}{dx}\)

 

\(=>\frac{dy}{dx}=\frac{d(-0.3x)}{dx}+\frac{d(1.5)}{dx}\)

 

\(=>\frac{dy}{dx}=-0.3\)

 

\(∴dB=\sqrt{1+(-0.3)^2}dx\)

 

\(∴Surface\ Area=\displaystyle\int\limits^{3.1}_{1.95}2\pi(-0.3x+1.5)\sqrt{1+(-0.3)^2}dx\)

 

\(=\displaystyle\int\limits^{3.1}_{1.95}\frac{\sqrt{109}\pi(1.5-0.3x)}{5}dx\)

 

\(=\frac{3\sqrt{109}\pi}{10}\displaystyle\int\limits^{3.1}_{1.95}1dx-\frac{3\sqrt{109}\pi}{50}\displaystyle\int\limits^{3.1}_{1.95}xdx\)

 

\(=\frac{3\sqrt{109}\pi}{10}[x]^{3.1}_{1.95}-\frac{3\sqrt{109}\pi}{50}\displaystyle\int\limits^{3.1}_{1.95}\big[\frac{x^2}{2}\big]^{3.1}_{1.95}\)

 

\(=\frac{3\sqrt{109}\pi}{10}[3.1-1.95]-\frac{3\sqrt{109}\pi}{100}[9.61-3.8025]\)

 

= 5.601

Figure 6
Figure 6

So, we can observe that the equation of curve for the above-mentioned limits, i.e., from x = 1.95 to x = 3.1.

 

y = - 0.23x + 0.92

 

From the equation, I can state that y = f(x). Thus, we can use the formula of Surface area as:

 

\(Surface\ Area=\displaystyle\int\limits^b_a2\pi ydB\)

 

\(dB=\sqrt{1+(\frac{dy}{dx})^2}dx\)

 

Therefore, by differentiating y with respect to x, we get,

 

\(\frac{dy}{dx}=\frac{d(-0.23x+0.92)}{dx}\)

 

\(=>\frac{dy}{dx}=\frac{d(-0.23x)}{dx}+\frac{d(0.92)}{dx}\).

 

\(=>\frac{dy}{dx}=-0.23\)

 

\(∴dB=\sqrt{1+(-0.23)^2}dx\)

 

\(∴Surface\ Area= \displaystyle\int\limits^{3.1}_{1.95}2\pi(-0.23x+0.92)\sqrt{1+(-0.23)^2}dx\)

 

\(= \displaystyle\int\limits^{3.1}_{1.95}\frac{\sqrt{10529}\pi(\frac{23}{25}-\frac{23}{100}x)}{50}dx\)

 

\(=\frac{23\sqrt{10529}\pi}{1250} \displaystyle\int\limits^{3.1}_{1.95}1dx-\frac{23\sqrt{10529}\pi}{5000}\displaystyle\int\limits^{3.1}_{1.95}xdx\)

 

\(=\frac{23\sqrt{10529}\pi}{1250}[x]^{3.1}_{1.95}-\frac{23\sqrt{10529}\pi}{5000}\displaystyle\int\limits^{3.1}_{1.95}[\frac{x^2}{2}]^{3.1}_{1.95}\)

 

\(=\frac{23\sqrt{10529}\pi}{1250}\) [3.1 - 1.95] - \(\frac{23\sqrt{10529}\pi}{10000}\) [9.61 - 3.8025]

 

\(=\frac{23\sqrt{10529}\pi}{1250}\) [3.1 - 1.95] - \(\frac{23\sqrt{10529}\pi}{10000}\) [9.61 - 3.8025]

 

= 2.515

 

Therefore, effective surface area = 5.601-2.515 = 3.086 sq. unit

 

Section 4:

Figure 7
Figure 7

So, we can observe that the equation of curve for the above - mentioned limits, i.e., from x = 3.1 to x = 3.65.

