Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 88 ppsx

27.2 Slicing to Find the Area Between Two Curves 851
x
y
y = 1
y = tan x
π
2
8. Write an integral (or the sum or difference of integrals) giving the area of the region
shaded below. (You need not evaluate.)
x
y
y = –x
2
+ 2
y = x
9. Which of the expressions below give the area of the shaded region? (Select all such
expressions.)
y =
x
π
2
y =
π
4
3
f(x) = arctan x
(a)

π/4
0
arctan xdx+


3
π/4
π
4
dx
(b)

1
0
arctan xdx+

3
1
π
4
dx
(c)

1
0
tan ydy
(d)

π/4
0
(3 − tan y) dy
(e)

π/4

0
3 dy −

π/4
0
tan ydy
(f)

3
0
arctan xdx−

3
1
(arctan x −
π
4
)dx
10. Find the area between the curve y =ln x and the x-axis for 1 ≤ x ≤ 10. Get an exact
answer. (Hint: Slice the area perpendicular to the y-axis so that the height of each slice
is y. Use this to arrive at an integral that you can evaluate exactly.)
11. Find (exactly) the area bounded by x =1/e, y = ln x, and y = 1.
852 CHAPTER 27 Applying the Definite Integral: Slice and Conquer
12. Find the area bounded below by the x-axis, and laterally by y = ln x, and the line
segment joining (e,1) to (2e,0).
13. Evaluate

1
0
arctan xdxby interpreting it as an area and slicing horizontally.

14. Evaluate

0.5
0
arcsin xdx.
15. The region A in the first quadrant is bounded by y = 2x, y =−3x + 10, and
y =−
1
9
(x
2
− 6x). It has corners at (0, 0), (2, 4), and (3, 1). Express the area of A
as the sum or difference of definite integrals. You need not evaluate.
28
CHAPTER
More Applications
of Integration
28.1 COMPUTING VOLUMES
Volumes by Slicing
We compute the signed area of a region in the plane using a divide-and-conquer technique.
To find the area under the graph of f we slice the region into n thin slices, each of
width x =
b−a
n
and approximate the area of each slice by the area of a rectangle. Let
x
i
= a +ix, for i =0, 1, , n.
x= ax=b
f(x)

Area ≈ ∑ f(x
i
) ∆x
n
i=1
Where we approximate
the height of the i
th
slice
by f(x
i
).
Figure 28.1
Summing the areas of the slices and taking the limit as the number of slices increases without
bound gives us the area in question.
area = lim
n→∞
n

i=1
f(x
i
)x =

b
a
f(x)dx
We’ll take a similar approach to calculating volume. Suppose we want to find the
volume of a loaf of bread. It could be a plain shape, like a typical loaf of rye bread, or
it could be a more complicated shape, like a braided loaf of challah. Whatever the loaf

looks like, put the whole thing in a bread slicer and ask for thin slices. The volume of the
ith slice, V
i
, can be approximated by multiplying the area of one of the faces, A(x
i
), by the
thickness of the slice, x. Suppose we have n slices each of thickness x.
853
854 CHAPTER 28 More Applications of Integration
x= ax=b
Volume = ∑ V
i
≈ ∑ A(x
i
) ∆x
n
i=1
n
i=1
A(x
i
)
x
∆x
}
Figure 28.2
The thinner the slices, the less error is involved in approximating the volume of a slice by
the area of a face times the thickness.
Let’s attach coordinates to the problem. Set the loaf down along the x-axis and denote
the positions of the ends of the loaf by x = a and x = b as shown in Figure 28.2. Slice

perpendicular to the x-axis, partitioning [a, b] into n equal pieces, each of length x =
b−a
n
.
Let x
i
= a + ix,soa = x
0
<x
1
<x
2
< ···<x
n
= b.We’ll refer to this as a standard
partition of [a, b]. Let A(x
i
) denote the cross-sectional area cut by a plane perpendicular to
the x-axis at x
i
. Then
volume = lim
n→∞
n

i=1
A(x
i
)x =


b
a
A(x) dx.
Naturally, this approach can be generalized from a loaf of bread to other solids.
◆
EXAMPLE 28.1 The Egyptians built the Great Pyramid of Cheops in Giza around 2600 b.c. Its original
height was about 481 feet and its cross sections are nearly perfect squares. The base is a
square whose sides measure about 756 feet.
1
Find its volume.
756 ft
Pyramid of Cheops
(a) (b) (c)
756
481
y
x
y
x
s
i
x
i
Figure 28.3
SOLUTION In your mind, pick up this massive pyramid and set it along the x-axis so the x-axis runs
along the height of the pyramid and pierces the square base at its center. Start chopping at
1
Facts from David Burton, The History of Mathematics: An Introduction, McGraw-Hill Companies, Inc., 1997, p 56.
28.1 Computing Volumes 855
x = 0 and stop at x = 481, slicing the interval [0, 481] into n equal subintervals of length

x. Let x
i
= ix.
volume =
n

i=1
V
i
≈
n

i=1
A(x
i
)x
Our job is to find A(x
i
). Let s
i
be the side of the squarecross section at x
i
. Then A(x
i
) =(s
i
)
2
.
s

i
is not constant; it varies with x. We use similar triangles to express s
i
in terms of x
i
.
s
i
x
i
=
756
481
s
i
=
756
481
x
i
so A(x
i
) ≈

