Deriving general conjectures for the surface area and volume of Torricelli’s Trumpet
Introduction and rationale:
As a passionate highschool mathematics student, I am often left unsatisfied by simply learning how to solve problems, or what to do when a specific type of problem comes up. Rather, I find that my passion for mathematics lies in proof; I want to know why and how every concept comes about in the first place.
While learning about volumes of revolution, a subtopic of integral calculus, I began to ask myself, “how did they think of this?” Upon further research, I stumbled upon an Italian physicist and mathematician by the name of Evangelista Torricelli. Torricelli thought of rotating the graph of f(x)=1/x upon the x-axis of a Cartesian plane, thus, obtaining what we now know as Torricelli’s Trumpet.
I personally found the concept of the Trumpet very interesting; according to Torricelli, its surface area is infinite, while its volume is definite, and this seemed very counter-intuitive to me. However, it is what sparked the initial idea behind my internal assessment. Throughout this assessment, I will be exploring the dimensions of Torricelli’s Trumpet, and eventually utilizing that knowledge in the derivation and application of two general conjectures detailing the surface area and volume of the Trumpet.
Proving Torricelli’s Trumpet’s infinite surface area and finite volume:
As previously mentioned, Torricelli’s Trumpet is formed by rotating the function f(x) by a measure of 2π radians about the x-axis of a Cartesian plane, when f(x)=1/x. This can be seen below;
In order to understand and prove the surface area and the volume of this shape, I will be using Cavilieri’s Principle. In the 17th century, Bonaventura Cavilieri deduced that when the cross-sectional areas of two different solids are equal, and the two solids have equivalent heights, their volumes, which would also be equal, are given by;
V=Bh
Here, the relationship between cross-sectional area (B), and height of the solids (h) is shown. This can be better understood in the following diagram;
According to Cavilieri’s Principle, all three of these shapes must have equal surface areas. (Image: Quintanilla, 2013)
Now, I will select a random point, given by (a,b), and inputting these coordinates into the previously defined function f(x) such that it is now defined by b=1/a . The point (a,b) on the graph will then be connected to the point (0,b) by a line segment, as shown below. The points – with the a-coordinate being chosen using an online randomizer and the y-coordinate subsequently deduced – will be (1/4,4).
*images to be added at a later date
Rotated about the x-axis by a measure of 2π radians, this line segment gives a cylindrical solid; hence, the following can be generalized about the process made above. In order to find the volume of that rotated solid, the following steps can be taken:
Choose a random point, (a,b).
Join the points (a,b) and (0,b) with a straight-line segment, parallel to the x-axis.
Rotate this line segment by a measure of 2π radians.
Hence, a solid cylindrical shape will be arrived at, with a height ‘h’, given by h=a, and a radius ‘r’, given by r=b. Using this information, as well as prior knowledge, the following can be deduced about the area of that solid cylinder (for the function b);
Area of the cylinder=2πrh=2πab, and b=1/a
∴Area of the cylinder=2π
We can here see that the curved surface area of the solid remains constant, independent of the values of a and b.
Although Torricelli’s Trumpet and the above cylinder are not necessarily linked by Cavilieri’s Principle, the concept can be used in order to make some preliminary deductions about the Trumpet. By proving that the area of the above cylinder is finite, the logic behind Cavilieri’s Principle would dictate that the surface area of Torricelli’s Trumpet is also infinite. This will now be explored.
Firstly, some variables must be defined:
x: the distance between an infinitesimally thin disk and a second disk, the closest one to y.
y: the radius of the circumference of the function’s rotated solid.
dx: a measure of the thickness of the rotated solid.
dy: the change in the already defined value of y, as a consequence of the change in x.
v: the volume of the rotated solid.
dv: the change in volume of the rotated solid.
