Volumes of Revolution

In this project, I found the volume of revolution of a margarita glass using a digital caliper in order to find themeasurements. I took measurements every 3mm apart in height, and measured the diameter of the glass at each

height. This is the data, shown {height, width

2}:

f = TableA990,80.44

2=, 93,

78.94

2=, 96,

60.19

2=, 99,

42.88

2=, 912,

25.03

2=, 915,

10.65

2=,

918,10.40

2=, 921,

10.42

2=, 924,

10.58

2=, 927,

10.72

2=, 930,

10.90

2=, 933,

11.00

2=,

936,11.02

2=, 939,

11.22

2=, 942,

11.40

2=, 945,

11.47

2=, 948,

11.67

2=, 951,

11.99

2=,

954,12.07

2=, 957,

12.11

2=, 960,

12.24

2=, 963,

12.62

2=, 966,

12.68

2=, 969,

12.93

2=,

972,12.97

2=, 975,

14.35

2=, 978,

17.20

2=, 981,

21.41

2=, 984,

25.58

2=, 987,

29.51

2=,

990,33.73

2=, 993,

37.25

2=, 996,

40.31

2=, 999,

43.66

2=, 9102,

46.07

2=,

9105,48.17

2=, 9108,

49.96

2=, 9111,

51.41

2=, 9114,

52.62

2=, 9117,

53.97

2=,

9120,55.61

2=, 9123,

56.78

2=, 9126,

67.60

2=, 9129,

77.09

2=, 9132,

84.53

2=,

9135,90.81

2=, 9138,

97.86

2=, 9141,

103.65

2=, 9144,

108.70

2=, 9147,

112.39

2=,

9150,114.57

2=, 9153,

116.80

2=, 9156,

118.80

2=, 9159,

120.14

2=, 9162,

121.29

2==E;

Trapezoidal Rule

The area for the trapezoid is: I b - a

nM@ f Hx0L + f Hx1L + f Hx2L + ... + f Hxn-1L + f HxnLD.

In this case, b=162, a=0, and n=54. But since we are trying to find the volume of the glass not just the area, wemust use the disk method, multiplying everything by Π and squaring each measurement. So,

In[4]:=162 Π

54* 9H80.44 � 2L2

+ H78.94 � 2L2+ H60.19 � 2L2

+ H42.88 � 2L2+ H25.03 � 2L2

+

H10.65 � 2L2+ H10.40 � 2L2

+ H10.42 � 2L2+ H10.58 � 2L2

+ H10.72 � 2L2+

H10.90 � 2L2+ H11.00 � 2L2

+ H11.02 � 2L2+ H11.22 � 2L2

+ H11.40 � 2L2+

H11.47 � 2L2+ H11.67 � 2L2

+ H11.99 � 2L2+ H12.07 � 2L2

+ H12.11 � 2L2+ H12.24 � 2L2

+

H12.62 � 2L2+ H12.68 � 2L2

+ H12.93 � 2L2+ H12.97 � 2L2

+ H14.35 � 2L2+ H17.20 � 2L2

+

H21.41 � 2L2+ H25.58 � 2L2

+ H29.51 � 2L2+ H33.73 � 2L2

+ H37.25 � 2L2+ H40.31 � 2L2

+

H43.66 � 2L2+ H46.07 � 2L2

+ H48.17 � 2L2+ H49.96 � 2L2

+ H51.41 � 2L2+ H52.62 � 2L2

+

H53.97 � 2L2+ H55.61 � 2L2

+ H56.78 � 2L2+ H67.60 � 2L2

+ H77.09 � 2L2+ H84.53 � 2L2

+

H90.81 � 2L2+ H97.86 � 2L2

+ H103.65 � 2L2+ H108.70 � 2L2

+ H112.39 � 2L2+

H114.57 � 2L2+ H116.80 � 2L2

+ H118.80 � 2L2+ H120.14 � 2L2

+ H121.29 � 2L2= �� N

Out[4]= 8453 080. <

The Trapezoidal Rule with the disk method gives us an approximated volume of 453, 080 mm3, or 453.08 cm3.

