FURTHER APPLICATIONS OF INTEGRATION

FURTHER APPLICATIONS FURTHER APPLICATIONS OF INTEGRATIONOF INTEGRATION

8

8.3Applications 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.

FURTHER APPLICATIONS OF INTEGRATION

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.

APPLICATIONS TO PHYSICS AND ENGINEERING

Deep-sea divers realize that water pressure increases as they dive deeper.

This is because the weight of the water above them increases.

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.


The fluid directly above the plate has volume

V = Ad

So, its mass is:

m = ρV = ρAd


Thus, the force exerted by the fluid on the plate is

F = mg = ρgAd

where g is the acceleration due to gravity.

HYDROSTATIC FORCE

The pressure P on the plate is defined to be the force per unit area:

HYDROSTATIC PRESSURE

FP gdA

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.


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:


3 21000kg/m 9.8m/s 2m19,600Pa19.6kPa

P gd

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.


Thus, the pressure in any direction at a depth d in a fluid with mass density ρ is given by:


P gd d

Equation 1

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.


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.

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.


We choose a vertical x-axis with origin at the surface of the water.


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].


The i th horizontal strip of the dam is approximated by a rectangle with height Δx and width wi


From similar triangles,

HYDROSTATIC F AND P

* *

*

1610 or 816 20 2 2

i i

i

x xa ax

Example 1

Hence,

*12

*

2(15 )

2(15 8 )

46

i

i

i

w a

x

x


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:

HYDROSTATIC F AND P

*(46 )i i iA w x x x

*1000i iP gx

Example 1

The hydrostatic force Fi acting on the i th

strip is the product of the pressure and the area:

HYDROSTATIC F AND P

* *1000 (46 )i i i

i i

F PA

gx x x

Example 1

Adding these forces and taking the limit as n → ∞, the total hydrostatic force on the dam is:

HYDROSTATIC F AND P

* *

1

16

0

16 2

0

1632 7

0

lim 1000 (46 )

1000 (46 )

1000(9.8) (46 )

9800 23 4.43 10 N3

n

i in i

F gx x x

gx x dx

x x dx

xx

Example 1

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.


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


As in Example 1, we divide the circular region into horizontal strips of equal width.


From the equation of the circle, we see that the length of the i th strip is:

So, its area is:

HYDROSTATIC F AND P

* 22 9 ( )iy

* 22 9 ( )i iA y y

Example 2

The pressure on this strip is approximately

So, the force on the strip is approximately

HYDROSTATIC F AND P

* * 262.5(7 )2 9 ( )i i i id A y y y

*62.5(7 )i id y

Example 2

We get the total force by adding the forces on all the strips and taking the limit:

HYDROSTATIC F AND P

* * 2

1

3 2

3

3 32 2

3 3

lim 62.5(7 )2 9 ( )

125 (7 ) 9

125 7 9 125 9

n

i in i

F y y y

y y dy

y dy y y dy

Example 2

The second integral is 0 because the integrand is an odd function.

See Theorem 7 in Section 5.5


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.


Thus,HYDROSTATIC F AND P

3 2

3

212

875 9

875 (3)7875

212,370lb

F y dy

Example 2

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

MOMENTS AND CENTERS OF MASS

This point is called the center of mass (or center of gravity) of the plate.

CENTERS OF MASS

We first consider the simpler situation illustrated here.

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.

CENTERS OF MASS

The rod will balance if:CENTERS OF MASS

1 1 2 2m d m d

Equation 2

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.

LAW OF THE LEVER

Now, suppose that the rod lies along

the x-axis, with m1 at x1 and m2 at x2

and the center of mass at .


x

Comparing the figures, we see that:

d1 = – x1 d2 = x1 –

x


x

CENTERS OF MASS

So, Equation 2 gives:

1 1 2 2( ) ( )m x x m x x

1 2 1 1 2 2m x m x m x m x

1 1 2 2

1 2

m x m xxm m

Equation 3

The numbers m1x1 and m2x2 are called

the moments of the masses m1 and m2

(with respect to the origin).

MOMENTS OF MASS

Equation 3 says that the center of mass is obtained by:

1. Adding the moments of the masses2. Dividing by the total mass m = m1 + m2

MOMENTS OF MASS

x

1 1 2 2

1 2

m x m xxm m

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.

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.

CENTERS OF MASS

1 1

1

n n

i i i ii in

ii

m x m xx

mm

Equation 4

The sum of the individual moments

is called the moment of the system about the origin.

MOMENT OF SYSTEM ABOUT ORIGIN

1

n

i ii

M m x

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.

MOMENT OF SYSTEM ABOUT ORIGIN

x

x

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.


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

MOMENT ABOUT AXES

1

n

y i ii

M m x

Equations 5 and 6

1

n

x i ii

M m y

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.

MOMENT ABOUT AXES

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.

CENTERS OF MASS

( , )x y

y xM Mx ym m

Equation 7

Since , the center of mass is the point where a single particle of mass m would have the same moments as the system.


and y xmx M my M ( , )x y

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.

