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Chapter Six
Applications of Integration
Section 6.1
More about Areas
Goals
◦ Use integrals to find areas of more general regions,
including
regions that lie between the graphs of two functions
regions enclosed by parametric curves
Area Between Curves
Let S be the region that lies…
◦ between two curves y = f(x) and y = g(x), and
◦ between the vertical lines x = a and x = b.
Here we assume that f and g are continuous
functions and that f(x) ≥ g(x) for all x in [a, b].
We want the area of S, as illustrated on the
next slide:
Areas (cont’d)
Riemann Sum
As earlier with areas under curves, we
◦ divide S into n strips of equal width, and
◦ approximate the ith strip by a rectangle of base ∆x
and height f(xi*) – g(xi*).
As the next slide shows, the Riemann sum
is an approximation to the area of S:
* *
1
n
i ii
f x g x x
Riemann Sum (cont’d)
Definition
Thus we define the area A of S to be the
limiting value of the sum of the areas of
these approximating rectangles:
We recognize this limit as the definite
integral of f – g :
* *
1
limn
i in
i
A f x g x x
Area Formula
In the special case where g(x) = 0,
◦ S is the region under the graph of f , and so
◦ our new formula reduces to our old one.
Example
Find the area of the region enclosed by the
parabolas y = x2 and y = 2x – x2.
Solution First, using algebra we find that the
points of intersection of these two curves are
(0, 0) and (1, 1).
From the figure on the next slide we identify
the top and bottom curves
yT = 2x – x2 and yB = x2 :
Solution (cont’d)
Solution (cont’d)
The area of a typical rectangle is
and the region lies between x = 0 and x = 1.
So the total area is
2 2 22 2 2T By y x x x x x x x x
Changing the Point of View
Some regions are best treated by
regarding x as a function of y.
The next slide shows a region bounded
by the curves x = f(y), x = g(y), y = c, and y
= d, where
◦ f and g are continuous and
◦ f(y) ≥ g(y) for c ≤ y ≤ d :
Point of View (cont’d)
Point of View (cont’d)
The area of the figure is given by
If we write xR for the right boundary and xL
for the left, then as the next slide shows,
Point of View (cont’d)
Example
Find the area enclosed by the line y = x –
1 and the parabola y2 = 2x + 6.
Solution By algebra we find that the
points of intersection are (–1, –2) and (5,
4).
We solve the equation of the parabola for
x ; as the next slide illustrates,
xL = (½)y2 – 3 and xR = y – 1
Solution (cont’d)
Solution (cont’d)
Thus
Parametric Curves
The area under a curve y = F(x) from a to b is
If a curve is traced out once by the parametric
equations x = f(t) and y = g(t),
α ≤ t ≤ β, then we can calculate an area
formula using the Substitution Rule:
, where 0.b
aA F x dx F x
Example
Find the area under one arch of the cycloid
x = r(θ – sinθ) , y = r(1 – cosθ).
Solution One arch of the cycloid (shown on
the next slide) is given by 0 ≤ θ ≤ 2π.
On the subsequent slide we apply the
Substitution Rule with
◦ y = r(1 – cosθ) and
◦ dx = r(1 – cosθ)dθ :
Solution (cont’d)
Solution (cont’d)
Review
Use of Riemann sums to define areas
between curves, both regarding…
◦ y as a function of x and
◦ x as a function of y
Areas enclosed by parametric curves
Section 6.2
Volumes
Goals
◦ Define volume of a solid
◦ Use integrals to find volumes of solids of known
cross-section
◦ Find volumes of solids of revolution, using the
methods of
disks/washers
cylindrical shells
What is Volume?
What do we mean by the volume of a
solid?
How do we know that the volume of a
sphere of radius r is 4πr3/3 ?
How can we give a precise definition of
volume?
Our Starting Point: Cylinders
A cylinder is bounded by a plane region B1
(the base) and a congruent region B2 in a
parallel plane
We define the volume
of the cylinder by V =
Ah , where A is the area
of the base and h is the
height
Familiar Cylinders
Circular cylinder
◦ Base is a circle of radius r , and
so has area πr2
◦ Thus V = πr2h
Rectangular box
◦ Base is a rectangle of area lw
◦ Thus V = lwh
The base of a cylinder need
not be circular!
What About Other Solids?
What if a solid S is not a cylinder?
We can use our knowledge of cylinders:
◦ Cut S into thin parallel slices
◦ Treat each piece as though it were a cylinder,
and add the volumes of the pieces
◦ The thinner we make the slices, the closer we
will be to the actual volume of S
This leads to the method of finding
volumes by cross-sections
The Method of Cross-Sections
Intersect S with a
plane Px
perpendicular to
the x-axis
Call the cross-
sectional area A(x)
A(x) will vary as x increases from a to b
Cross-Sections (cont’d)
Divide S into “slabs” of equal width ∆x
using planes at x1, x2,…, xn
◦ Like slicing a loaf of bread!
