further applications of integration
DESCRIPTION
8. FURTHER APPLICATIONS OF INTEGRATION. 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. - PowerPoint PPT PresentationTRANSCRIPT
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FURTHER APPLICATIONS FURTHER APPLICATIONS OF INTEGRATIONOF INTEGRATION
8
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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
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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
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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
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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.
HYDROSTATIC FORCE AND PRESSURE
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The fluid directly above the plate has volume
V = Ad
So, its mass is:
m = ρV = ρAd
HYDROSTATIC FORCE AND PRESSURE
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Thus, the force exerted by the fluid on the plate is
F = mg = ρgAd
where g is the acceleration due to gravity.
HYDROSTATIC FORCE
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The pressure P on the plate is defined to be the force per unit area:
HYDROSTATIC PRESSURE
FP gdA
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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.
HYDROSTATIC PRESSURE
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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:
HYDROSTATIC PRESSURE
3 21000kg/m 9.8m/s 2m19,600Pa19.6kPa
P gd
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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.
HYDROSTATIC PRESSURE
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Thus, the pressure in any direction at a depth d in a fluid with mass density ρ is given by:
HYDROSTATIC PRESSURE
P gd d
Equation 1
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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.
HYDROSTATIC FORCE AND PRESSURE
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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
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Find the force on the dam due to hydrostatic pressure if the water level is 4 m from the top of the dam.
HYDROSTATIC F AND P Example 1
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We choose a vertical x-axis with origin at the surface of the water.
HYDROSTATIC F AND P Example 1
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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].
HYDROSTATIC F AND P Example 1
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The i th horizontal strip of the dam is approximated by a rectangle with height Δx and width wi
HYDROSTATIC F AND P Example 1
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From similar triangles,
HYDROSTATIC F AND P
* *
*
1610 or 816 20 2 2
i i
i
x xa ax
Example 1
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Hence,
*12
*
2(15 )
2(15 8 )
46
i
i
i
w a
x
x
HYDROSTATIC F AND P Example 1
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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
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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
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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
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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.
HYDROSTATIC F AND P Example 2
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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
HYDROSTATIC F AND P Example 2
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As in Example 1, we divide the circular region into horizontal strips of equal width.
HYDROSTATIC F AND P Example 2
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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
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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
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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
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The second integral is 0 because the integrand is an odd function.
See Theorem 7 in Section 5.5
HYDROSTATIC F AND P Example 2
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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.
HYDROSTATIC F AND P Example 2
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Thus,HYDROSTATIC F AND P
3 2
3
212
875 9
875 (3)7875
212,370lb
F y dy
Example 2
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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
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This point is called the center of mass (or center of gravity) of the plate.
CENTERS OF MASS
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We first consider the simpler situation illustrated here.
CENTERS OF MASS
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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
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The rod will balance if:CENTERS OF MASS
1 1 2 2m d m d
Equation 2
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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
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Now, suppose that the rod lies along
the x-axis, with m1 at x1 and m2 at x2
and the center of mass at .
MOMENTS AND CENTERS OF MASS
x
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Comparing the figures, we see that:
d1 = – x1 d2 = x1 –
x
MOMENTS AND CENTERS OF MASS
x
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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
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The numbers m1x1 and m2x2 are called
the moments of the masses m1 and m2
(with respect to the origin).
MOMENTS OF MASS
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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
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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
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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
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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
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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
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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.
MOMENTS AND CENTERS OF MASS
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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
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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
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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
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Since , the center of mass is the point where a single particle of mass m would have the same moments as the system.
MOMENTS AND CENTERS OF MASS
and y xmx M my M ( , )x y
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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
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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
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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
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Thus, the center of mass is:MOMENTS & CENTERS OF MASS
1415(1 ,1)
Example 3
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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
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In doing so, we use the following physical principles.
CENTROIDS
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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
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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
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Also, the moment of the union of two non-overlapping regions should be the sum of the moments of the individual regions.
CENTROIDS
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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
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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
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This determines the polygonal approximation to R shown below.
CENTROIDS
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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
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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
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Therefore,CENTROIDS
( ) ( )
( )y i i i
i i
M R f x x x
x f x x
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Adding these moments, we obtain the moment of the polygonal approximation to R .
CENTROIDS
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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
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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
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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
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Just as for systems of particles, the center of mass of the plate is defined so that
CENTROIDS
andy xmx M my M
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However, the mass of the plate is the product of its density and its area:
CENTROIDS
( )b
a
m A
f x dx
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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
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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
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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
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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
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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
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The center of mass is located at the point (0, 4r/(3π)).
CENTERS OF MASS Example 4
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Find the centroid of the region bounded by the curves
y = cos x, y = 0, x = 0, x = π/2
CENTERS OF MASS Example 5
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The area of the region is:CENTERS OF MASS
2 2
00cos sin
1
A xdx x
Example 5
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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:
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The centroid is ((π/2) – 1, π/8).CENTROIDS Example 5
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Suppose the region R lies between two curves y = f(x) and y = g(x), where f(x) ≥ g(x).
CENTROIDS
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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
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Find the centroid of the region bounded by the line x = y and the parabola y = x2.
CENTROIDS Example 6
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The region is sketched here.
We take f(x) = x, g(x) = x2, a = 0, and b = 1 in Formulas 9.
CENTROIDS Example 6
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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
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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
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The centroid is:CENTROIDS
1 2,2 5
Example 6
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We end this section by showing a surprising connection between centroids and volumes of revolution.
CENTROIDS
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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
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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
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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
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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
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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
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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
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Compare the method of Example 7 with that of Exercise 63 in Section 6.2
THEOREM OF PAPPUS