Mathematician A
Matematikprojekt
Sculptor
A sculptor has been given the idea that he will erect a sculpture composed of marble slabs, the
geometric shape of which can all be described relatively simply. The marble slabs from which the
parts are cut are 8cm thick. The density of marmor is 3250 kg/m3.
As well as x y in the funktion pressures are measured in meters.
Due to the intended location of the sculpture, it is necessary to know its exact mass. The results are
also given in 3 significant digits.
a) Calculate the mass of a plate whose rounded part is in the shape of
a parabola with the function f ( x)  1  x 2 (See Figure 1)
2
r(x)  1  x
2
r(x)  1  x
Figur 1
2
r( x)
1
2
1
0
1
2
x
Lets call
8
𝑚
100
𝜌 = 3250 𝑘𝑔/𝑚3
𝐴 = 𝑠𝑢𝑟𝑓𝑎𝑐𝑒; 𝑉 = 𝑣𝑜𝑙𝑢𝑚𝑒,
𝑀 = 𝑚𝑎𝑠𝑠
𝛿 = 8𝑐𝑚 =
𝜌=
The units of surface are square meters (𝑚2 )
𝑀
𝑉
1
1
1
1
−1
−1
−1
−1
𝐴 = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ (−𝑥 2 + 1)𝑑𝑥 = − ∫ 𝑥 2 𝑑𝑥 + ∫ 𝑑𝑥
The results of the indefinite integrals are
Mathematician A
1
𝐹(𝑥) = − 𝑥 3 + 𝑥 + 𝐶
3
Where C is any real constant and can be set to zero (as we will do from now on this work). Thus the
value of the area is given by the fundamental theorem of calculus:
1
1
4
𝐴 = 𝐹(1) − 𝐹(−1) = − (13 − (−1)3 ) + (1 − (−1)) = − (2) + 2 = 𝑚2
3
3
3
The volume will be
4
8
8 3
𝑉 = 𝛿𝐴 = ( 𝑚2 ) (
𝑚) =
𝑚
3
100
75
Now we can determine the mass
𝑀 = 𝑉𝜌 = (
8 3
𝑘𝑔
1040
𝑚 ) (3250 3 ) =
𝑘𝑔 ≈ 346.667 𝑘𝑔
75
𝑚
3
Another part must be prepared so that it has the shape of two parabolas with a hole. The hole has
the shape of a circle with radius r = 0.3m.
Both parabolas go through the points (-1.0) and (1.0). The small parabola passes through the point
(0 ; 0.8) and the large parabola passes through the point (0 ; 1.6)
The dimensions of the figure are in meters.
b) Calculate the mass of this part. (See Figure 2)
2
px()  0.8 0.8x
2
hx()  1.6 1.6x
1
 
6 1
2
j(x)    9  100x
5 10
2
Mathematician A
1
l
(
x
)


6
5

1
1
0


9

1
0
0


2
x
2
2
p(x)  0.8 0.8x
Figur 2
2
1
h( x)
l( x)
p ( x)
2
1
0
1
2
j( x)
1
2
x
2
h(x)  1.6 1.6x
The radius of the circle is 0.3𝑚 = 3𝑚/10 . The area of the circle will be
𝐴𝑐𝑖𝑟𝑐𝑙𝑒
2
3
9𝜋 2
= 𝜋 ( 𝑚) =
𝑚
10
100
The little parabola has an equation of the form
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Its vertex is (0, 0.8) this means that
0.8 = 𝑎(0)2 + 𝑏(0) + 𝑐
Therefore 𝑐 = 0.8
Also the x-coordinate of the vertex, 𝑥𝑣 , is given by
𝑥𝑣 = −
