Hint: Slice the area perpendicular to the y-axis so that the height of each slice is y.. More Applications of Integration 28.1 COMPUTING VOLUMES Volumes by Slicing We compute the signed
Trang 127.2 Slicing to Find the Area Between Two Curves 851
x
y
y = 1
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 = –x2 + 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 x dx+3
π/4
π
4 dx
(b) 1
0 arctan x dx+3
1
π
4 dx
(c)1
0 tan y dy
(d) π/4
0 (3− tan y) dy
(e)π/4
0 3 dy−π/4
0 tan y dy
(f )3
0 arctan x dx−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.
Trang 2852 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 Evaluate1
0 arctan x dx by interpreting it as an area and slicing horizontally.
14 Evaluate0.5
0 arcsin x dx.
15 The region A in the first quadrant is bounded by y = 2x, y = −3x + 10, and
9(x2− 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
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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 + ix, for i = 0, 1, , n.
x = a x = b
f(x)
Area ≈ ∑ f(x i) ∆x
n i=1
Where we approximate
the height of the ith 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
i th slice, Vi , can be approximated by multiplying the area of one of the faces, A(xi ), by the
thickness of the slice, x Suppose we have n slices each of thickness x.
853
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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 xi = a + ix, so a = x0< x1 < x2 < · · · < x n = b We’ll refer to this as a standard partition of [a, b] Let A(xi )denote the cross-sectional area cut by a plane perpendicular to
the x-axis at xi Then
volume= lim
n→∞
n
i=1
b a A(x) dx
Naturally, this approach can be generalized from a loaf of bread to other solids
height was about 481 feet and its cross sections are nearly perfect squares The base is a square whose sides measure about 756 feet.1Find its volume
756 ft
Pyramid of Cheops
756 481
y
x
y
x
s i
x i
Figure 28.3
along the height of the pyramid and pierces the square base at its center Start chopping at
1Facts from David Burton, The History of Mathematics: An Introduction, McGraw-Hill Companies, Inc., 1997, p 56.
Trang 528.1 Computing Volumes 855
x = 0 and stop at x = 481, slicing the interval [0, 481] into n equal subintervals of length
x Let xi = ix.
volume=
n
i=1
V i≈
n
i=1
A(x i )x
Our job is to find A(xi ) Let si be the side of the square cross section at xi Then A(xi ) = (s i )2
s i is not constant; it varies with x We use similar triangles to express si in terms of xi
s i
x i =756 481
s i=756
481x i so A(xi )≈
756
481x i
2
volume= lim
n→∞
n
i=1
A(x i )x=
481
0
A(x) dx
volume=
481 0
756
481x
2
dx
=
756 481
2 481
0
x2dx
=
756 481
2
x3
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?
(10, 3)
y
x
r i
x i
Figure 28.4
0 A(x) dx , where A(xi )is the
cross-sectional area of a slice at xi produced by a standard partition of [0, 10] into n equal pieces A(xi )=1π r2because the cross sections are semicircles We need to express riin
Trang 6856 CHAPTER 28 More Applications of Integration
terms of xi, so we use similar triangles
r i
10
r i= 3
10x i
A(x i )=1
2π
3
10x i
2
= 9π
200x
2
i
Therefore the volume is
10
0
9π
200x
2dx= 9π 200
x3
3
10
0
=3000π
200 = 15π.
The volume is 15π ft3, or approximately 47.12 cubic feet
REMARK Instead of taking slices we could have noticed that this tent is half of a right circular cone with height 10 and radius 3 Volume=1
2
1
3π(32)· 10=1
230π = 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
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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 = b
x = a
y = f(x)
x = b
x = a
R
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 (xi ) or f (xi ) − g(x i ), as indicated,
and the base is denoted by x.
y = g(x)
y = g(x) x
r
r
R
∆x
∆x
or
Disk or coin
x
h
y
∆x
∆x
}
∆x
}
Volume
of disk
≈ πr2 ∆x
≈ π[ f(x i)] 2 ∆x
(a)
Volume
of annulus
≈ [ πR2 – πr2 ] ∆x
≈ π[[ f(x i)] 2 – [g(x i) 2 ]] ∆x
(b)
Volume of cylindrical shell
≈ 2 πr h ∆x
≈ 2 πx i [ f(x i ) – g(x i)] ∆x
(c)
Annulus or bagel chip
or washer
Hollow cylinder or cylindrical shell
Figure 28.7
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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 (xi ) and the little radius, r, corresponds
to g(xi )
f(x)
g(x)
Figure 28.8
CAUTION π(R2− r2) = π(R − r)2
In case (c), r corresponds to the distance from the y-axis and is therefore given by xi;
h corresponds to the height of the rectangle and is given by f (xi ) − g(x i ) To see why
the volume of the cylindrical shell is approximately 2π rhx, 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
2 πr
Figure 28.9
◆ EXAMPLE 28.3 Show that the volume of a sphere of radius R is43π R3
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 ≈ πr2
i x
≈ π R2− x2
i
2
x
≈ πR2− x2
i
x
y
R
R
x2 + y2 = R2
y = √R2 – x2
r
disk
Figure 28.10
Trang 928.1 Computing Volumes 859
volume= 2 lim
n→∞
n
i=1
π(R2− x2
i )x
volume= 2
R
0
π(R2− x2) dx
= 2π
R2x−x3 3
R
0
= 2π
R3−R3 3
= 2π · 2R3
3
=4π R3 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 = x2, 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?
a standard partition Revolving a representative slice around the y-axis gives a cylindrical
shell.2
y
x
y = x2
∆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.
Trang 10860 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 = x2
i volume of ith cylindrical shell ≈ 2πx i x2i x = 2πx3
i x
So the volume=2
0 2π x3dx = 2π x4
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 = x2
∆y
∆y
{
{
(2, 4)
2
R r
Figure 28.12
volume of ith annulus = π[R2
i − r2
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, we want x in terms of y.
The curve is x = √y, so x i = √y i
r i=√y i volume of ith annulus ≈ π22− (√y i )2
y = π[4 − y i]y
volume=
4
0
π[4− y] dy = π
4y−y2
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
... 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.”... 2: Slice A along the y-axis as shown Partition [0, 4] into n equal pieces using astandard partition Revolving a representative slice gives an annulus
y... “annulus.” In other words, in our language
we will treat pseudorectangles as rectangles.
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