Mathematically modelling the surface area of
an ancient vase
Index
Introduction
Despite their seeming contrasts, mathematics and art have a lot in common, especially when it comes to
their focus on design, balance, and structure. By numerically simulating the surface area of an old vase, this
work investigates the harmony between its geometric and artistic components. The vase, a complex item of
Indian origin, is a prime example of cultural craftsmanship that skilfully combines practicality and beauty.
This object has particular meaning for me as an Indian since it not only symbolises my cultural background
but also provides a special chance to unite mathematics and art. This study attempts to identify the
mathematical ideas that may have influenced the vase's design by closely analysing its geometry and
precisely measuring its surface area. By doing this, I hope to demonstrate how geometry and optimisation
can be used practically to understand and evaluate culturally meaningful designs.
Aim
During my research, I learnt how to use integral calculus to find the volume of a solid of revolution in the
interval [�, �]. The formula is as follows:
𝑏
V=∫𝑏
𝑏𝑏2 𝑏𝑏
However, my IA revolves around the surface area, and I wondered if there was a similar method in which
the surface area of a solid of revolution could be found using calculus. So, I found a formula that could
allow me to calculate the surface area, A, of the ancient vase in the interval [�, �]. The formula is as
follows:
𝑏
2𝑏𝑏 𝑏𝑏
A=∫𝑏
Where, y = f(x) > 0, a ≤ x ≤ b and ds = √1 + (
𝑏𝑏 2
) 𝑏𝑏
𝑏𝑏
This formula is only applicable to solids or revolution, which matches my chosen Vase. This is
because it has a circular cross section as shown in the figure below. Hence, the aim of my IA will
be to learn the mathematics for deriving the formula for the surface area for a solid of revolution
and then apply it to approximation for the surface area of the vase.
Exploring the surface area formula
A solid of revolution is made by rotating a continuous function y= f(x) about the x-axis in the
interval (a,b). The solid of revolution can be divided into an infinite number of frustums, which
are created by involving a line segment and rotating it along the x-axis, with equal width Δ dx.
The number of frustums is taken as infinite as this allows the solid to be modelled as closely as
possible.
To derive the equation for the surface area of a solid of revolution, I have to start with the
formula of a frustum which is given by:
𝑏𝑏𝑏
A= 2
Figure: Diagram of a frustum
Where � = �1+�2 as the frustum has sides of different radii and � is the slant height of the
frustum as shown in the figure above. L can be equated to ds which is the curve length of the
function. This gives us
� = 2���
= 2𝑏(
𝑏1+𝑏2
2
)𝑏𝑏
r1 = (fxi -1 ) and r2= (fxi). As ∆ is small and f(x) is a continuous function, f(xi-1) and f(xi) can
both be approximated to be f(xi*). These values can then be substituted back into the original
formula for the surface area of a frustum. So the surface area can be of a frustum in the interval
(xi-1, xi) is:
A=2𝑏 (
𝑏(𝑏𝑏∗)+𝑏(𝑏𝑏∗)
2
)𝑏𝑏=2𝑏𝑏(𝑏𝑏 ∗)𝑏𝑏
Since the definite integral of f(x) in the interval (a,b) is defined as
𝑏
∫𝑏
𝑏(𝑏)𝑏𝑏 = 𝑏𝑏𝑏 ∑𝑏
𝑏=1
𝑏→∞
𝑏(𝑏𝑏)𝑏𝑏
The surface area of a solid of revolution with an infinite number of frustums can therefore be
approximated to be:
A= 𝑙𝑖𝑚 ∑𝑏
𝑏
𝑏→∞
𝑏
2𝑏𝑏(𝑏𝑏 ∗)𝑏𝑏 = ∫𝑏
2𝑏𝑏(𝑏)𝑏𝑏
Consequently, I then need to find ds. Since dx and dy are small, the curve length, ds, can be
taken into a straight line; the curved solid of revolution can then be said to comprise an infinite
number of straight lines. ds can then be said to be consisting of an infinite number of straight
lines. ds is equated to l which is the slant height of the frustum. To find l and by extension ds, the
Pythagorean Theorem can be used. dx and dy in relation to ds can be shown through a right
angled triangle.