756
481
x
i

2

volume = lim
n→∞
n

i=1
A(x
i
)x =

481
0
A(x) dx
volume =

481
0

756
481
x

2
dx
=

756
481

2


481
0
x
2
dx
=

756
481

2
x
3
3




481
0
=

756
481

2
(481)
3
3
= (756)

2
·
481
3
≈ 91,636,272
The volume is approximately 91,636,272 cubic feet.
◆
◆
EXAMPLE 28.2 A tent has a base that is an isosceles triangle. The mouth of the tent is 6 feet wide and 3 feet
high; the length is 10 feet. The cross sections perpendicular to the base are all semicircles.
What is the volume of the tent?
6 ft
10
(10, 3)
y
x
r
i
x
i
Figure 28.4
SOLUTION Position the x-axis as shown in Figure 28.4. Volume =

10
0
A(x) dx, where A(x
i
) is the
cross-sectional area of a slice at x
i

produced by a standard partition of [0, 10] into n equal
pieces . A(x
i
) =
1
2
πr
2
i
because the cross sections are semicircles. We need to express r
i
in
856 CHAPTER 28 More Applications of Integration
terms of x
i
, so we use similar triangles.
r
i
x
i
=
3
10
r
i
=
3
10
x
i

A(x
i
) =
1
2
π

3
10
x
i

2
=
9π
200
x
2
i
Therefore the volume is

10
0
9π
200
x
2
dx =
9π
200

x
3
3




10
0
=
3000π
200
= 15π.
The volume is 15πft
3
, or approximately 47.12 cubic feet.
REMARK Instead of taking slices we could have noticed that this tent is half ofa right circular
cone with height 10 and radius 3. Volume =
1
2

1
3
π(3
2
) · 10

=
1
2

30π = 15π. ◆
Knowing the area of a circle, we can derive the formulas for the volume of a right
circular cone and the volume of a sphere using techniques similar to those employed in
Examples 28.1 and 28.2. Alternatively, we can treat these objects as volumes of revolution,
a viewpoint that sometimes amounts to cross-sectional slicing, as you will see.
Volumes of Revolution
Many familiar objects can be thought of as solids of revolution. A sphere, a cone, a bead,
an ellipsoid, and a bagel all can be constructed geometrically by revolving a region in the
plane about some axis of revolution. (See Figure 28.5.)
y
x
y
x
y
x
y
x
y
x
y
x
⇒⇒
⇒
(a) A sphere
(c) An egg stand
(b) A right circular cone
(d) A bagel
y
x
y

x
Figure 28.5 Volumes of Revolution
28.1 Computing Volumes 857
In the case of the sphere or the right circular cone positioned as pictured above, we can
think of slicing the object just like a loaf of bread, perpendicular to the x-axis . Each slice
will have a circular cross-sectional area. The same is true in the case of the volume of the
vase pictured in Figure 28.6.
y = f(x)
x = bx = a
y = f(x)
x = bx = a
yy
R
xx
Figure 28.6
An alternative, but equivalent, point of view is to consider the volume as generated by
revolving the region R bounded by y = f(x)from x =a to x = b about the x-axis . We can
chop up R into n rectangular-like strips as indicated in Figure 28.6, by rotating each strip
about the x-axis, and then summing the resultant volumes. We’ll see that this viewpoint,
chopping the planar region, revolving each “rectangle” to get a volume, and summing the
volumes, is quite versatile; it can be used in ways that are not equivalent to slicing bread.
Let’s consider the different situations that can arise when rotating a rectangular strip
around a vertical or horizontal axis. Three possibilities are given below. In Figure 28.7 the
height of the shaded “rectangle” is approximated by f(x
i
) or f(x
i
) − g(x
i
), as indicated,

and the base is denoted by x.
y = f(x)
y = f(x)
y = f(x)
y = g(x)
y = g(x)
x
r
r
r
r
R
∆x
∆x
or
Disk or coin
x
h
y
}
∆x
}
∆x
∆x
}
}
}
∆x
}
Volume

of disk
≈ πr
2
∆x
≈ π[ f(x
i
)]
2
∆x
(a)
Volume
of annulus
≈ [πR
2
– πr
2
] ∆x
≈ π[[f(x
i
)]
2
– [g(x
i
)
2
]] ∆x
(b)
Volume of
cylindrical shell
≈ 2πr h ∆x