we have that 𝑥𝑣 = 0
𝑏
2𝑎
𝑏
=0
2𝑎
Thus we know that 𝑏 = 0. Finally we know that the parabola passes through the point (1, 0). If we
substituthe this in the general equation of the parabola we will have
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑎𝑥 2 + 0.8
0 = 𝑎(1)2 + 0.8
𝑎 = −0.8
We can write the equation of the little parabola as
𝑙𝑖𝑡𝑡𝑙𝑒(𝑥) = −0.8 𝑥 2 + 0.8 (x is measured in meters)
To calculate its area (measured in square meters) we must integrate between -1 and 1
−
Mathematician A
1
1
−1
−1
𝐴𝑙𝑖𝑡𝑡𝑙𝑒 = ∫ 𝑙𝑖𝑡𝑡𝑙𝑒(𝑥)𝑑𝑥 = ∫ (−0.8𝑥 2 + 0.8)𝑑𝑥
The result of the indefinite integral is
𝐿(𝑥) = −
0.8 3
𝑥 + 0.8𝑥 + 𝐶
3
Where C is just a real constant. The value of area will be then
0.8
16 2
(2) + 0.8(2) =
𝐴𝑙𝑖𝑡𝑡𝑙𝑒 = 𝐿(1) − 𝐿(−1) = −
𝑚
3
15
Finally we need to know the area of the big parabola. First we determine its general equation,
having the form
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Substituting the coordinates of its vertex (0, 1.6) we get 𝑐 = 1.6. Also since −𝑏/2𝑎 = 0 we can
also say that 𝑏 = 0. Now we substitute the value of its point (1, 0):
𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 𝑎𝑥 2 + 1.6
0 = 𝑎(1)2 + 1.6
−1.6 = 𝑎
Thus we write the equation of the big parabola as
𝑏𝑖𝑔(𝑥) = −1.6 𝑥 2 + 1.6
With x measured in meters. The area measured in square meters will be
1
1
−1
−1
𝐴𝑏𝑖𝑔 = ∫ 𝑏𝑖𝑔(𝑥)𝑑𝑥 = ∫ (−1.6𝑥 2 + 1.6)𝑑𝑥
Here the antiderivative is
𝐵(𝑥) = −
1.6 3
𝑥 + 1.6𝑥
3
thus
𝐴𝑏𝑖𝑔 = 𝐵(1) − 𝐵(−1) = −
1.6
32 2
(2) + 1.6(2) =
𝑚
3
15
The total area of this slab will be
𝐴𝑠𝑙𝑎𝑏 = 𝐴𝑏𝑖𝑔 − 𝐴𝑙𝑖𝑡𝑡𝑙𝑒 − 𝐴𝑐𝑖𝑟𝑐𝑙𝑒
32 2 16 2 9𝜋 2 16 2 9𝜋 2
𝐴𝑠𝑙𝑎𝑏 =
𝑚 − 𝑚 −
𝑚 =
𝑚 −
𝑚
15
15
100
15
100
1600 − 135𝜋 2 320 − 27𝜋 2
𝐴𝑠𝑙𝑎𝑏 =
𝑚 =
𝑚
1500
300
The volume of this slab is
𝑉 = 𝐴𝑠𝑙𝑎𝑏 𝛿 = (
And its mass will be
320 − 27𝜋 2
8
320 − 27𝜋 3
𝑚 )(
𝑚) =
𝑚
300
100
3750
320 − 27𝜋 3
𝑘𝑔
𝑀 = 𝑉𝜌 = (
𝑚 ) (3250 3 ) = 203.820 𝑘𝑔
3750
𝑚
Mathematician A
The sculptor wants to apply the parabola to several parts of the sculpture. Figure 3 shows half of the
lower parabola from Figure 2, which is part to be cut from the sculpture in Figure 2.
In order not to have to integrate each time to determine the area of the plate in Figure 3, the sculptor
wishes to derive a finished (integrated) formula for calculating the area mentioned, expressed in b.
Figur 3
1
2
px()  0.8 0.8x
h
x 0 0.1 1
p ( x) 0.5
0
0.5
1
x
b
c) What expression should the sculptor come up with if he calculates correct?