ds
dy
θ
dx
Hence, by utilizing the pythagorean theorem, I can find ds:
ds^2= dx^2 + dy^2
Then i square both sides to find ds
�� = √��2 + ��2
Then the next step is to factorise dx^2 from the terms inside the square root and simplify to find
the equation of ds:
2
𝑏 𝑏
ds=√𝑏𝑏2 (1 + 𝑏𝑏2 )
𝑏𝑏
=√1 + (𝑏𝑏)2 dx
Finally, the formula for the surface area of a solid of revolution in the interval (a,b) is shown to
be:
𝑏
A=∫𝑏
𝑏
2𝑏𝑏 𝑏𝑏 = ∫𝑏
𝑏𝑏
2𝑏𝑏 √1 + (𝑏𝑏)2 𝑏𝑏
Graphing the Vase:
The formula indicates that the curve's equation must be found to calculate the surface area of a
vase. Before finding the functions of the graph of a vase, I had to first record the coordinates of
several points on the vase in order to plot it on the graph. This was initially done by graphing the
vase on desmos and finding specific points. After finding the circumferences, the formula for the
area of a circle was used to find the radius of each circular cross-section:
Top Radius= r2 = 0.35 cm
������������� �� � ������ = 2��
The radius of the circle, which then corresponds to its positive �-coordinate on the graph, is
then given by:
For example, the circumference of the bottom cross-section of the ceramic pot gives a radius of:
� =3.7/2
≈ 0.6 �� (1 �. �. )
The radii of the bottom and top cross sections of the ceramic pot are recorded below:
Circumference(cm)
Radius (cm)
Bottom
3.7
0.5
Top
2.2
0.35
Table 1: Circumference & radius of bottom and top tips
The values of the radii, 0.6 cm and 0.35 cm, give the upper and lower bounds of the domain of the graph
respectively. I used Desmos to plot the vase along with the coordinates and positioned and recorded the
coordinates accurately. Finding the radii values of the bottom and top cross sections of the vase allowed
me to scale the image of the pot accurately onto the math grid. The rest of the points were then recorded
using the graph rather than by hand using the tape measure which reduced the margin for human error.
Using Desmos, the graph is being plotted based on the vase symmetry
In total, 13 points were chosen to derive a piecewise function that will allow me to calculate the total
surface area of the vase. I chose points located at the bottom/top tips of the vase or where the shape of the
vase curves noticeably to form points of inflection. The x-coordinates and y-coordinates of the points were
then plotted onto the graph of the ceramic pot as shown above.
The x-coordinates and y-coordinates of the points chosen are shown in the Table below.
Coordinate
x
y
1
0
0.5
2
0.2
0.54
3
0.4
0.57
4
1.5
0.9
5
3
0.59
6
3.5
0.7
7
4
0.5
8
4.5
0.38
7
5.4
0.25
8
5.5
0.26
9
5.6
0.24
10
6.4
0.4
11
6.7
0
There will be five different function equations to be found, divided into the bottom, middle and top section
vase. The bottom section is represented by the orange colored curve line, the middle section represented by
the green colored curve line and the top section by the pink colored curve line. Each section consists of
three coordinates that will be used to find their respective curve equations.
Finding the functions
Functions are formed based on each of the 5 sections & corresponding coordinates are as follows :
1)
Bottom :
Coordinate
x
y
0
0
0.5
1
0.2
0.54
2
0.4
0.57
To find the equation of the parabola passing through the points (0,0.5), (0.2,0.54) and (0.4,0.57,) we start
with the general quadratic equation y = ax^2+bx+c and substitute the points one by one.