 

y = 0.6

 

From the equation, I can state that y = f(x). Thus, we can use the formula of Surface area as:

 

Surface Area = \(\displaystyle\int\limits^b_a2\pi ydB\)

 

\(dB=\sqrt{1+(\frac{dy}{dx})^2}dx\)

 

Therefore, by differentiating y with respect to x, we get,

 

\(\frac{dy}{dx}=\frac{d(0.6)}{dx}\)

 

\(=>\frac{dy}{dx}=0\)

 

\(∴dB=\sqrt{1+0}dx=dx\)

 

∴ Surface Area = \(\displaystyle\int\limits^{3.65}_{3.1}2\pi(0.6)dx\)

 

\(=\displaystyle\int\limits^{3.65}_{3.1}2\pi(0.6)dx\)

 

\(=2π(0.6)\displaystyle\int\limits^{3.65}_{3.1}dx\)

 

\(=2π(0.6)[x]^{3.65}_{3.1}\)

 

= 2π(1.15)[3.65-3.1]

 

= 2.07

Section 5

Figure 8
Figure 8

So, we can observe that the equation of curve for the above - mentioned limits, i.e., from x = 3.6 to x = 8.5.

 

y = 0.014x- 0.24x + 1.189

 

From the equation, I can state that y = f(x). Thus, we can use the formula of Surface area as:

 

Surface Area = \(\displaystyle\int\limits^b_a2\pi ydB\)

 

dB =\(\sqrt{1+(\frac{dy}{dx})^2}dx\)

 

Therefore, by differentiating y with respect to x, we get,

 

\(\frac{dy}{dx}=\frac{d(0.014x^2-0.24x+1.189)}{dx}\)

 

\(=>\frac{dy}{dx}=\frac{d(0.014x^2)}{dx}+\frac{d(-0.24x)}{dx}+\frac{d(1.189)}{dx}\)

 

\(=> \frac{dy}{dx}\) = 0.028x - 0.24

 

\(∴dB=\sqrt{1+(0.028x-0.24)^2}dx\)

 

∴Surface Area = \(\ \displaystyle\int\limits^{8.5}_{3.6}2\pi(0.014x^2-0.24x+1.189)\)

 

\(\sqrt{1+(0.028x-0.24)^2}\ dx\)

 

\(\displaystyle\int2\pi\sqrt{\bigg(\frac{7x}{250}-\frac{6}{25}\bigg)^2+1}\bigg(\frac{7x^2}{500}-\frac{6x}{25}+\frac{1189}{1000}\bigg)dx\)

 

\(=\displaystyle\int\bigg(\frac{7\pi x^2\sqrt{(7x-60)^2+62500}}{62500}\)

 

\(-\frac{6\pi x\sqrt{(7x-60)^2+62500}}{3125}\)

 

\(+\frac{1189\pi\sqrt{(7x-60)^2+62500}}{125000}\bigg)dx\)

 

\(=\frac{7\pi}{62500}\displaystyle\int x^2\sqrt{(7x-60)^2+62500}dx\)

 

\(-\frac{6\pi}{3125}\displaystyle\int x\sqrt{(7x-60)^2+62500}dx\)

 

\(+\frac{1189\pi}{125000}\displaystyle\int\sqrt{(7x-60)^2+62500}dx\)

 

Now solving

 

\(\displaystyle\int x^2\sqrt{(7x-60)^2+62500}dx\)

 

\(​​​​​​​Let, u=7x-60 \bigg|\frac{du}{dx}=7\rightarrow dx=\frac{1}{7}du\)

 

\(x^2=\frac{(4+60)^2}{49}\)

 

\(=\frac{1}{343}\displaystyle\int(u+60)^2\sqrt{u^2+62500}du\)

 

\(=\displaystyle\int\bigg(u^2\sqrt{u^2+62500}+120u\sqrt{u^2+62500}\)

 

\(+3600\sqrt{u^2+62500}\bigg)du\)

 

