Volumes of Solids of Revolution12

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John McCarthy
SACE Stage 2 Mathematical Studies
Volumes of Solids of Revolution
John McCarthy
Introduction:
Throughout our studies of integration, it has been shown that it is possible to use the processes of
integral calculus to calculate the volume of a solid. Much in the same way the area under a two
dimensional function can be found by adding the infinite set of infinitesimal rectangles that make up the
area under the function, the volume of a three dimensional object can be found by adding the infinite
set of infinitesimal cross sections that make up the volume of the object.
To calculate this integral however, the area of any cross section of the object must be expressed in
terms of 𝑥 value of the profile of the object on a plane. In the example case of a cone, the area of any
cross section is given by the function 𝐴 = 𝜋𝑟 2 . When the profile of the cone is plotted onto a set of axes,
and then rotated around the 𝑥 axis to form the cone in question, it is found that the radius of the cross
section is equal to the 𝑦 value of the profile.
𝑦
𝑝𝑟𝑜𝑓𝑖𝑙𝑒
𝑟
𝑥
ℎ
In the above example, this 𝑦 value for each cross section is found using the formula for the profile,
∆𝑦
𝑟
which is 𝑦 = 𝑚𝑥 where 𝑚 = ∆𝑥 = ℎ, resulting in the area of each cross section being 𝐴 = 𝜋𝑟 2 = 𝜋𝑦 2 =
𝑟
2
𝜋 (ℎ 𝑥) .
To hence find the volume of the object, as stated above, the volume of each of these infinite number of
cross sections with an infinitely small width 𝑑𝑥 must be added together:
ℎ
𝑟 2
𝑉 = ∑ 𝜋 ( 𝑥) 𝑑𝑥
ℎ
𝑦=0
Or in the form of an integral:
Page 1
John McCarthy
SACE Stage 2 Mathematical Studies
ℎ
𝑟 2
𝑉 = ∫ 𝜋 ( 𝑥) 𝑑𝑥
ℎ
0
For example, in the case of a 4cm tall cone with a base radius of 2cm, the volume would be given by 𝑉 =
4
2
2
∫0 𝜋 (4 𝑥) 𝑑𝑥 =
1
𝜋
3
16
𝜋
3
1
cm3. Using the already known formula for the volume of a cone, 𝑉 = 3 𝜋𝑟 2 ℎ =
× 22 × 4, we get the same result of
16
𝜋
3
cm3 .
In general, this process demonstrates the relationship that the volume of any solid when rotated around
𝑏
the 𝑥 axis is given by 𝑉 = ∫𝑎 𝐴(𝑥) 𝑑𝑥 , where 𝑎 and 𝑏 are the starting and ending points of the object on
the profile function and 𝐴(𝑥) is the area.
Calculating the volume of a vase:
In this investigation, the processes described above in conjunction with mathematical modeling were
used to estimate the volume of a vase with a circular cross section.
The vase in question
Page 2
John McCarthy
SACE Stage 2 Mathematical Studies
The vase in question was laid on a piece of A4 lined grid paper with the 𝑥 axis aligned with the center of
the vase. A pencil was then used to trace the profile of the vase onto the paper, and using a ruler as well
as the grid provided, the values of the radius of the cross section at regular intervals along the profile
were recorded to an accuracy of 1 mm.