Choose a sample point in each *ix 1 ,i ix x
Cross-Sections (cont’d)
The ith slab (or, slice of bread) is roughly a
cylinder with volume
◦ Here is the base of the “cylinder”
and is its height
Adding the volumes of the individual slabs
gives an approximation to the volume of S:
*
1
n
ii
V A x x
*iA x x
*iA x
x
Precise Definition of Volume
Now we use more and more slabs
◦ This corresponds to letting n ∞
The larger the value of n, the closer each slab
becomes to an actual cylinder
◦ In other words, our approximation becomes better and better as n ∞
So we define the volume of S by
*
1
limn b
i ani
V A x x A x dx
Example: Volume of a Sphere
Here
So
(as the usual formula predicts!)
2 2 2A x y r x
2 2 34
3
r
rV r x dx r
A Special Case
A solid of revolution is formed by rotating a
plane region about an axis
About Solids of Revolution
The slabs are always in one of two shapes:
◦ Either the shape of a disk…
…as was the case in the sphere problem above
◦ or the shape of a washer
A washer is the region between two concentric circles
Let’s look at two examples of each type
◦ First, an example of disks, using the solid in the
preceding slide
First Example of Disks
Problem Find the volume of the solid
obtained by rotating about the x-axis the
region under the curve from 0
to 1.
◦ See the diagram two slides back
Solution Each cross-section is a disk, so
Here the radius is , so
y x
2radius of cross-sectionA x
x
2
A x x x
First Example of Disks (cont’d)
So the volume of the solid of revolution is
In general: Whenever we use the disk
method, we will start with the formula
121
002 2
xV xdx
2radius of cross-sectionA x
Second Example of Disks
Problem Find the volume of the solid obtained by rotating the region bounded by y = x3 , y = 8 , and x = 0 about the y-axis
Here the slices are perpendicular to the y-axis, rather than the x-axis
Second Example of Disks (cont’d)
Solution The cross –
section is given by
So the volume is
2 2 / 33A y y y
8 8 2 / 3
0 0
85 / 3
0
3 96
5 5
V A y dy y dy
y
First Example of Washers
Problem Find the volume of the solid
obtained by rotating about the x-axis the
region enclosed by the curves y = x and
y = x2
On the next slide are shown the region and
the solid
First Example of Washers (cont’d)
First Example of Washers (cont’d)
Solution To find A(x) we
subtract the area of the
inner disk from that of the
outer disk:
22 2
2 4
A x x x
x x
First Example of Washers (cont’d)
Therefore
In general: Whenever we use the washer
method, we will start with the formula
13 51 8 2 4
0 00
2
3 5 15
x xV A x dx x x dx
2 2outer radius inner radiusA
A Bigger Picture
The formula
can be applied to any solid for which the
cross-sectional area A(x) can be found
This includes solids of revolution, as
shown above…
…but includes many other solids as well
b
aV A x dx
Example Problem A solid has a circular base of radius
1 . Parallel cross-sections perpendicular to the
base are equilateral triangles. Find the volume
of the solid.
Example (cont’d)
Solution Here the cross-sections are
equilateral triangles with
2 2
2
12 3
21
2 1 3 12
3 1
A x y y
x x
x
Example (cont’d)
So
At right is shown
the completed solid.
1
14 3 / 3V A x dx
Review
Use of Riemann sums to define volume
◦ Cylinders
Volumes of solids with known cross-
section
Solids of revolution
◦ Method of disks/washers
6.3 Volume by Cylindrical Shells
Sometimes the disk/washer method is difficult for a solid of revolution
An alternative is to divide the solid into concentric circular cylinders
This leads to the method of cylindrical shells
Volumes by Cylindrical Shells
Let’s consider the problem of finding the volume of the
solid obtained by rotating about the y-axis the region bo
unded by y = 2x2 – x3 and y = 0. (See Figure 1.)
If we slice perpendicular to the y-axis, we get a washer.
Figure 1
Volumes by Cylindrical Shells
But to compute the inner radius and the outer radius of the washer
, we’d have to solve the cubic equation y = 2x2 – x3 for x in terms of
y; that’s not easy.
Fortunately, there is a method,
called the method of cylindrical
shells, that is easier to use in
such a case. Figure 2 shows a
cylindrical shell with inner radius r1,
outer radius r2, and height h.
Figure 2
Volumes by Cylindrical Shells
Its volume V is calculated by subtracting the volume V1 of the inner
cylinder from the volume V2 of the outer cylinder:
V = V2 – V1
= r22h – r1
2h = (r22 – r1
2)h
= (r2 + r1) (r2 – r1)h
= 2 h(r2 – r1)
Volumes by Cylindrical Shells
If we let r = r2 – r1 (the thickness of the shell) and r = (r2 + r1)
(the average radius of the shell), then this formula for the volume of
a cylindrical shell becomes
and it can be remembered as
V = [circumference] [height] [thickness]
Volumes by Cylindrical Shells
Now let S be the solid obtained by rotating about the y-axis the region bounded by
y = f (x) [where f (x) 0], y = 0, x = a and x = b, where b > a 0. (See Figure 3.)