≈ 2πx
i
[ f(x
i
) – g(x
i
)] ∆x
(c)
Annulus or bagel chip
or washer
Hollow cylinder or
cylindrical shell
Figure 28.7
858 CHAPTER 28 More Applications of Integration
Observations
Suppose a solid can be thought of as generated by revolving an area between the graph
of f(x)and the x-axis or between f(x)and g(x) around the x-axis. Cases (a) and (b)
in Figure 28.7 are equivalent to slicing perpendicular to the x-axis.
In case (b), the big radius, R, corresponds to f(x
i
) and the little radius, r, corresponds
to g(x
i
).
f(x)
g(x)
Figure 28.8
CAUTION π(R
2
− r

2
) = π(R −r)
2
In case (c), r corresponds to the distance from the y-axis and is therefore given by x
i
;
h corresponds to the height of the rectangle and is given by f(x
i
) − g(x
i
). To see why
the volume of the cylindrical shell is approximately 2πrhx, picture constructing the
shell from a sheet of paper. Then unroll the paper. We essentially want the volume of
the sheet. x corresponds to the paper’s thickness.
r
hh
2πr
Figure 28.9
◆
EXAMPLE 28.3 Show that the volume of a sphere of radius R is
4
3
πR
3
.
SOLUTION Rotate the region in the first quadrant bounded by y =
√
R
2
− x

2
and the coordinate axes
around the x-axis and double the result.
Chop along the x-axis. Partition [0, R] into n equal pieces, creating a standard partition.
Revolving a representative slice gives a “disk.”
volume of ith “disk” = V
i
≈ πr
2
i
x
≈ π


R
2
− x
2
i

2
x
≈ π

R
2
− x
2
i


x
y
R
R–R
x
x
2
+ y
2
= R
2
y = √R
2
– x
2
r
disk
Figure 28.10
28.1 Computing Volumes 859
volume =2 lim
n→∞
n

i=1
π(R
2
− x
2
i
)x

volume =2

R
0
π(R
2
− x
2
)dx
= 2π

R
2
x −
x
3
3

R
0
= 2π

R
3
−
R
3
3

= 2π ·

2R
3
3
=
4πR
3
3
This solution is like slicing a hemisphere as one would slice a bread loaf.
◆
◆
EXAMPLE 28.4 Model a hard-boiled egg holder as follows. Let A be the region in the first quadrant bounded
above by y = x
2
, below by the x-axis, and laterally by x = 2. Revolve A about the y-axis
to generate the egg holder. What is its volume?
SOLUTION Option 1: Slice A along the x-axis as shown. Partition [0, 2] into n equal pieces using
a standard partition. Revolving a representative slice around the y-axis gives a cylindrical
shell.
2
y
x
y = x
2
∆x
{
(2, 4)
2
Cylindrical shell
Figure 28.11
2

From this point on we will refer to the geometric object obtained by revolving a “rectangle” about a vertical or horizontal
line as a disk, cylindrical shell, or annulus as opposed to “disk,”“cylindrical shell,” and “annulus.” In other words, in our language
we will treat pseudorectangles as rectangles.
860 CHAPTER 28 More Applications of Integration
volume of ith cylindrical shell ≈ 2πr
i
h
i
x
r
i
= x
i
= distance from the y-axis
h
i
≈ y
i
= x
2
i
volume of ith cylindrical shell ≈ 2πx
i
x
2
i
x = 2πx
3
i
x

So the volume =

2
0
2πx
3
dx = 2π
x
4
4
|
2
0
=
π
2
· 16 = 8π.
Option 2: Slice A along the y-axis as shown. Partition [0, 4] into n equal pieces using a
standard partition. Revolving a representative slice gives an annulus.
y
x
y = x
2
∆y
∆y
{
{
(2, 4)
2
R

r
Figure 28.12
volume of ith annulus = π [R
2
i
− r
2
i
]y
R
i
= 2 (It is constant)
r
i
varies. It is given by the x-coordinate of the curve.
Because we have y,wewantx in terms of y.
The curve is x =
√
y,sox
i
=
√
y
i
.
r
i
=
√
y

i
volume of ith annulus ≈ π

2
2
− (
√
y
i
)
2

y = π[4 − y
i
]y
volume =

4
0
π[4 −y] dy = π

4y −
y
2
2






4
0
= π

16 −
16
2

= 8π
Notice that we generally have two options on how to slice a region. How we slice the region
is not predetermined by how we plan to rotate it.
◆
In Example 28.4, option (2) is equivalent to using a bread slicer along the y-axis. Option
(1), however, does not correspond to taking cross sectional slices. It is more along the lines
of coring an apple repeatedly, each time readjusting the size of the core.