2D Calculus Analyzation of Revolving Objects

7/28/2019 2D Calculus Analyzation of Revolving Objects

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Riaz, Burhan

000536140

Mathematics SL

2D Calculus Analysation of Revolving Objects

The problem that is being faced in this paper involves a block of wood that has anunidentifiable curved shape which also makes the surface area immeasurable to simple

instruments such as a pen or ruler. To measure this piece of wood, it will be first traced

on graph paper (4mm2

ruling). After this initial step, the surface area will be evaluatedusing estimation, geometry, and integration. Then, the volume of the solid of revolution

will be determined using geometry and integration. The results of this method may be

useful to such industries that utilize different shapes of wood or other materials. One

example is the lamination of wooden materials that can only be created knowing theexact surface area of the piece of material. Laser printing on wooden objects is another

example of needing to know the exact area of the surface. Unlike the methods that will be

used in this paper to calculate the surface area and volume of the sold of revolution,industries depend on computer imaging techniques that simplify the process.

Nevertheless, these longer techniques were once used widespread in industries and their

uses still exist today.

Traced Curve of the Wooden Block (4mm2

ruling)

Part 1: Area:


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Estimation: The quickest method to calculate the surface area would be to tally thenumber of units which do not intersect with the curve and multiply by the area of a singleunit which is 2mm x 2mm= 4mm since each segment of the unit is 2mm and the area of a

square is length x width.

2mm 4mm2

2mm

There were a total of 2646 units counted that didnt intersect with the curve. The next

step would be to multiply the number of units with the area of a single unit.

2646 x 4mm2

= 10584mm2

Next, since the area of the units intersected by the curve are impossible to measure, it isbest to try to visually estimate the area and multiply by the area of a single unit. My

estimated number was 30.2 units.

30.2 x 4mm2

= 120.8mm2

Now, to complete my estimation it is necessary to add the two areas together.

10584.0mm2

+ 120.8mm2

= 10704.8mm2

Thus, my estimated surface area of the bock of wood is 10704.8mm2. This method of

estimation is limited due to human error and cannot solely be depended on whencalculating the area.

Geometry:Another way to find area of a planar region is by measuring areas ofcircumscribed and inscribed rectangles that are tangent to the curve and then finding their

averages which will estimate the total amount of area. In this case, I will be using 20rectangles to calculate the surface area. Each rectangle will be spaced 10mm apart.

Left Endpoints:

First, the height or y coordinates were measured for each rectangle according to wherethe left side of the rectangle first became tangent with the curve. The heights were then

multiplied by 10mm which is the width of the rectangle. Since length x width equals areain rectangles, the sum of all 20 rectangles is a feasible estimate of the total area.

Left Endpoints


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Box # Height(mm) Area(mm2)

1 74 740

2 78 780

3 81 810

4 82 820

5 82 820

6 81 810

7 78 780

8 72 720

9 66 660

10 58 580

11 51 510

12 46 460

13 45 450

14 44 440

15 42 420

16 40 400

17 36 360

18 30 300

19 21 210

20 10 100

Total Area 11170

Circumscribed Rectangles (4mm2

ruling)

Right Endpoints:

First, the height or y coordinates were measured for each rectangle according to wherethe right side of the rectangle first became tangent with the curve. The heights were then


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multiplied by 10mm which is the width of the rectangle. Since length x width equals area

in rectangles, the sum of all 20 rectangles is a feasible estimate of the total area.

Right Endpoints

Box # Height(mm) Area(mm2)

1 78 7802 80 800

3 82 820

4 82 820

5 81 810

6 78 780

7 72 720

8 66 660

9 58 580

10 50 500

11 47 470

12 45 450

13 44 440

14 43 430

15 40 400

16 36 360

17 30 300

18 21 210

19 10 100

20 0 0

Total Area 10430

Inscribed Rectangles (4mm2

ruling)


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Thus, the total sum the circumscribed rectangles is 11170mm2

and the total sum

of the inscribed rectangles is 10430mm2. This must mean that the actual area is between

10430mm2

and 11170mm2. If we find the average of these two sums by adding them

together and dividing by 2, the area is 10800mm2. The actual area is most likely to be

close to this number but a better way to measure geometrically is by adding more boxes

so that the chance of error is reduced.