\(=\displaystyle\int u^2\sqrt{u^2+62500}du\)

 

\(​​​​​​​Let, u=250\ tan(\theta)\rightarrow\theta=arctan\bigg(\frac{u}{250}\bigg)\)

 

\(du=250\ sec^2\theta\ {d\theta}\)

 

\(=\displaystyle\int15625000\ sec^2\theta\ tan^2\theta\ d\theta\)

 

Applying previous results

 

\(=488281250\ In\bigg(\sqrt{\frac{u^2}{62500}+1+}\frac{u}{250}\bigg)\)

 

\(+3906250u\bigg(\frac{u^2}{62500}+1\bigg)^{\frac{3}{2}}\)

 

\(-1953125u\sqrt{\frac{u^2}{62500}+1}\)

 

Now solving:

 

\(\displaystyle\int u\sqrt{u^2+62500}\ du\)

 

Substitute:

 

\(\displaystyle\int u\sqrt{u^2+62500}du\)

 

\(\frac{d\theta}{du}=2u \rightarrow du=\frac{1}{2u}d\theta\)

 

\(\theta=u^2+62500\rightarrow\frac{d\theta}{du}=2u\)

 

\(du=\frac{1}{2u}d\theta\)

 

\(=\frac{1}{2}\displaystyle\int\sqrt{\theta}d\theta\)

 

Using previous solution

 

\(=\frac{\theta^{\frac{2}{3}}}{3}\)

 

\(=\frac{(u^2+62500)^{\frac{3}{2}}}{3}\)

 

Now solving:

 

\(\displaystyle\int\sqrt{u^2+62500}du\)

 

Substitute,

 

\(u=250\ tan\theta\)

 

\(\theta=arctan\bigg(\frac{u}{250}\bigg)\)

 

\(du=250\ sec^2\theta\ d\theta\)

 

\(=\displaystyle\int250\ sec^2\theta\sqrt{62500tan^2\theta\ +\ 62500}\ d\theta\)

 

\(=62500\displaystyle\int sec^3\theta\ d\theta\)

 

\(=62500\times\bigg(In(tan\theta\ +\ sec\theta)+(31250\ sec\theta\ tan\theta)\bigg)\)

 

\(=31250\ In\bigg(\sqrt{\frac{u^2}{62500}+1+}\frac{u}{250}\bigg)+125u\sqrt{\frac{u^2}{62500}+1}\)

 

Plug in solved integral:

 

\(\displaystyle\int u^2\sqrt{u^2+62500}\ du+120\displaystyle\int u\sqrt{u^2+62500}\ du\)

 

\(+3600\displaystyle\int\sqrt{u^2+62500}\ du\)

 

\(=-375781250\ In\bigg(\sqrt{\frac{u^2}{62500}}+1+\frac{u}{250}\bigg)\)

 

\(+40\big(u^2+62500\big)^{\frac{3}{2}}+\)

 

\(390625u\bigg(\frac{u^2}{62500}+1\bigg)^{\frac{3}{2}}-1503125u\sqrt{\frac{u^2}{62500}+1}\)

 

Plug in solved integral:

 

\(\frac{1}{343}\displaystyle\int(u+60)^2\sqrt{u^+62500}\ du\)

 

\(=-\frac{375781250\ In\ \bigg(\sqrt{\frac{u^2}{62500}+1}+\frac{u}{250}\bigg)}{343}+\frac{40(u^2+62500)^{\frac{3}{2}}}{343}\)

 

\(+\frac{3906250u \bigg(\frac{u^2}{62500}+1\bigg)^{\frac{3}{2}}}{343}-\frac{1503125u\sqrt{{\frac{u^2}{62500}+1}}}{343}\)

 

\(=\frac{375781250\ In\ \bigg(\sqrt{\frac{(7x-60)^2}{62500}+1}+\frac{7x-60}{250}\bigg)}{343}+\frac{40((7x-60)^2+62500)^{\frac{3}{2}}}{343}\)