Lets consider the parabola with vertex (0, ℎ) that passes throught the point (𝑏, 0). Its general
equation will be
𝑦(𝑥) = 𝛼𝑥 2 + 𝛽𝑥 + 𝛾
Substituting the vertex:
ℎ=𝛾
also since
𝛽
−
=0
2𝛼
we get 𝛽 = 0.
Substituting the root (b, 0) in the equation
𝑦(𝑥) = 𝛼𝑥 2 + 𝛽𝑥 + 𝛾 = 𝛼𝑥 2 + ℎ
0 = 𝛼𝑏 2 + ℎ
ℎ
𝛼=− 2
𝑏
Finally we will have the general equation of this kind of parabola
𝑦 = 𝑓(𝑥) = −
ℎ 2
𝑥 +ℎ
𝑏2
To know the area we must integrate between 0 and b:
𝑏
𝑏
𝐴𝑔𝑒𝑛𝑒𝑟𝑎𝑙 = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ (−
0
The antiderivative
𝐹(𝑥) = −
0
ℎ 2
𝑥 + ℎ) 𝑑𝑥
𝑏2
ℎ 3
𝑥 + ℎ𝑥 + 𝐶
3𝑏 2
Mathematician A
Where is a real constant. Therefore
ℎ
1
𝐴𝑔𝑒𝑛𝑒𝑟𝑎𝑙 = 𝐹(𝑏) − 𝐹(0) = − 2 𝑏 3 + ℎ𝑏 = − ℎ𝑏 + ℎ𝑏
3𝑏
3
2
𝐴𝑔𝑒𝑛𝑒𝑟𝑎𝑙 = ℎ𝑏
3
d) Check the result in question c using the result in question b.
For the big parabola we have ℎ = 1.6; 𝑏 = 1. Its area will be
2
2 16
16
32 2
𝐴𝑏𝑖𝑔 = 2 ( (1.6𝑚)(1𝑚)) = 2 ( × ) 𝑚2 = 2 ( ) 𝑚2 =
𝑚
3
3 10
15
15
For the little parabola, ℎ = 0.8; 𝑏 = 1
2
2 8
8
16 2
𝐴𝑙𝑖𝑡𝑡𝑙𝑒 = 2 ( (0.8𝑚)(1𝑚)) = 2 ( × ) 𝑚2 = 2 ( ) 𝑚2 =
𝑚
3
3 10
15
15
As for the circle we know
𝐴𝑐𝑖𝑟𝑐𝑙𝑒
2
3
9𝜋 2
= 𝜋 ( 𝑚) =
𝑚
10
100
Now the total area of the slab is
𝐴𝑠𝑙𝑎𝑏 = 𝐴𝑏𝑖𝑔 − 𝐴𝑙𝑖𝑡𝑡𝑙𝑒 − 𝐴𝑐𝑖𝑟𝑐𝑙𝑒
It is easy to check that
𝐴𝑠𝑙𝑎𝑏 =
Exactly like in question b).
320 − 27𝜋 2
𝑚
300
Mathematician A
The sculpture part corresponding to the plate part shown in Figure 3 shall be cut up in such a way
that a vertical line is cut as shown in Figure 4.
Figur 4
1
g( x)
x
0.5
x p
The area under the graph from x = 0 to x = p should be 0.40m2.
d) What is the value for p where the blue line intersects?
