1. FURTHER APPLICATIONS OF INTEGRATION9
FURTHER APPLICATIONS OF INTEGRATION

2. 8.3 Applications to Physics and Engineering
FURTHER APPLICATIONS OF INTEGRATION
8.3 Applications to Physics and Engineering
In this section, we will learn about:
The applications of integral calculus to
force due to water pressure and centers of mass.

3. APPLICATIONS TO PHYSICS AND ENGINEERING
As with our previous applications to geometry (areas, volumes, and lengths) and to work, our strategy is:
Break up the physical quantity into small parts.
Approximate each small part.
Add the results.
Take the limit.
Then, evaluate the resulting integral.

4. HYDROSTATIC FORCE AND PRESSURE
Deep-sea divers realize that water pressure increases as they dive deeper.
This is because the weight of the water above them increases.

5. HYDROSTATIC FORCE AND PRESSURE
Suppose that a thin plate with area A m2 is submerged in a fluid of density ρ kg/m3 at a depth d meters below the surface of the fluid.

6. HYDROSTATIC FORCE AND PRESSURE
The fluid directly above the plate has volume
V = Ad
So, its mass is:
m = ρV = ρAd

7. Thus, the force exerted by the fluid on the plate is F = mg = ρgAd
HYDROSTATIC FORCE
Thus, the force exerted by the fluid on the plate is F = mg = ρgAd
where g is the acceleration due to gravity.

8. The pressure P on the plate is defined to be the force per unit area:
HYDROSTATIC PRESSURE
The pressure P on the plate is defined to be the force per unit area:

9. HYDROSTATIC PRESSURE
The SI unit for measuring pressure is newtons per square meter—which is called a pascal (abbreviation: 1 N/m2 = 1 Pa).
As this is a small unit, the kilopascal (kPa) is often used.

10. HYDROSTATIC PRESSURE
For instance, since the density of water is ρ = 1000 kg/m3, the pressure at the bottom of a swimming pool 2 m deep is:

11. HYDROSTATIC PRESSURE
An important principle of fluid pressure is the experimentally verified fact that, at any point in a liquid, the pressure is the same in all directions.
This is why a diver feels the same pressure on nose and both ears.

12. HYDROSTATIC PRESSURE
Equation 1
Thus, the pressure in any direction at a depth d in a fluid with mass density ρ is given by:

13. HYDROSTATIC FORCE AND PRESSURE
This helps us determine the hydrostatic force against a vertical plate or wall or dam in a fluid.
This is not a straightforward problem.
The pressure is not constant, but increases as the depth increases.

14. A dam has the shape of the trapezoid shown below.
HYDROSTATIC F AND P
Example 1
A dam has the shape of the trapezoid shown below.
The height is 20 m.
The width is 50 m at the top and 30 m at the bottom.

15. HYDROSTATIC F AND P
Example 1
Find the force on the dam due to hydrostatic pressure if the water level is 4 m from the top of the dam.

16. We choose a vertical x-axis with origin at the surface of the water.
HYDROSTATIC F AND P
Example 1
We choose a vertical x-axis with origin at the surface of the water.

17. The depth of the water is 16 m.
HYDROSTATIC F AND P
Example 1
The depth of the water is 16 m.
So, we divide the interval [0, 16] into subintervals of equal length with endpoints xi.
We choose xi* [xi–1, xi].

18. HYDROSTATIC F AND P
Example 1
The i th horizontal strip of the dam is approximated by a rectangle with height Δx and width wi

19. From similar triangles,
HYDROSTATIC F AND P
Example 1
From similar triangles,

20. HYDROSTATIC F AND P
Example 1
Hence,

21. If Ai is the area of the strip, then
HYDROSTATIC F AND P
Example 1
If Ai is the area of the strip, then
If Δx is small, then the pressure Pi on the i th strip is almost constant, and we can use Equation 1 to write:

22. HYDROSTATIC F AND P
Example 1
The hydrostatic force Fi acting on the i th strip is the product of the pressure and the area:

23. HYDROSTATIC F AND P
Example 1
Adding these forces and taking the limit as n → ∞, the total hydrostatic force on the dam is:

24. HYDROSTATIC F AND P
Example 2
Find the hydrostatic force on one end of a cylindrical drum with radius 3 ft, if the drum is submerged in water 10 ft deep.

25. HYDROSTATIC F AND P
Example 2
In this example, it is convenient to choose the axes as shown—so that the origin is placed at the center of the drum.
Then, the circle has a simple equation: x2 + y2 = 9

26. HYDROSTATIC F AND P
Example 2
As in Example 1, we divide the circular region into horizontal strips of equal width.

27. HYDROSTATIC F AND P
Example 2
From the equation of the circle, we see that the length of the i th strip is:
So, its area is:

28. The pressure on this strip is approximately
HYDROSTATIC F AND P
Example 2
The pressure on this strip is approximately
So, the force on the strip is approximately

29. HYDROSTATIC F AND P
Example 2
We get the total force by adding the forces on all the strips and taking the limit:

30. The second integral is 0 because the integrand is an odd function.
HYDROSTATIC F AND P
Example 2
The second integral is 0 because the integrand is an odd function.
See Theorem 7 in Section 5.5

31. HYDROSTATIC F AND P
Example 2
The first integral can be evaluated using the trigonometric substitution y = 3 sin θ.
However, it’s simpler to observe that it is the area of a semicircular disk with radius 3.