Position along 𝒙 axis (cm)
0
0.5
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Radius of cross section (cm ±ðŸmm)
2.5
3.5
3.7
3.9
4.1
4.3
4.5
4.6
4.7
4.9
5
5.1
5.2
5.3
5.4
5.4
5.2
5
4.5
4
3.7
3.7
4
4.8
6
5
4
Radius of
3
cross section
y = 4E-07x6 + 1E-05x5 - 0.0014x4 + 0.0293x3 - 0.2494x2 + 1.0195x + 2.7527
2
1
0
0
5
10
15
20
25
Position along x axis
Page 3
John McCarthy
SACE Stage 2 Mathematical Studies
As seen in the graph above, due to the irregular nature of the profile of the vase, a single line of best fit
was unable to pass inside all the error bars, and hence could not be taken as an accurate model of the
profile of the vase. As such, the vase was separated into three component parts to allow for a model to
be accurately produced:
Model of the profile of the vase
6
y = 2x + 2.5
y = 0.01389x3 - 0.65357x2 + 9.61825x - 38.3667
5
4
First component
Radius of
3
cross section
Second component
Third component
2
y = -0.0000156x5 + 0.0005014x4 - 0.00549x3 + 0.01734x2 + 0.19252x + 3.48928
1
0
0
5
10
15
20
25
Position along x axis
As seen in this new model, three models, one linear, and two polynomials, were used to model separate
components of the profile. Each of these three components now pass through the error bars of each
point of ±1mm and smoothly join up to each other to accurately represent the profile of the vase.
To find the total volume of the vase, the volumes of the three solids that result from the rotation of the
three models around the 𝑥 axis must be added together. To find where each of these three models start
and end, the intersections between each line of best fit must be found.
Page 4
John McCarthy
SACE Stage 2 Mathematical Studies
The intersection between the first and second components occurs when
2𝑥 + 2.5 = −0.0000156𝑥 5 + 0.0005014𝑥 4 − 0.00549𝑥 3 + 0.01734𝑥 2 + 0.19252𝑥 + 3.48928
−0.0000156𝑥 5 + 0.0005014𝑥 4 − 0.00549𝑥 3 + 0.01734𝑥 2 + −1.80748𝑥 + 0.98928 = 0
𝑥 = 0.550
The intersection between the second and third components occurs when
−0.0000156𝑥 5 + 0.0005014𝑥 4 − 0.00549𝑥 3 + 0.01734𝑥 2 + 0.19252𝑥 + 3.48928
= 0.01389𝑥 3 − 0.65357𝑥 2 + 9.61825𝑥 − 38.3667
−0.0000156𝑥 5 + 0.0005014𝑥 4 − 0.01938𝑥 3 + 0.67091𝑥 2 − 9.42573𝑥 + 41.856 = 0
𝑥 = 8.67395, 16.3896, 17.2046
Between the second and third components we have three intersection points. The first point 𝑥 =
8.67395 is not part of the model, however the second and third points represent where the two
functions have overlapped slightly. As such, the average of the two points
16.3896+17.2046
2
= 16.7971
will be used. The final point, where the third component of the profile ends, is at the point 𝑥 = 22.5
where the original plot of the vase ended.
Knowing these intersection points, the volumes of the three components can hence be calculated to
give the total volume of the vase. To calculate the volumes using the equation found earlier, 𝑉 =
𝑏
∫𝑎 𝐴(𝑥) 𝑑𝑥, we must know the area of cross sections of each component, 𝐴(𝑥). For a circle, the area is
given by the expression 𝐴 = 𝜋𝑟 2 but it was shown earlier that the radius is given by the 𝑦 value of the
profile, or 𝐴 = 𝜋𝑦 2 . Using this knowledge, we can hence calculate the total volume of the vase by
adding the volumes of each of the three components:
0.550
𝑉=∫
𝜋(2𝑥 + 2.5)2 𝑑𝑥
0
16.7971
𝜋(−0.0000156𝑥 5 + 0.0005014𝑥 4 − 0.00549𝑥 3 + 0.01734𝑥 2
+∫
0.550
+ 0.19252𝑥 + 3.48928)2 𝑑𝑥
22.5
+∫
𝜋(0.01389𝑥 3 − 0.65357𝑥 2 + 9.61825𝑥 − 38.3667)2 𝑑𝑥
16.7971
When evaluated, 𝑉 is found to be 1501.74 𝑐𝑚3 .
Page 5
John McCarthy
SACE Stage 2 Mathematical Studies
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