First, I used the point (0,0.5). Substituting x = 0 and y = 0.5 into the equation, I got:
0.5 = a(0)^2+b(0)+c
This made it clear that c = 0.5. So now the equation becomes y = ax^2+bx+0.5
Next, I substituted the second point, (0.2,0.54). Plugging x = 0.2 and y = 0.54 into the equation, I got:
0.54=a(0.2)^2+b(0.2)+0.5
Since 0.2^2 = 0.04 this became:
0.54 = 0.04a+0.2b+0.5
After subtracting 0.5 from both sides, I was left with:
0.04a+0.2b=0.04(Equation 1)
Then, I used the third point, (0.4,0.57). Substituting x=0.4 and y=0.57 into the equation gave:
0.57=a(0.4)^2+b(0.4)+0.5
Since 0.4^2=0.16, this simplified to:
0.57=0.16a+0.4b+0.5
Subtracting 0.5 from both sides, I got:
0.16a+0.4b=0.07(Equation 2
Now I had two equations to solve:
1. 0.04a+0.2b=0.04
2. 0.16a+0.4b=0.07
To make the equations easier to work with, I multiplied the first equation by 4 so that the coefficient of aa
matched in both equations:
0.16a+0.8b=0.16(Equation 3)
Next, I subtracted Equation (2) from Equation (3):
(0.16a+0.8b)−(0.16a+0.4b) = 0.16−0.07
This left me with:
0.4b=0.090
Dividing both sides by 0.4 gave:
b=0.225
With b=0.225, I substituted it back into Equation (1) to find a:
0.04a+0.2(0.225)=0.04
Simplifying, I got:
0.04a+0.045=0.04
Subtracting 0.045 from both sides gave:
0.04a = −0.005
So:
a=−0.125
Now that I had a=−0.125, b=0.225, and c=0.5, I could write the final equation of the parabola:
y=−0.125x2+0.225x+0.5
This equation represents the parabola that passes through the three given points.
Bottom
Middle Body 1
Middle Body 2
Coordinate
x
y
0
0
0.5
1
0.2
0.54
2
0.4
0.57
Coordinate
x
y
0
0.4
0.57
1
1.5
0.9
2
3
0.59
Coordinate
x
y
0
3
0.59
1
3.5
0.7
2
4
0.5
Equation of the Function
y=−0.125x2+0.225x+0.5
Equation of the Function
y=−0.1949x2+0.671x+0.3328
Equation of the Function
y = -0.62x² + 4.25x - 6.58
Middle Body 3
Neck
Neck
Coordinate
x
y
0
4
0.5
1
4.5
0.38
2
5.4
0.25
Coordinate
x
y
0
5.4
0.25
1
5.5
0.26
2
5.6
0.24
Coordinate
x
y
0
5.6
0.24
1
6.4
0.4
Equation of the Function
y=−0.1949x2+0.671x+0.3328
Equation of the Function
y = -1.5x² + 16.45x - 44.84
Equation of the Function
y = -1.5x² + 16.45x - 44.84
Stopper
2
6.7
0
Coordinate
x
y
0
5.6
0.24
1
6.4
0.4
2
6.7
0
Equation of the Function
y= -1.39x^2 + 16.88x -50.70
2) Middle Body 1:
Coordinate
x
y
0
0.4
0.57
1
1.5
0.9
2
3
0.59
To find the equation of the parabola passing through the points (0.4, 0.57), (1.5, 0.9), and (3, 0.59), we
use the general quadratic equation y=ax^2+bx+c and substitute each point into the equation to form a
system of equations.
First, I used the points (0.4, 0.57). Substituting x=0.4 and y=0.57into the equation gave:
0.57=a(0.4)2+b(0.4)+c
Since 0.42=0.16, this became:
0.57=0.16a+0.4b+c (Equation 1)
Next, I substituted the point (1.5, 0.9). Plugging x=1.5 and y=0.9 into the equation gave:
0.9=a(1.5)2+b(1.5)+c
Since 1.52=2.25 this simplified to:
0.9=2.25a+1.5b+c (Equation 2)
Finally, I used the point (3, 0.59). Substituting x=3 and y=0.59 into the equation gave:
0.59=a(3)2+b(3)+c
Since 3^2=9, this became:
0.59=9a+3b+c
To eliminate c, I subtracted Equation (1) from Equation (2):
(2.25a+1.5b+c)−(0.16a+0.4b+c)=0.9−0.57
Simplifying:
2.09a+1.1b=0.33 (Equation 4)
Next, I subtracted Equation (2) from Equation (3):
(9a+3b+c)−(2.25a+1.5b+c)=0.59−0.9
Simplifying:
6.75a+1.5b=−0.31
From Equation (4), I solved for b in terms of a:
Next, I substituted bb from Equation (6) into Equation (5):
Simplifying the terms:
Combining like terms:
3.9a=−0.76
Substituting a=−0.1949 into Equation (6) gave:
Finally, I substituted a=−0.1949 and b=0.671 into Equation (1):
0.57=0.16(−0.1949)+0.4(0.671)+c
0.57=−0.0312+0.2684+c
c=0.57−0.2372
c=0.3328
With a=−0.1949, b=0.671, and c=0.3328, the equation of the parabola is:
y=−0.1949x2+0.671x+0.3328
3) Middle Body 2:
Coordinate
x
y
0
3
0.59
1
3.5
0.7
2
4
0.5
To derive the equation of a parabola passing through the points (3, 0.59), (3.5, 0.7), and (4.0, 0.5), we
start with the general quadratic equation:
y = ax² + bx + c
I substitute each of the three given points into the quadratic equation to form a system of equations.