32. HYDROSTATIC F AND P
Example 2
Thus,

33. MOMENTS AND CENTERS OF MASS
Our main objective here is to find the point P on which a thin plate of any given shape balances horizontally as shown.

34. CENTERS OF MASS
This point is called the center of mass (or center of gravity) of the plate.

35. We first consider the simpler situation illustrated here.
CENTERS OF MASS
We first consider the simpler situation illustrated here.

36. CENTERS OF MASS
Two masses m1 and m2 are attached to a rod of negligible mass on opposite sides of a fulcrum and at distances d1 and d2 from the fulcrum.

37. The rod will balance if:
CENTERS OF MASS
Equation 2
The rod will balance if:

38. LAW OF THE LEVER
This is an experimental fact discovered by Archimedes and called the Law of the Lever.
Think of a lighter person balancing a heavier one on a seesaw by sitting farther away from the center.

39. MOMENTS AND CENTERS OF MASS
Now, suppose that the rod lies along the x-axis, with m1 at x1 and m2 at x2 and the center of mass at .

40. MOMENTS AND CENTERS OF MASS
Comparing the figures, we see that:
d1 = – x1
d2 = x1 –

41. CENTERS OF MASS
Equation 3
So, Equation 2 gives:

42. MOMENTS OF MASS
The numbers m1x1 and m2x2 are called the moments of the masses m1 and m2 (with respect to the origin).

43. Equation 3 says that the center of mass is obtained by:
MOMENTS OF MASS
Equation 3 says that the center of mass is obtained by:
Adding the moments of the masses
Dividing by the total mass m = m1 + m2

44. CENTERS OF MASS
Equation 4
In general, suppose we have a system of n particles with masses m1, m2, , mn located at the points x1, x2, , xn on the x-axis.
Then, we can show where the center of mass of the system is located—as follows.

45. The center of mass of the system is located at
CENTERS OF MASS
Equation 4
The center of mass of the system is located at
where m = Σ mi is the total mass of the system.

46. MOMENT OF SYSTEM ABOUT ORIGIN
The sum of the individual moments
is called the moment of the system about the origin.

47. MOMENT OF SYSTEM ABOUT ORIGIN
Then, Equation 4 could be rewritten as:
m = M
This means that, if the total mass were considered as being concentrated at the center of mass , then its moment would be the same as the moment of the system.

48. MOMENTS AND CENTERS OF MASS
Now, we consider a system of n particles with masses m1, m2, , mn located at the points (x1, y1), (x2, y2) , (xn, yn) in the xy-plane.

49. and the moment of the system about the x-axis as
MOMENT ABOUT AXES
Equations 5 and 6
By analogy with the one-dimensional case, we define the moment of the system about the y-axis as
and the moment of the system about the x-axis as

50. My measures the tendency of the system to rotate about the y-axis.
MOMENT ABOUT AXES
My measures the tendency of the system to rotate about the y-axis.
Mx measures the tendency of the system to rotate about the x-axis.

51. where m = ∑ mi is the total mass.
CENTERS OF MASS
Equation 7
As in the one-dimensional case, the coordinates of the center of mass are given in terms of the moments by the formulas
where m = ∑ mi is the total mass.

52. MOMENTS AND CENTERS OF MASS
Since , the center of mass is the point where a single particle of mass m would have the same moments as the system.

53. MOMENTS & CENTERS OF MASS
Example 3
Find the moments and center of mass of the system of objects that have masses 3, 4, and 8 at the points (–1, 1), (2, –1) and (3, 2) respectively.

54. MOMENTS & CENTERS OF MASS
Example 3
We use Equations 5 and 6 to compute the moments:

55. MOMENTS & CENTERS OF MASS
Example 3
As m = = 15, we use Equation 7 to obtain:

56. MOMENTS & CENTERS OF MASS
Example 3
Thus, the center of mass is:

57. CENTROIDS
Next, we consider a flat plate, called a lamina, with uniform density ρ that occupies a region R of the plane.
We wish to locate the center of mass of the plate, which is called the centroid of R .

58. In doing so, we use the following physical principles.
CENTROIDS
In doing so, we use the following physical principles.

59. CENTROIDS
The symmetry principle says that, if R is symmetric about a line l, then the centroid of R lies on l.
If R is reflected about l, then R remains the same so its centroid remains fixed.
However, the only fixed points lie on l.
Thus, the centroid of a rectangle is its center.

60. CENTROIDS
Moments should be defined so that, if the entire mass of a region is concentrated at the center of mass, then its moments remain unchanged.

61. CENTROIDS
Also, the moment of the union of two non-overlapping regions should be the sum of the moments of the individual regions.