 

\(+\frac{3906250(7x-60)\bigg(\frac{(7x-60)^2}{62500}+1\bigg)^{\frac{3}{2}}}{343}-\frac{1503125(7x-60)\sqrt{\frac{(7x-60^2}{62500}+1}}{343}\)

 

Now:

 

\(\displaystyle\int x\sqrt{(7x-60)^2+62500}\ dx\)

 

\(=\frac{1}{7}\displaystyle\int(7x-60)\sqrt{(7x-60)^2+62500}\ dx\)

 

\(+\frac{60}{7}\displaystyle\int\sqrt{(7x-60)^2+62500}\ dx\)

 

\(=\frac{1875000\ In\bigg(\sqrt{\frac{(7x-60)^2}{62500}+1}+\frac{7x-60}{250}\bigg)}{49}+\frac{\big((7x-60)^2+62500\big)^{\frac{3}{2}}}{147}\)

 

\(+\frac{7500(7x-60)\sqrt{\frac{(7x-60)^2}{62500}+1}}{49}\)

 

Now

 

\(\displaystyle\int\sqrt{(7x-60)^2+62500}\ dx\)

 

\(=\frac{31250\ In\bigg(\sqrt{\frac{(7x-60)^2}{62500}+}1+\frac{7x-60}{250}\bigg)}{7}+\frac{1250(7x-60)\sqrt{\frac{(7x-60)^2}{62500}+1}}{7}\)

 

Plugging in the integrals:

 

\(=-\frac{30127\pi\ In\bigg(\sqrt{\frac{(7x-60)^2}{62500}+1+}\frac{7x-60}{250}\bigg)}{196}+\frac{125\pi(7x-60)\bigg(\frac{(7x-60)^2}{62500}+1\bigg)}{98}^{\frac{3}{2}}\)

 

\(-\frac{30127\pi(7x-60)\sqrt{\frac{(7x-60)^2}{62500}+1}}{49000}\)

 

= 8.5789

 

Surface Area = 8.579 sq. unit

 

Surface Area of each Quadrant = 0.54 + 0.86 + 0.46 + 0.33 + 1.35 = 24.394 sq. unit

 

Surface Area of Eiffel Tower without the Head of the tower = 24.394 X 4 = 97.576 sq. unit

 

Surface Area of Eiffel Tower in metric system = \(\frac{97.576}{8.5}\) × 300 = 3443.85 sq. m

 

Surface Area of the Head of the Eiffel Tower = 5.5 sq. m

 

Total Surface Area of Eiffel Tower = 3443.85 + 5.5 = 3449.35 sq.m

 

But, as there are multiple rods in the structure of Eiffel Tower, it is not possible theoretically to find the value of the total surface area of the Eiffel Tower. In fact, the surface area of the Eiffel Tower is 4415 sq. m.​​​​​​​

Requirement of paint

Total Surface Area = 4415 sq. m

 

Therefore, required quantity of paint \(\frac{4415}{11}\) = 402 liter

 

Therefore, 402 litre of paint is required to coat the Eiffel Tower just once.

 

Therefore, cost of paint will range between:

 

Lower Limit = $(30 X 402) = $12060

 

Upper Limit = $(120 X 402) = $48240.

Conclusion

In this IA, we have tried to find the total surface area of the Eiffel Tower. But we have come into a conclusion that as the Eiffel Tower is cone shaped but it is not uniform throughout. Moreover, the face of each side of Eiffel is not walled up, i.e., there is gaps in between the rods and the structure is more like a tuft of rods which extensively increases the surface area of the tower which cannot be calculated theoretically. The calculated value of surface area of Eiffel Tower is 1005 square meter whereas the actual surface area of the Eiffel Tower is 4415 square meter. The calculation has been done using desmos and an integration calculator. On the other hand, cost of paint required for single coating the Eiffel Tower may range between $12060 and $48240.

Bibliography