Now, the change in the volume of this solid can be given by the following. This particular definition becomes clearer as it is later integrated, giving a formula for the volume of rotation itself;
dv=πy^2 dx
Hence, the volume of the solid can be given by this definite sum, and subsequently, a definite integral, from the previously defined values of a and b;
v=lim┬(dx→0)∑_(x=a)^(x=b)▒dv=lim┬(dx→0)〖∑_(x=a)^(x=b)▒〖πy^2 〗 dx〗
v=π∫_a^b▒〖(f(x))〗^2 dx
∴v=π∫_a^b▒(1/x)^2 dx
As we know, this is the formula for the volume of rotation of 1/x, or in other words, the volume of Torricelli’s Trumpet. Remembering that the domain of the function f(x) is given by x≥1, we can now give both a and b numerical values:
v=π∫_1^∞▒〖(1/x)^2 dx〗=π∫_1^∞▒〖1/x^2 dx〗=π∫_1^∞▒〖x^(-2) dx〗
Since by definition, ∫▒〖z^n dx〗=z^(n+1)/(n+1)+c,
v=π lim┬(n→∞)〖[-1/x]_1^n 〗=π lim┬(n→∞)(-1/n+1)
=-π(lim┬(n→∞)(1/n-1) )
Now, solving the equation within its limits,
∴v=-π(1/∞-1)=-π(0-1)=π
Hence, the volume of Torricelli’s Trumpet is finite and definite, independent of any variables. The solid has a constant volume of π, no matter what.
Next, I will calculate the surface area of Torricelli’s Trumpet. Again, I will be defining the function y=f(x); x_1=a,x_2=b, which will be rotated about the x-axis at a measure of 2π radians. Similar to before, a few terms must be defined before the surface area is found. These terms will be used to find this area, alongside some of the terms defined previously.
β: area (specifically, the surface area of the rotating solid).
y: the radius of the circumference of the solid, as it was previously defined.
dβ: the arc-length of the solid, or its maximum circumference.
Following on from the definition of arc-length in the solid, we can now mathematically define its symbol, using previously defined variables;
dβ=√(dx^2+〖dy〗^2 )=√(1+(dy/dx)^2 )
Hence, the curved surface area of the solid can be given by integrating dβ;
∴β=2π∫_a^b▒(y)dβ
=2π∫_a^b▒〖y√(1+(dy/dx)^2 )〗 dx
Again, remembering that the domain of the function f(x)=1/x is given by x≥1,
⇒2π∫_1^∞▒〖1/x √(1+(-1/x)^2 )〗 dx
=2π∫_1^∞▒〖1/x √(1+1/x^4 )〗 dx
Since it is given that x≥1, any number given by 1/x^4 must abide by 1/x^4 ≥0. Thus, the above integral – with its limits in mind – must give us a new definition for the number 1/x^4 ,
1/x^4 ≥√(1+0)≥1
⟹β=lim┬(n→∞)〖2π∫_1^n▒〖1/x dx〗〗
=2π lim┬(n→∞)〖[ln|x| ]_1^n 〗=2π lim┬(n→∞) (lnn-ln1 )
Now, resolving the above equation within its limits,
=2π(ln∞-ln1 )=∞
Hence, the function lim┬(n→∞)〖2π∫_1^n▒〖1/x dx〗〗 is divergent. Using the comparison test, it can thus also be deduced that 2π∫_1^∞▒〖1/x √(1+1/x^4 )〗 dx is also divergent, giving an infinite value for the surface area of Torricelli’s Trumpet ( sincelim┬(n→∞)〖2π∫_1^n▒〖1/x dx〗〗<2π∫_1^∞▒〖1/x √(1+1/x^4 )〗 dx).
Deriving general conjectures for the surface area and volume of Torricelli’s Trumpet:
Now, I will find two general conjectures; one for the volume of different variations of Torricelli’s Trumpet, and one for their surface areas. Following the above calculations, a given function f(x)=k/x; (k≥1) must have an infinite surface area and a finite volume when rotated. This will loosely be used below, in order to form these conjectures. Any variables used here are defined as they were previously (unless stated otherwise).
Firstly, I will derive a general conjecture for the volume of Torricelli’s Trumpet. For this, I will let a new function, g(x) be defined by g(x)=1/x^p . This function is to be rotated about the x-axis, by a measure of 2π radians. For the solid formed, we can say,
v=π∫_1^∞▒〖(1/x^p )^2 dx〗
=π∫_1^∞▒〖x^(-2p) dx〗
Integrating the function, we get,
=π[x^(-2p+1)/(-2p+1)]_1^∞=π/(1-2p) [1/x^(2p-1) ]_1^∞
And now, solving it within its limits,
=π/(1-2p) (1/∞^(2p-1) -1/1^(2p-1) )
=π/(1-2p) (0-1)
∴v=π/(2p-1)
Hence, as p tends to 1/2 , the volume tends to infinity. Otherwise, the volume of the solid is finite, and given by the above conjecture.