MOMENTS & CENTERS OF MASS Example 3

We use Equations 5 and 6 to compute the moments:

MOMENTS & CENTERS OF MASS

3( 1) 4(2) 8(3) 29

3(1) 4( 1) 8(2) 15

y

x

M

M

Example 3

As m = 3 + 4 + 8 = 15, we use Equation 7 to obtain:

MOMENTS & CENTERS OF MASS Example 3

29 15 115 15

y xM Mx ym m

Thus, the center of mass is:MOMENTS & CENTERS OF MASS

1415(1 ,1)

Example 3

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 .

CENTROIDS

In doing so, we use the following physical principles.

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.

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.

CENTROIDS

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

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

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

CENTROIDS

ix

ix

This determines the polygonal approximation to R shown below.

CENTROIDS

The centroid of the i th approximating

rectangle Ri is its center .

Its area is: f( ) ∆x

So, its mass is:

CENTROIDS

12( , ( ))i i iC x f x

( )if x x

ix

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 .

CENTROIDS

ix

Therefore,CENTROIDS

( ) ( )

( )y i i i

i i

M R f x x x

x f x x

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

CENTROIDS

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

CENTROIDS

1

lim ( )

( )

n

y i in i

b

a

M x f x x

x f x dx

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:

CENTROIDS

12

212

( ) ( ) ( )

( )

x i i i

i

M R f x x f x

f x x

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

CENTROIDS

212

1

212

lim ( )

( )

n

x in i

b

a

M f x x

f x dx

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

CENTROIDS

andy xmx M my M

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

CENTROIDS

( )b

a

m A

f x dx

Thus,

Notice the cancellation of the ρ’s. The location of the center of mass is independent

of the density.

CENTROIDS

( ) ( )

( ) ( )

b b

y a ab b

a a

xf x dx xf x dxMx

m f x dx f x dx

2 21 12 2( ) ( )

( ) ( )

b b

x a ab b

a a

f x dx f x dxMy

m f x dx f x dx

In summary, the center of mass of the plate (or the centroid of R ) is located at the point , where

CENTROIDS

( , )x y

212

1 ( )

1 ( )

b

a

b

a

x xf x dxA

y f x dxA

Formula 8

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.

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 .

CENTERS OF MASS

x

0x

Example 4

The area of the semicircle is A = ½πr2.

Thus,

CENTERS OF MASS

212

22 21

2212

32 2 2

2 200

3

2

1 ( )

1

2 23

2 2 43 3

r

r

r

r

rr

y f x dxA

r x dxr

xr x dx r xr r

r rr

Example 4

The center of mass is located at the point (0, 4r/(3π)).


Find the centroid of the region bounded by the curves

y = cos x, y = 0, x = 0, x = π/2


The area of the region is:CENTERS OF MASS

2 2

00cos sin

1

A xdx x

Example 5

CENTROIDS

/ 2 212

/ 2 212

/ 214

/ 21 14 2 0

1 ( )

cos

1 cos 2

sin 2

8

a

a

a

y f x dxA

x dx

x dx

x x

Example 5

/ 2

/ 2

0

2

0

2

0

1 ( )

cos

sin

sin

12

ax xf x dx

A

x x dx

x x

x dx

So, Formulas 8 give:

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

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

CENTROIDS

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

CENTROIDS

( , )x y

2 212

1 ( ) ( )

1 ( ) ( )

b

a

b

a

x x f x g x dxA

y f x g x dxA

Formula 9

Find the centroid of the region bounded by the line x = y and the parabola y = x2.

CENTROIDS Example 6

The region is sketched here.

We take f(x) = x, g(x) = x2, a = 0, and b = 1 in Formulas 9.

CENTROIDS Example 6

First, we note that the area of the region is:

CENTROIDS

1 2

0

12 3

0

( )

2 2

16

A x x dx

x x

Example 6

Therefore, CENTROIDS

1

0

1 21 06

1 2 3

0

13 4

0

1 ( ) ( )

1 ( )

6 ( )

163 4 2

x x f x g x dxA

x x x dx

x x dx

x x

Example 6

1 2 2120

1 2 4121 0

6

13 5

0

1 ( ) ( )

1 ( )

33 5

25

y f x g x dxA

x x dx

x x

The centroid is:CENTROIDS

1 2,2 5

Example 6

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

CENTROIDS

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 .

THEOREM OF PAPPUS

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.

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.

THEOREM OF PAPPUS

2 ( ) ( )

2 ( ) ( )

2 ( ) (Formulas 9)(2 )

b

a

b

a

V x f x g x dx

x f x g x dx

xAx A

Ad

Proof

2d x

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.

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.

THEOREM OF PAPPUS Example 7

Therefore, by the Theorem of Pappus, the volume of the torus is:

THEOREM OF PAPPUS

2

2 2

(2 )( )

2

V Ad

R r

r R

Example 7

Compare the method of Example 7 with that of Exercise 63 in Section 6.2

THEOREM OF PAPPUS

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