Using the point (3, 0.59):
Substituting x = 3 and y = 0.59 into the equation:
0.59 = a(3)² + b(3) + c
Since 3² = 9, this simplifies to:
0.59 = 9a + 3b + c (Equation 1)
Using the point (3.5, 0.7):
Substituting x = 3.5 and y = 0.7 into the equation:
0.7 = a(3.5)² + b(3.5) + c
Since 3.5² = 12.25, this simplifies to:
0.7 = 12.25a + 3.5b + c (Equation 2)
Using the point (4.0, 0.5):
Substituting x = 4.0 and y = 0.5 into the equation:
0.5 = a(4.0)² + b(4.0) + c
Since 4.0² = 16, this simplifies to:
0.5 = 16a + 4b + c (Equation 3)
Now we have the following system of equations:
1. 9a + 3b + c = 0.59
2. 12.25a + 3.5b + c = 0.7
3. 16a + 4b + c = 0.5
We can eliminate c by subtracting equations from each other.
Subtract Equation (1) from Equation (2):
(12.25a + 3.5b + c) - (9a + 3b + c) = 0.7 - 0.59
3.25a + 0.5b = 0.11 (Equation 4)
Subtract Equation (2) from Equation (3):
(16a + 4b + c) - (12.25a + 3.5b + c) = 0.5 - 0.7
3.75a + 0.5b = -0.2 (Equation 5)
Now, subtract Equation (4) from Equation (5):
(3.75a + 0.5b) - (3.25a + 0.5b) = -0.2 - 0.11
0.5a = -0.31
Solving for a:
a = -0.62
Now substitute a = -0.62 into Equation (4):
3.25(-0.62) + 0.5b = 0.11
-2.015 + 0.5b = 0.11
0.5b = 2.125
Solving for b:
b = 4.25
Now that I have a = -0.62 and b = 4.25, we substitute these values into Equation (1) to solve for c:
0.59 = 9(-0.62) + 3(4.25) + c
0.59 = -5.58 + 12.75 + c
Simplifying:
0.59 = 7.17 + c
Solving for c:
c = 0.59 - 7.17 = -6.58
Now that I have the coefficients a = -0.62, b = 4.25, and c = -6.58, the equation of the parabola is:
y = -0.62x² + 4.25x - 6.58
4) Middle Body 3:
Coordinate
x
y
0
4
0.5
1
4.5
0.38
2
5.4
0.25
To derive the equation of the parabola passing through the points (4, 0.5), (4.5, 0.38), and (5.4, 0.25), we
start with the general quadratic equation of a parabola:
y = ax² + bx + c
We substitute each of the three given points into the quadratic equation to form a system of equations.
Using the point (4, 0.5):
0.5 = a(4)² + b(4) + c
Since 4² = 16, this simplifies to:
0.5 = 16a + 4b + c (Equation 1)
Using the point (4.5, 0.38):
0.38 = a(4.5)² + b(4.5) + c
Since 4.5² = 20.25, this simplifies to:
0.38 = 20.25a + 4.5b + c (Equation 2)
Using the point (5.4, 0.25):
0.25 = a(5.4)² + b(5.4) + c
Since 5.4² = 29.16, this simplifies to:
0.25 = 29.16a + 5.4b + c (Equation 3)
Now we have the following system of equations:
1. 16a + 4b + c = 0.5
2. 20.25a + 4.5b + c = 0.38
3. 29.16a + 5.4b + c = 0.25
We eliminate c by subtracting equations from each other.