Integration: The final method that I will use to estimate the area under the curve involvesa graphing calculator with regression capability. Before creating a regression, a scatter

plot needs to be created by inserting the x and y coordinates. I will use 20 points that arespaced 10mm away.

Scatter Plot Coordinates

X Y

0 74

10 78

20 81

30 82

40 82

50 81

60 78

70 72

80 66

90 58

100 51

110 46

120 45

130 44

140 42

150 40

160 36

170 30

180 21

195 0

Scatter Plot

90

0 200


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Next, the correct regression must be chosen from the calculator. The quartic regression

best fits the scatter plot. This is to be expected since quartic functions usually often

contain 3 critical points.Quartic Regression

90

0 200Quartic Equation from Calculator

The quartic equation is f(x) = ax4

+ bx3

+ cx2

+ dx + e

The definite integral for this function is:

195

0

[(-5.617E-7)x4

+ (2.267E-4)x3

+ (-.031)x2

+(1.221)x + (70.482)] dx

By using the Finite Integral function on the calculator and pasting the quartic

equation into this function, I received the area under the curve; 10609.963mm2.This

method of estimation works very well since the regression is close to the actual datapoints. We know this because the R

2value is .990 which is close to 1. We can safely say

that the actual area is near 10609.963mm2.

All methods that were used unsurprisingly lead to similar estimations of the totalarea. What was surprising was that the geometric method was more off than the

estimation method. This of course can be counteracted by increasing the number ofrectangles that are measured. The estimation method only presented .9% error while the

geometric method presented 1.8% error in relation to the area found by using the

integration method. The integration was most successful since the regression was close tothe actual curve.


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Part 2: Volume of a Solid of Revolution

Solid of Revolution of Wooden Block (4mm2

ruling)

Geometry: The equation for finding the volume of a disk is R2w. In other words,(Height

210mm). The summation of all 20 disks results in the total volume of the 3D

figure. Since I already have the y coordinates (height) for 20 bars, I will simply square it,

multiply it by 10mm, and multiply .

Example: Disk 1

[(74mm)2(10mm)]

[(5476mm2)(10mm)]

54760mm3


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Volume of Disks

Disk # Height(mm) Height2(mm2) Volume(mm3)

1 74 5476 54760

2 78 6084 60840

3 81 6561 65610

4 82 6724 67240

5 82 6724 67240

6 81 6561 65610

7 78 6084 60840

8 72 5184 51840

9 66 4356 43560

10 58 3364 33640

11 51 2601 26010

12 46 2116 2116013 45 2025 20250

14 44 1936 19360

15 42 1764 17640

16 40 1600 16000

17 36 1296 12960

18 30 900 9000

19 21 441 4410

20 10 100 1000

Total Volume 718970

Thus, by adding the volume of all of the disks together, we have 718970mm3 as the

volume of the entire solid of revolution. The best way to decrease the percent error is by

adding more disks.

We can find the definite integral of the volume by inputting the quartic equation into theformula of the disk method.

195

0

[(-5.617E-7)x4

+ (2.267E-4)x3

+ (-.031)x2

+(1.221)x + (70.482)]2

dx

By inputting the entire definite integral into the finite integral function, the calculator

displays 680831.572mm3

as the total volume of the solid of revolution. This is quitedifferent from the geometric estimation of718970mm

3. There is 5.3% percent error in the

geometric method. The geometric method was probably off since it contained only 20disks and the error continued to grow because of squaring the height and multiplying by

10mm. The integration was most successful since the regression was close to the actual

curve (R2=.990). To check whether this information is correct, one can fill this object

with water and extract the water into a beaker. The beaker will be able to show the

volume.

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2D Calculus Analyzation of Revolving Objects