The area of the left side of the cut, using the expression derived in the previous question
𝑝
𝑝
ℎ
𝐴𝑐𝑢𝑡 = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ (− 2 𝑥 2 + ℎ) 𝑑𝑥
𝑏
0
0
ℎ
As we already know, the antiderivative is − 3𝑏2 𝑥 3 + ℎ𝑥. Therefore
𝐴𝑐𝑢𝑡 = −
Solving for p
ℎ 3
ℎ 3
𝑝
+
ℎ𝑝
=
−
𝑝 + ℎ𝑝 = 0.40𝑚2
3𝑏 2
3𝑏 2
ℎ 3
4 2
𝑝
−
ℎ𝑝
=
−
𝑚
3𝑏 2
10
ℎ
2 2
3
2
(𝑝
−
3𝑏
𝑝)
=
−
𝑚
3𝑏 2
5
2
3𝑏 2
6
𝑏2
𝑝3 − 3𝑏 2 𝑝 = − ( 𝑚2 )
= − ( 𝑚2 )
5
ℎ
5
ℎ
This leads to the cubic equation
6𝑏 2
𝑝3 − 3𝑏 2 𝑝 +
=0
5ℎ
With p, b and h measured in meters.
Now let’s use this expression for the little parabola, i.e. 𝑏 = 1; ℎ = 0.8 = 4/5
𝑝3 − 3(1)2 𝑝 +
6(1)2
=0
4
5( )
5
Mathematician A
𝑝3 − 3𝑝 +
6
=0
4
𝑝3 − 3𝑝 +
3
=0
2
The graph of this equation is depicted below
We can see that only one root is between 0 and 1, so this is the only one that makes sense. We can
approximate this value using the Newton-Raphson method, obtaining
𝑝 ≈ 0.558
Measured in meters.
The area of the hole in the sculpture can easily be calculated as the area of a circle. But since our
sculptor is a bit of a math enthusiast, he or she will also calculate the area in a different way.
e) Determine the area of the hole by integral calculus.
The radius of the circle is 𝑟 = 0.3𝑚 = 3/10 𝑚. The general equation of a circle of radius 𝑟
centered in the point (𝑥0 , 𝑦0 ) is
(𝑥 − 𝑥0 )2 + (𝑦 − 𝑦0 )2 = 𝑟 2
Since we only wish to know its area we can assume that the circle is centered in the origin for
simplicity, thus the equation will be
𝑥2 + 𝑦2 = 𝑟2
Solving for y
𝑦2 = 𝑟2 − 𝑥2
𝑦 = ±√𝑟 2 − 𝑥 2
Thus we will have two functions corresponding each to half a circle:
Mathematician A
The total area of the circle will be the sum of the areas of the upper and lower halves but since they
have equal areas, we can say that the total area will be twice the area of the upper half.
𝑟
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 2 ∫ √𝑟 2 − 𝑥 2 𝑑𝑥
−𝑟
Or, for simplicity’s sake, we can just say that the total area will be four times the area of a quarter of
the circle:
𝑟
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 4 ∫ √𝑟 2 − 𝑥 2 𝑑𝑥
0
Solving the indefinite integral:
∫ √𝑟 2 − 𝑥 2 𝑑𝑥 = ∫ √𝑟 2 (1 −
Now we make the substitution
𝑥
= sin 𝜃 ;
𝑟
𝑥2
)
𝑟2
𝑑𝑥 = 𝑟 ∫ √1 −
𝑥2
𝑟2
𝑑𝑥
𝑑𝑥 = 𝑟 cos 𝜃 𝑑𝜃
Obtaining
𝑟 ∫ √1 − sin2 𝜃 𝑟 cos 𝜃 𝑑𝜃 = 𝑟 2 ∫ √cos2 𝜃 cos 𝜃 𝑑𝜃 = 𝑟 2 ∫(cos 𝜃) cos 𝜃 𝑑𝜃 = 𝑟 2 ∫ cos 2 𝜃 𝑑𝜃
Using the identity
cos2 𝜃 =
1 + cos 2𝜃
2
Mathematician A
We get
𝑟 2 ∫ cos2 𝜃 𝑑𝜃 =
𝑟2
𝑟2
𝑟2
∫(1 + cos 2𝜃) 𝑑𝜃 = ∫ 𝑑𝜃 + ∫ cos 2𝜃 𝑑𝜃