Subtract Equation (1) from Equation (2):
(20.25a + 4.5b + c) - (16a + 4b + c) = 0.38 - 0.5
4.25a + 0.5b = -0.12 (Equation 4)
Subtract Equation (2) from Equation (3):
(29.16a + 5.4b + c) - (20.25a + 4.5b + c) = 0.25 - 0.38
8.91a + 0.9b = -0.13 (Equation 5)
Now, subtract Equation (4) from Equation (5):
(8.91a + 0.9b) - (4.25a + 0.5b) = -0.13 - (-0.12)
4.66a + 0.4b = -0.01
Multiply Equation (4) by 9 to align the coefficients of b:
9(4.25a + 0.5b) = 9(-0.12)
38.25a + 4.5b = -1.08 (Equation 6)
Subtract Equation (5) from Equation (6):
(38.25a + 4.5b) - (8.91a + 0.9b) = -1.08 - (-0.13)
29.34a + 3.6b = -0.95
Now we solve for a and b:
4.25a + 0.5b = -0.12
Solving for a, we get:
a ≈ -0.0324, b ≈ 0.0354
Now that we know a ≈ -0.0324 and b ≈ 0.0354, we substitute these values into Equation (1) to solve for c:
0.5 = 16(-0.0324) + 4(0.0354) + c
0.5 = -0.5184 + 0.1416 + c
0.5 = -0.3768 + c
c = 0.5 + 0.3768 = 0.8768
Now that we have the coefficients a ≈ -0.0324, b ≈ 0.0354, and c ≈ 0.8768, the equation of the parabola
is:
y = -0.0324x² + 0.0354x + 0.8768
5) Neck :
Coordinate
x
y
0
5.4
0.25
1
5.5
0.26
2
5.6
0.24
To derive the equation of a parabola passing through the points (5.4, 0.25), (5.5, 0.26), and (5.6, 0.24), we
start with the general quadratic form:
y = ax² + bx + c
I substitute each point into the equation y = ax² + bx + c:
For (5.4, 0.25):
Substituting x = 5.4 and y = 0.25 into the equation:
0.25 = a(5.4)² + b(5.4) + c
0.25 = 29.16a + 5.46 + c (Equation 1)
For (5.5, 0.26):
Substituting x = 5.5 and y = 0.26 into the equation:
0.26 = a(5.5)² + b(5.5) + c
0.26 = 30.25a + 5.56 + c (Equation 2)
For (5.6, 0.24):
Substituting x = 5.6 and y = 0.24 into the equation:
0.24 = a(5.6)² + b(5.6) + c
0.24 = 31.36a + 5.66 + c (Equation 3)
I now solve by eliminating c.
Subtract Equation (1) from Equation (2):
(30.25a + 5.56 + c) - (29.16a + 5.46 + c) = 0.26 - 0.25
1.09a + 0.16 = 0.01 (Equation 4)
Subtract Equation (2) from Equation (3):
(31.36a + 5.66 + c) - (30.25a + 5.56 + c) = 0.24 - 0.26
1.11a + 0.16 = -0.02 (Equation 5)
Subtract Equation (4) from Equation (5):
(1.11a + 0.16) - (1.09a + 0.16) = -0.02 - 0.01
0.02a = -0.03
Solving for a:
a = -1.5
Substitute a = -1.5 into Equation (4):
1.09(-1.5) + 0.16 = 0.01
-1.635 + 0.16 = 0.01
0.16 = 1.645
So:
b = 16.45
Now that we have a = -1.5 and b = 16.45, we substitute these values into Equation (1) to solve for c:
0.25 = 29.16(-1.5) + 5.4(16.45) + c
0.25 = -43.74 + 88.83 + c
0.25 = 45.09 + c
Solving for c:
c = 0.25 - 45.09 = -44.84
Now that we have the coefficients a = -1.5, b = 16.45, and c = -44.84, the equation of the parabola is:
y = -1.5x² + 16.45x - 44.84
6) Stopper:
Coordinate
x
y
0
5.6
0.24
1
6.4
0.4
2
6.7
0
To derive the equation of a parabola passing through the points ( 5.6,0.24),(6.4,0.4) and (6.7,0) , we start
with the general quadratic equation:
𝑏 = 𝑏𝑏2 + 𝑏𝑏 + 𝑏
For (5.6,0.24):
0.24 = 𝑏(5.6)2 + 𝑏(5.6) + 𝑏
0.24 = 31.36𝑏 + 5.6𝑏 + 𝑏
. For (6.4,0.4):
0.4 = 𝑏(6.4)2 + 𝑏(6.4) + 𝑏
0.4 = 40.96𝑏 + 6.4𝑏 + 𝑏
For (6.7,0):
0 = 𝑏(6.7)2 + 𝑏(6.7) + 𝑏
0 = 44.89𝑏 + 6.7𝑏 + 𝑏
1. 31.36𝑏 + 5.6𝑏 + 𝑏 = 0.24
2.