We divide the interval [a, b] into n subintervals [xi – 1, xi] of equal
width x and let be the midpoint of the i th subinterval.
Figure 3
Volumes by Cylindrical Shells
If the rectangle with base [x i – 1, xi] and height is rotated
about the y-axis, then the result is a cylindrical shell with average
radius , height , and thickness x (see Figure 4), so by
Formula 1 its volume is
Figure 4
Volumes by Cylindrical Shells
Therefore an approximation to the volume V of S is give
n by the sum of the volumes of these shells:
This approximation appears to become better as n
. But, from the definition of an integral, we know that
Volumes by Cylindrical Shells
Thus the following appears plausible:
The best way to remember Formula 2 is to think of a typical shell, c
ut and flattened as in Figure 5, with radius x, circumference 2x, hei
ght f (x), and thickness x or dx:
This type of reasoning will be helpful in other situations, such as wh
en we rotate about lines other than the y-axis.
Volumes by Cylindrical Shells
Figure 5
Example 1 – Using the Shell Method
Find the volume of the solid obtained by rotating
about the y-axis the region bounded by y = 2x2 – x3 and
y = 0.
Solution:
From the sketch in Figure 6 we see that a typical shell h
as radius x, circumference 2x, and height f (x) = 2x2 – x3
Figure 6
Example 1 – Solution
So, by the shell method, the volume is
It can be verified that the shell method gives the same a
nswer as slicing
cont’d
Review
Solids of revolution
◦ Method of cylindrical shells
Section 6.4
Arc Length
Goals
◦ Define arc length of a smooth curve given
parametrically
◦ Give an integral formula for arc length
Introduction
What do we mean by the length of a
curve?
If the curve is a polygon, then the length is
simply the sum of the lengths of the line
segments that form the polygon.
Our approach to the length of a general
curve C is shown on the next slide:
Introduction (cont’d)
Introduction (cont’d)
We define the length by…
◦ first approximating it by a polygon, and then
◦ taking a limit as the number of segments of
the polygon is increased.
We will assume that C is described by the
parametric equations
x = f(t), y = g(t), a ≤ t ≤ b
Introduction (cont’d)
We also assume that C is smooth, that is, that the derivatives f (t) and g (t) are
◦ continuous, and
◦ not simultaneously zero, a < t < b.
This ensures that C has no sudden change
in direction.
Inscribed Polygon
We divide the parameter interval [a, b] into n
subintervals of equal width ∆t.
The preceding figure shows the points
P0, P1, …, Pn corresponding to the endpoints of
the above subintervals.
The polygon with these vertices approximates
C, and so the length of this polygon approaches that of C as n ∞
Inscribed Polygon (cont’d)
Thus we define the length of C to be the
limit of the lengths of these inscribed
polygons:
Now we work toward a more convenient
expression for L :
11
limn
i in
i
L P P
Riemann Sum
If we let ∆xi = xi – xi-1 and ∆yi = yi – yi-1,
then the length of the ith line segment of
the polygon is
But from the definition of a derivative we know that f (ti) ≈ ∆xi/∆t if ∆t is small.
Therefore ∆xi ≈ f (ti)∆t and ∆yi ≈ g (ti)∆t.
22
1i i i iP P x y
Riemann Sum (cont’d)
Therefore
so that
Riemann Sum (cont’d)
This is a Riemann sum for the function
suggests that
With the restriction that no portion of the
curve is traced out more than once, this is the
correct arc length formula:
2 2
, and so our argumentf t g t
2 2b
aL f t g t dt
Arc Length Formula
As an example we find the length of the arc of
the curve x = t2, y = t3 that lies between the
points (1, 1) and (4, 8), shown on the next
slide:
Example (cont’d)
Solution
We note that the given portion of the
curve corresponds to the parameter
interval 1 ≤ t ≤ 2, so the formula gives
Solution (cont’d)
Substituting u = 4 + 9t2 leads to
This number is just slightly larger than the
distance joining the endpoints (1, 1) and (4,
8)—as the figure would suggest.
A Special Case
If we are given a curve y = f(x), a ≤ t ≤ b,
then we can regard x as a parameter.
With the parametric equations x = x,
y = f(x), our formula becomes
Example
Find the length of one arch of the cycloid
x = r(θ – sinθ) , y = r(1 – cosθ).
Solution We know from earlier work that
one arch is described by the parameter
interval 0 ≤ θ ≤ 2π. Since
Solution (cont’d)
we have
A computer algebra system gives the value of
8r for this integral.
◦ Thus the length of one arch of a cycloid is eight
times the radius of the generating circle.
Review
Use of Riemann sums to define arc length
using polygonal approximation
Integral formula for arc length