2
2
2
𝑟2
𝑟2 1
= 𝜃 + ( sin 2𝜃) + 𝐶, 𝐶 ∈ ℝ
2
2 2
Using the identity
sin 2𝜃 = 2 sin 𝜃 cos 𝜃
We get
𝑟2
𝑟2 1
𝑟2
𝑟2
𝜃 + ( sin 2𝜃) + 𝐶 = 𝜃 + sin 𝜃 cos 𝜃 + 𝐶
2
2 2
2
2
Since
𝑥
= sin 𝜃
𝑟
then
cos 𝜃 = √1 −
And
𝑥2
𝑟2
𝑥
𝜃 = arcsin ( )
𝑟
Finally we get the antiderivative that we will call Z(𝑥)
𝑟2
𝑟2
𝑟2
𝑥
𝑥
𝑥2
√
𝑍(𝑥) = 𝜃 + sin 𝜃 cos 𝜃 + 𝐶 = [ arcsin ( ) +
1− 2 ]+𝐶
2
2
2
𝑟
𝑟
𝑟
Substituting this in the expression for the area:
𝑟
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 4 ∫ √𝑟 2 − 𝑥 2 𝑑𝑥 = 4 [𝑍(𝑟) − 𝑍(0)]
0
4𝑟 2
𝑟
𝑟
𝑟2
0
0
02
[arcsin ( ) + √1 − 2 − arcsin ( ) + √1 − 2 ]
2
𝑟
𝑟
𝑟
𝑟
𝑟
𝑟
𝜋
= 2𝑟 2 [arcsin(1) + 1√1 − 1 − arcsin(0)] = 2𝑟 2 ( + 0 − 0)
2
𝜋
2
2
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 2𝑟 ( ) = 𝜋𝑟
2
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 =
𝐴𝑐𝑖𝑟𝑐𝑙𝑒
And we just derived the general formula for the area of a circle of radius r. In our case we have 𝑟 =
0.3𝑚, thus
2
3
9𝜋 2
2
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋(0.3𝑚) = 𝜋 ( 𝑚) =
𝑚
10
100
Mathematician A
On the occasion of a science festival, the sculptor has been asked to make a proposal for a
decoration that draws the mind to geometric shapes. The sculptor decides that he will use a
sculpture like the one on Figure 4, but he will expand the sculpture with a rectangle that in the same
x-range has the same area as the parabola part.
f) Determine the height of the rectangle in the same interval as the parabola part, which has the
same area as the parabola part, and draw both shape parts in the same coordinate system.
The domain of the rectangle is therefore [0, 𝑝], this means one of his sides has length 𝑝. Let’s
call 𝜅 the other side, its area will be given by
𝐴𝑅 = 𝑝𝜅
Since it has the same area of the chunk of parabola
2
𝑝𝜅 = 0.4𝑚2 = 𝑚2
5
Therefore
2
1
𝜅 = ( 𝑚2 )
5
𝑝
We previously determined the approximate value 𝑝 = 0.558 𝑚. Substituting this we get
2
1
2
1000 1
200
𝜅 = ( 𝑚2 ) (
) = ( 𝑚2 ) (
)( ) =
𝑚 ≈ 0.717 𝑚
5
0.558 𝑚
5
558
𝑚
279
Mathematician A
g) Determine the intersection of the parabola part and the top side of the rectangle in the new
sculpture.
If the function that describes the parabola is 𝑓(𝑥) = −0.8𝑥 2 + 0.8, and we call 𝑔(𝑥) = 𝜅
the constant function we calculated above. We can evaluate the intersection:
𝑓(𝑥) = 𝑔(𝑥)
−0.8𝑥 2 + 0.8 = 𝜅
Solving for x
−0.8𝑥 2 = 𝜅 − 0.8
0.8 − 𝜅
𝑥2 =
0.8
0.8 − 𝜅
0.8 − 𝜅
𝑥=√
=√
≈ 0.322
0.8
0.8
Measured in meters.
Thus the intersection will occur at the point (0.322 , 0.717)