40.96𝑏 + 6.4𝑏 + 𝑏 = 0.4
3.
44.89𝑏 + 6.7𝑏 + 𝑏 = 0
Eliminate c by subtracting equations
1. Subtract Equation (1) from Equation (2):
(40.96𝑏 + 6.4𝑏 + 𝑏)-(31.36𝑏 + 5.6𝑏 + 𝑏)=0.4-0.24
9.6𝑏 + 0.8𝑏 = 0.16
(4)
2. Subtract Equation (2) from Equation (3):
(44.89𝑏 + 6.7𝑏 + 𝑏)-(40.96𝑏 + 6.4𝑏 + 𝑏)=0-0.4
3.93𝑏 + 0.3𝑏 = −0.4
Solve for a & b
Multiply Equation (4) by 3 to align the coefficient of b:
3(9.6𝑏 + 0.8𝑏) = 3(0.16)
28.8𝑏 + 2.4𝑏 = 0.48
Multiply Equation (5) by 8 to align the coefficient of b:
8(3.93𝑏 + 0.3𝑏) = 8(−0.4)
31.44𝑏 + 2.4𝑏 = −3.2
.Subtract Equation (6) from Equation(7) :
(31.44a+2.4b)-(28.8a+2.4b)=-3.2-0.48
2.64a=-3.68
−3.68
𝑏 = 2.64 = −1.39
.Substitute a=-1.39 into the Equation(4):
9.6(−1.39) + 0.8𝑏 = 0.16
−13.344 + 0.8𝑏 = 0.16
0.8b=13.504
b=16.88:
Substitute a=-1.39 and b=16.88 into the equation (1):
31.36(−1.39) + 5.6(16.88) + 𝑏 = 0.24
-43.5904+94.528+c = 0.24
c=0.24-50.9376
c=-50.70
Therefore the equation is
y= -1.39x^2 + 16.88x -50.70
Calculating the surface area of the Vase
The surface area of the vase was then calculated with the formula for the surface area of a solid of
revolution I derived :
𝑏=∫
=∫
2𝑏𝑏 𝑏𝑏
𝑏𝑏
2𝑏𝑏√1 + (𝑏𝑏)2 dx
Bottom Section :
To find ds , the derivative of the equation of the function 𝑦 = −0.125𝑦2 + 0.225𝑦 + 0.5 has to be
obtained.
𝑏𝑏
That is given by 𝑏𝑏 = −0.25𝑏 + 0.225
𝑏𝑏
Substituting y and 𝑏𝑏 in the equation for Area ,
𝑏
0.4
𝑏𝑏𝑏𝑏𝑏𝑏
=∫
𝑏𝑏 2
2𝑏𝑏√1 + (
) 𝑏𝑏
𝑏𝑏
(2𝑏(−0.125𝑏2 + 0.225𝑏 + 0.5)(√1 + (−0.25𝑏 + 0.225)2 )𝑏𝑏
=∫0
Using GDC , we can evaluate the value of the area and it is given by
𝑏
= 1.83
𝑏𝑏𝑏𝑏𝑏𝑏
Middle Body 1 :
To find ds , the derivative of the equation of the function 𝑦 = −0.1949𝑦2 + 0.671𝑦 + 0.3328 has to be
obtained.
𝑏𝑏
That is given by 𝑏𝑏 = −0.3898𝑏 + 0.671
𝑏𝑏
Substituting y and 𝑏𝑏 in the equation for Area ,
𝑏
3
=∫0.4
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 1
=∫
𝑏𝑏 2
2𝑏𝑏√1 + (
) 𝑏𝑏
𝑏𝑏
(2𝑏(−0.1949𝑏2 + 0.671𝑏 + 0.3328)(√1 + (−0.3898𝑏 + 0.671)2 )𝑏𝑏
Using GDC , we can evaluate the value of the area and it is given by
𝑏
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 1
= 13.47
Middle Body 2 :
To find ds , the derivative of the equation of the function 𝑦 = −0.62𝑦2 + 4.25𝑦 − 6.58 has to be
obtained.