Simpson’s Rule

The area for Simpson’s Rule is: Hb-aL3 n

@ f Hx0L + 4 f Hx1L + 2 f Hx2L + 4 f Hx3L + ... + 4 f Hxn-1L + f HxnLD.In this case, b=162, a=0, and n=54. But since we are trying to find the volume not just the area, we must use thedisk method, multiplying everything by Π and squaring each measurement. So,

Simpson’s Rule

The area for Simpson’s Rule is: Hb-aL3 n

@ f Hx0L + 4 f Hx1L + 2 f Hx2L + 4 f Hx3L + ... + 4 f Hxn-1L + f HxnLD.In this case, b=162, a=0, and n=54. But since we are trying to find the volume not just the area, we must use thedisk method, multiplying everything by Π and squaring each measurement. So,

In[2]:=162 Π

162*

9H80.44 � 2L2+ 4 H78.94 � 2L2

+ 2 H60.19 � 2L2+ 4 H42.88 � 2L2

+ 2 H25.03 � 2L2+ 4 H10.65 � 2L2

+

2 H10.40 � 2L2+ 4 H10.42 � 2L2

+ 2 H10.58 � 2L2+ 4 H10.72 � 2L2

+ 2 H10.90 � 2L2+

4 H11.00 � 2L2+ 2 H11.02 � 2L2

+ 4 H11.22 � 2L2+ 2 H11.40 � 2L2

+ 4 H11.47 � 2L2+

2 H11.67 � 2L2+ 4 H11.99 � 2L2

+ 2 H12.07 � 2L2+ 4 H12.11 � 2L2

+ 2 H12.24 � 2L2+

4 H12.62 � 2L2+ 2 H12.68 � 2L2

+ 4 H12.93 � 2L2+ 2 H12.97 � 2L2

+ 4 H14.35 � 2L2+

2 H17.20 � 2L2+ 4 H21.41 � 2L2

+ 2 H25.58 � 2L2+ 4 H29.51 � 2L2

+ 2 H33.73 � 2L2+

4 H37.25 � 2L2+ 2 H40.31 � 2L2

+ 4 H43.66 � 2L2+ 2 H46.07 � 2L2

+ 4 H48.17 � 2L2+

2 H49.96 � 2L2+ 4 H51.41 � 2L2

+ 2 H52.62 � 2L2+ 4 H53.97 � 2L2

+ 2 H55.61 � 2L2+

4 H56.78 � 2L2+ 2 H67.60 � 2L2

+ 4 H77.09 � 2L2+ 2 H84.53 � 2L2

+ 4 H90.81 � 2L2+

2 H97.86 � 2L2+ 4 H103.65 � 2L2

+ 2 H108.70 � 2L2+ 4 H112.39 � 2L2

+ 2 H114.57 � 2L2+

4 H116.80 � 2L2+ 2 H118.80 � 2L2

+ 4 H120.14 � 2L2+ H121.29 � 2L2= �� N

Out[2]= 8428 315.<

Simpson’s Rule with the disk method gives us an approximated volume of 428, 315 mm3, or 428.32 cm3.

This graph shows the data points, representing the glass turned sideways and split in half. We need to split theglass in half symetrically so that it can be rotated around the x-axis in order to create the 3D object, shown furtherdown, and to find the volume of revolution.

ListLinePlot@fD

50 100 150

10

20

30

40

50

60

And here is a 3D representation of the object:

g = Interpolation@f, InterpolationOrder ® 1D

InterpolatingFunction@880., 162.<<, <>D

2 Untitled-3.nb

ParametricPlot3D@8x, g@xD * Cos@tD, g@xD * Sin@tD<, 8x, 0, 162<,

8t, 0, 2 Π<, AxesOrigin ® 80, 0<, Boxed ® False, AxesLabel ® 8x, y, z<D

0

50

100

150

x-50

0

50

y

-50

0

50

z

Untitled-3.nb 3