Finally, I will form a general conjecture for the surface area of Torricelli’s Trumpet. This, again will be done using the defined function g(x). This function is to be rotated about the x-axis, by a measure of 2π radians. In terms of this solid’s surface area, β, we can say,
β=2π∫_1^∞▒〖y√(1+(dy/dx)^2 ) dx〗
Now, differentiating y, which is given by y=1/x^p =x^(-p),
dy/dx=-px^(-(p+1) )=-p/x^(p+1)
According to their definitions, and the surface area proof presented above, this gives us the change in the radius of the circumference of the trumpet, with respect to its thickness. Subsequently,
⇒β=2π∫_1^∞▒〖1/x^p √(1+p^2/x^(2p+2) ) dx〗
From the above solution of dy/dx, we get that,
β=2π lim┬(n→∞)∫_1^n▒〖1/x^p (1/x^p +1/x^(2p+1) )dx〗
=2π lim┬(n→∞)∫_1^n▒(1/x^2p +1/x^(3p+1) )dx
=2π 〖 lim┬(n→∞) ∫_1^n▒(x^(-2p)+x^(-3p-1) ) 〗dx
Now, integrating the function,
2π lim┬(n→∞)〖[x^(-2p+1)/(-2p+1)+x^(-3p-1+1)/(-3p)]_1^n 〗
2π lim┬(n→∞)〖[1/((1-2p)∙x^(2p-1) )-1/(3p∙x^3p )]_1^n 〗
2π lim┬(n→∞)(1/((1-2p)∙n^(2p-1) )-1/(3p∙n^p )-1/(1-2p)+1/3p)
Now, this simplifies to the following, allowing us to finally apply the function’s limits,
=2π lim┬(n→∞)(1/(1-2p) (1/n^(2p-1) -1)-1/3p (1/n^3p -1))
=2π((1/(1-2p) (1/∞^(2p-1) -1)-1/3p (1/∞^3p -1)))
Hence, we now know that the surface area of the rotating function y=1/x^p between x=1 and x¬=∞ is given by,
=2π(0-1)(1/(1-2p)-1/3p)=-2π((5p-1)/3p(1-2p) )
Which thus, simplifies to give the following,
∴β=2π((1-5p)/(3p(1-2p)))
Thus, as p tends to 0, the value of the solid’s surface area tends to infinity. Otherwise, the surface area of the solid is given by the above conjecture.
Evaluating the accuracy of the derived conjectures in a real-life situation:
Although these formulae have been derived mathematically, they are rendered useless if they cannot be applied in the real world. While visiting Kolkata in India last winter, I came across a musical instrument that I had never seen before. As a trumpet player myself, I was intrigued by the sounds that the Indian shehnai produced. It had the visual appearance of a trumpet yet was wooden and produced a sound so different to what I was used to. I was amazed!
While reflecting on my trip later on that year, I began to realize a similarity between the shape of the shehnai and Torricelli’s Trumpet. Thus, I decided to use this instrument as a way to test my derived conjectures for the volume and surface area of Torricelli’s Trumpet.
There are two primary flaws in terms of the shehnai’s shape. Firstly, there is a slight ridge between the wide, sound producing trumpet and the body of the instrument. This ridge adds about 0.4 cm to the shape of the shehnai from every side. Also, the bottom end of the instrument has three small ridges, each of which add around 0.1 cm to the shape of the shehnai.
These ‘imperfections’ in the shape must be ignored in order to test the derived formulae on the shehnai. Although this will lead to some inaccuracy in the final results, this is mandatory, as considering these ridges will add a whole new level of complexity to this calculation.
In order to test these conjectures, I must first obtain values for the real volume and surface area of a shehnai instrument. For this, I will be using the average measurements of a shehnai. These are given in the diagram below:
These measurements are particularly useful for finding the surface area of the instrument. Although, due to the imperfect nature of the shehnai, some more approximations must be made; calculating its surface area manually would be very difficult, time-consuming, and expensive. Thus, I decided to approximate the shape of the shehnai to that of a right-cone. Albeit a major approximation, it is not one that has a significant impact on the overall surface area of the shehnai. Hence, the surface area of the shehnai, keeping in mind that the area of the circular top of the shehnai is to be deducted, was given by:
A_shehnai=πr(r+√(r^2+h^2 ))-πr^2=πr√(r^2+h^2 );h=28cm,r=4.0cm
A_shehnai=4π(4+√(4^2+〖28〗^2 ))=405.70〖cm〗^2
Finding the volume of the shehnai, on the other hand, is a far more accurate process. Luckily, I had bought a souvenir of the instrument when I visited Kolkata. The shehnai that I bought possessed the same dimensions as that represented above, and so presented an obvious method for measuring its volume. To obtain this value, I filled a two-liter container with one liter of water. I then submerged the shehnai in the water completely, measuring how the difference in water levels between the initial liter of water and the water with the shehnai submerged in it. The value this yielded was a volume of ~202.56cm^3.