𝑏𝑏
That is given by 𝑏𝑏 = −1.24𝑏 + 4.25
𝑏𝑏
Substituting y and 𝑏𝑏 in the equation for Area ,
𝑏
4
=∫3
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 2
=∫
𝑏𝑏 2
2𝑏𝑏√1 + (
) 𝑏𝑏
𝑏𝑏
(2𝑏(−0.62𝑏2 + 4.25𝑏 − 6.58)√1 + (−1.24𝑏 + 4.25)2 )𝑏𝑏
Using GDC , we can evaluate the value of the area and it is given by
𝑏
= 4.31
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 2
Middle Body 3 :
To find ds , the derivative of the equation of the function 𝑦 = −0.0324𝑦2 + 0.0354𝑦 + 0.8768 has to
be obtained.
𝑏𝑏
That is given by 𝑏𝑏 = −0.0648𝑏 + 0.0354
𝑏𝑏
Substituting y and 𝑏𝑏 in the equation for Area ,
𝑏
5.4
=∫4
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 3
=∫
𝑏𝑏 2
2𝑏𝑏√1 + (
) 𝑏𝑏
𝑏𝑏
(2𝑏(−0.0324𝑏2 + 0.0354𝑏 + 0.8768)√1 + (−0.0648𝑏 + 0.0354)2 )𝑏𝑏
Using GDC , we can evaluate the value of the area and it is given by
𝑏
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 3
Neck :
=2.85
To find ds , the derivative of the equation of the function 𝑦 = −1.5𝑦2 + 16.45𝑦 − 44.84 has to be
obtained.
𝑏𝑏
That is given by 𝑏𝑏 = −3𝑏 + 16.45
𝑏𝑏
Substituting y and 𝑏𝑏 in the equation for Area ,
𝑏
5.6
𝑏𝑏𝑏𝑏
=∫
𝑏𝑏 2
2𝑏𝑏√1 + (
) 𝑏𝑏
𝑏𝑏
(2𝑏(−1.5𝑏2 + 16.45𝑏 − 44.84)√1 + (−3𝑏 + 16.45)2 )𝑏𝑏
=∫5.4
Using GDC , we can evaluate the value of the area and it is given by
𝑏
= 0.325
𝑏𝑏𝑏𝑏
Stopper :
To find ds , the derivative of the equation of the function 𝑦 = −1.39𝑦2 + 16.88𝑦 − 50.70 has to be
obtained.
That is given by
𝑏𝑏
= −2.78𝑏 + 16.88
𝑏𝑏
𝑏𝑏
Substituting y and 𝑏𝑏 in the equation for Area ,
𝑏
6.7
𝑏𝑏𝑏𝑏𝑏𝑏𝑏
=∫
𝑏𝑏 2
2𝑏𝑏√1 + (
) 𝑏𝑏
𝑏𝑏
(2𝑏(−1.39𝑏2 + 16.88𝑏 − 50.70)√1 + (−2.78𝑏 + 16.88 )𝑏𝑏
=∫5.6
Using GDC , we can evaluate the value of the area and it is given by
𝑏
𝑏𝑏𝑏𝑏𝑏𝑏𝑏
=3.36
Total Surface Area:
The value for the total surface area of the vase is given by adding up the values of the surface area of the
stopper , neck, middle body(1,2,3) & bottom sections together .
This gives me : 𝑏
𝑏
𝑏𝑏𝑏𝑏𝑏 = 𝑏
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 3 + 𝑏
𝑏𝑏𝑏𝑏𝑏𝑏 + 𝑏
𝑏𝑏𝑏𝑏 + 𝑏
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 1 + 𝑏
𝑏𝑏𝑏𝑏𝑏𝑏 𝑏𝑏𝑏𝑏 2 +
𝑏𝑏𝑏𝑏𝑏𝑏𝑏
=....1.83......+.....13.47....+......4.31..+.....2.85...+..0.325......+.3.36........
=......26.145.....𝑏𝑏2 (3 s.f)
Thus the total surface area of the vase can be approximated as 26.145…..𝑏𝑏2
𝑏=