62. Suppose that the region R is of the type shown here.
CENTROIDS
Suppose that the region R is of the type shown here.
That is, R lies:
Between the lines x = a and x = b
Above the x-axis
Beneath the graph of f, where f is a continuous function

63. CENTROIDS
We divide the interval [a, b] into n subintervals with endpoints x0, x1, , xn and equal width ∆x.
We choose the sample point xi* to be the midpoint of the i th subinterval.
That is, = (xi–1 + xi)/2

64. This determines the polygonal approximation to R shown below.
CENTROIDS
This determines the polygonal approximation to R shown below.

65. The centroid of the i th approximating rectangle Ri is its center .
CENTROIDS
The centroid of the i th approximating rectangle Ri is its center
Its area is: f( ) ∆x
So, its mass is:

66. CENTROIDS
The moment of Ri about the y-axis is the product of its mass and the distance from Ci to the y-axis, which is .

67. CENTROIDS
Therefore,

68. CENTROIDS
Adding these moments, we obtain the moment of the polygonal approximation to R .

69. CENTROIDS
Then, by taking the limit as n → ∞, we obtain the moment of R itself about the y-axis:

70. CENTROIDS
Similarly, we compute the moment of Ri about the x-axis as the product of its mass and the distance from Ci to the x-axis:

71. CENTROIDS
Again, we add these moments and take the limit to obtain the moment of R about the x-axis:

72. CENTROIDS
Just as for systems of particles, the center of mass of the plate is defined so that

73. CENTROIDS
However, the mass of the plate is the product of its density and its area:

74. Thus, CENTROIDS Notice the cancellation of the ρ’s.
The location of the center of mass is independent of the density.

75. CENTROIDS
Formula 8
In summary, the center of mass of the plate (or the centroid of R ) is located at the point , where

76. Find the center of mass of a semicircular plate of radius r.
CENTERS OF MASS
Example 4
Find the center of mass of a semicircular plate of radius r.
To use Equation 8, we place the semicircle as shown so that f(x) = √(r2 – x2) and a = –r, b = r.

77. Here, there is no need to use the formula to calculate .
CENTERS OF MASS
Example 4
Here, there is no need to use the formula to calculate .
By the symmetry principle, the center of mass must lie on the y-axis, so

78. The area of the semicircle is A = ½πr2. Thus,
CENTERS OF MASS
Example 4
The area of the semicircle is A = ½πr2.
Thus,

79. The center of mass is located at the point (0, 4r/(3π)).
CENTERS OF MASS
Example 4
The center of mass is located at the point (0, 4r/(3π)).

80. CENTERS OF MASS
Example 5
Find the centroid of the region bounded by the curves y = cos x, y = 0, x = 0, x = π/2

81. The area of the region is:
CENTERS OF MASS
Example 5
The area of the region is:

82. CENTROIDS
Example 5
So, Formulas 8 give:

83. The centroid is ((π/2) – 1, π/8).
CENTROIDS
Example 5
The centroid is ((π/2) – 1, π/8).

84. CENTROIDS
Suppose the region R lies between two curves y = f(x) and y = g(x), where f(x) ≥ g(x).

85. CENTROIDS
Formula 9
Then, the same sort of argument that led to Formulas 8 can be used to show that the centroid of R is , where

86. CENTROIDS
Example 6
Find the centroid of the region bounded by the line x = y and the parabola y = x2.

87. The region is sketched here.
CENTROIDS
Example 6
The region is sketched here.
We take f(x) = x, g(x) = x2, a = 0, and b = 1 in Formulas 9.

88. First, we note that the area of the region is:
CENTROIDS
Example 6
First, we note that the area of the region is:

89. CENTROIDS
Example 6
Therefore,

90. CENTROIDS
Example 6
The centroid is:

91. CENTROIDS
We end this section by showing a surprising connection between centroids and volumes of revolution.

92. of R and the distance d traveled by the centroid of R .
THEOREM OF PAPPUS
Let R be a plane region that lies entirely on one side of a line l in the plane.
If R is rotated about l, then the volume of the resulting solid is the product of the area A
of R and the distance d traveled by the centroid of R .

93. THEOREM OF PAPPUS
Proof
We give the proof for the special case in which the region lies between y = f(x) and y = g(x) as shown and the line l is the y-axis.

94. By the cylindrical shells method (Section 6.3), we have:
THEOREM OF PAPPUS
Proof
By the cylindrical shells method (Section 6.3), we have:
is the distance traveled by the centroid during one rotation about the y-axis.

95. Find the volume of the torus.
THEOREM OF PAPPUS
Example 7
A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R(> r) from the center of the circle.
Find the volume of the torus.

96. The circle has area A = πr2.
THEOREM OF PAPPUS
Example 7
The circle has area A = πr2.
By the symmetry principle, its centroid is its center.
So, the distance traveled by the centroid during a rotation is d = 2πR.

97. Therefore, by the Theorem of Pappus, the volume of the torus is:
Example 7
Therefore, by the Theorem of Pappus, the volume of the torus is:

98. THEOREM OF PAPPUS
Compare the method of Example 7 with that of Exercise 63 in Section 6.2