Next, I plotted the above diagram of the shehnai on a graphing software. This essentially allowed me to ‘model’ the shehnai graphically, using a function in the form of S(x)=1/x^p . The plotted graph can be found below:
The equation of this graph is given by the equation;
S(x)=1/x^0.509
The scale being used on the graph (one small square represents on centimeter2) is used in order to accurately depict the scale of measurements present in the shehnai, while also giving surface area and volume measurements in cm2 and cm3, respectively. While this graph can be seen as an accurate depiction of the shehnai’s model, there are some limitations that must be addressed;
The shehnai’s horn radius is not infinite; the graph plotted depicts otherwise. However, due to the increasingly small impact this section of the graph has on its volume and surface area of rotation, it will be neglected.
The shehnai is not of infinite length; the graph depicts otherwise. However, as the graph gradually tends towards the asymptotic line of y=0, the surface area and volume of rotation begin to increase at a negligibly small rate. Thus, this will also be neglected.
With these assumptions being addressed and an equation for the graph given, the graph is now rotated, thus, giving a ‘trumpet’ in the shape of the shehnai. Using the above equation, we know that p=0.509. Hence, through the previously derived formula, the volume and surface area of the shehnai can be found;
V_shehnai=π/(2(0.509)-1)=500/9 π=174.53 〖cm〗^3 (5 s.f)
A_shehnai=2π((1-5(0.509))/(3(0.509)(1-2(0.509))))=353.18 〖cm〗^2 (5 s.f)
Although these results are not extremely farfetched relative to the originally found values, there is clearly a non-negligible difference between the two values of surface area and volume of the shehnai. More precisely, the formula-obtained value for the volume of the shehnai differs from that obtained experimentally by a factor of 0.191, or 19.1%, while its surface area differed by a value of 0.129, or 12.9%. In order to help improve the methods of the application of the derived formulae, it is vital that I acknowledge, understand and consolidate the systematic error present.
The first source of this error was surely the large amount of assumptions being made prior to calculations. For example, the shehnai’s shape was approximated to that of a right-cone. This is a huge assumption, adding significant magnitude to the value for the surface area of the instrument. The assumption was initially made in order to help find the surface area of the shehnai in an accessible, and thus, highlights a larger issue with what object/solid is used to test the validity of these conjectures.
In reality, it is impossible to create a solid with a finite volume and an infinite surface area; thus, it becomes very, very difficult to test these conjectures. However, it is possible to do so with a certain degree of accuracy, as was seen here. This degree of accuracy must always be acknowledged, no matter how great or minute it may be.
Conclusion:
Over the course of this exploration, I was able to understand the dimensions of Torricelli’s Trumpet, derive equations for its surface area and volume and eventually, apply them in real life. The intense paradox that became apparent between the two measurements, to me, highlighted the immense difficulty that Torricelli himself must have had in initially understanding his own creation.
In terms of the exploration, I believe that there were both positives and negatives present. Mathematically, the derived equations are concise, complete and consistent. When tested with shapes consistent with that of the function (variations of Torricelli’s Trumpet), it proved to be accurate and never failed.
However, when it came to applying the formula to a real-life situation, it failed. Although the difference in results was not substantial, it was clearly non-negligible. This, again, reaffirms the paradox present with this shape. It came across as very confusing to me at first, and it still is to this point. While I understood the mathematics behind its composition, my logical thinking refuses to accept an object with an infinite surface area and a finite volume.
As an aspiring mechanical engineer, this taught me a very important lesson. While formulae and conjectures may appear to be useful for real-life application at first glance, it is very important that one considers every single factor before deciding to apply them. Mechanical engineers may tend to neglect small, seemingly useless information; however, this exploration has taught me that regardless of how insignificant it may initially appear to be.
In conclusion, I feel that I have accomplished what I initially set out to achieve. I proved the finite volume and infinite surface area of Torricelli’s Trumpet, derived two different conjectures applicable to variations of the trumpet in general, and ensured to apply this theory all towards determining the dimensions of the Indian shehnai.