Download Applications of the Definite Integral in Geometry, Science, and Engineering and more Exercises Mathematics in PDF only on Docsity!
Courtesy NASA
Calculus is essential for the computations required to land an astronaut on the moon.
In the last chapter we introduced the definite integral as the limit of Riemann sums in the context of finding areas. However, Riemann sums and definite integrals have applications that extend far beyond the area problem. In this chapter we will show how Riemann sums and definite integrals arise in such problems as finding the volume and surface area of a solid, finding the length of a plane curve, calculating the work done by a force, finding the center of gravity of a planar region, finding the pressure and force exerted by a fluid on a submerged object, and finding properties of suspended cables. Although these problems are diverse, the required calculations can all be approached by the same procedure that we used to find areas—breaking the required calculation into “small parts,” making an approximation for each part, adding the approximations from the parts to produce a Riemann sum that approximates the entire quantity to be calculated, and then taking the limit of the Riemann sums to produce an exact result.
APPLICATIONS OF THE
DEFINITE INTEGRAL IN
GEOMETRY, SCIENCE,
AND ENGINEERING
6.1 AREA BETWEEN TWO CURVES
In the last chapter we showed how to find the area between a curve y = f(x) and an
interval on the x -axis. Here we will show how to find the area between two curves.
A REVIEW OF RIEMANN SUMS
Before we consider the problem of finding the area between two curves it will be helpful to
review the basic principle that underlies the calculation of area as a definite integral. Recall
that if f is continuous and nonnegative on [a, b], then the definite integral for the area A
under y = f(x) over the interval [a, b] is obtained in four steps (Figure 6.1.1):
a b
x
y y^ =^ f^ ( x ) Δ xk
x*k
f ( x*k )
Figure 6.1.
- Divide the interval^ [a, b]^ into^ n^ subintervals, and use those subintervals to divide the
region under the curve y = f(x) into n strips.
- Assuming that the width of the^ kth strip is^ x^ k , approximate the area of that strip by
the area f(x∗ k )x k of a rectangle of width xk and height f(x∗ k ), where x∗ k is a point
in the kth subinterval.
- Add the approximate areas of the strips to approximate the entire area^ A^ by the
Riemann sum:
A ≈
∑^ n
k= 1
f(x∗ k )x k
414 Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering
- Take the limit of the Riemann sums as the number of subintervals increases and all
their widths approach zero. This causes the error in the approximations to approach
zero and produces the following definite integral for the exact area A:
A = lim
max xk → 0
∑^ n
k= 1
f(x∗ k )xk =
∫ b
a
f(x) dx
b
a
f ( x ) dx
f ( x^ * (^) k ) Δ x (^) k k = 1
n
Effect of the limit process on the Riemann sum Figure 6.1.
Figure 6.1.2 illustrates the effect that the limit process has on the various parts of the
Riemann sum:
- The quantity x∗ k in the Riemann sum becomes the variable x in the definite integral.
- The interval width xk in the Riemann sum becomes the dx in the definite integral.
- The interval [a, b], which is the union of the subintervals with widths x 1 , x 2 ,... ,
xn, does not appear explicitly in the Riemann sum but is represented by the upper
and lower limits of integration in the definite integral.
AREA BETWEEN y = f ( x ) AND y = g ( x )
We will now consider the following extension of the area problem.
6.1.1 first area problem Suppose that f and g are continuous functions on an
interval [a, b] and
f(x) ≥ g(x) for a ≤ x ≤ b
[This means that the curve y = f(x) lies above the curve y = g(x) and that the two can
touch but not cross.] Find the area A of the region bounded above by y = f(x), below
by y = g(x), and on the sides by the lines x = a and x = b (Figure 6.1.3 a ).
Figure 6.1.
a b
A x
y
x
y y = f ( x )
y = g ( x )
a b
y = f ( x )
y = g ( x )
Δ x (^) k
f ( x*k ) – g ( x*k )
( a ) ( b )
x*k
To solve this problem we divide the interval [a, b] into n subintervals, which has the
effect of subdividing the region into n strips (Figure 6.1.3 b ). If we assume that the width of
the kth strip is x k , then the area of the strip can be approximated by the area of a rectangle
of width x k and height f(x k∗ ) − g(x k∗ ), where x k∗ is a point in the kth subinterval. Adding
these approximations yields the following Riemann sum that approximates the area A:
A ≈
∑^ n
k= 1
[f(x k∗ ) − g(x k∗ )]x k
Taking the limit as n increases and the widths of all the subintervals approach zero yields
the following definite integral for the area A between the curves:
A = lim
max xk → 0
∑^ n
k= 1
[f(x∗ k ) − g(x k∗ )]x k =
∫ b
a
[f(x) − g(x)] dx
416 Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering
From (1) with f(x) = x + 6 , g(x) = x^2 , a = − 2 , and b = 3 , we obtain the area
A =
− 2
[(x + 6 ) − x^2 ] dx =
[
x^2
+ 6 x −
x^3
] 3
− 2
In the case where f and g are nonnegative on the interval [a, b], the formula
A =
∫ b
a
[f(x) − g(x)] dx =
∫ b
a
f(x) dx −
∫ b
a
g(x) dx
states that the area A between the curves can be obtained by subtracting the area under
y = g(x) from the area under y = f(x) (Figure 6.1.7).
a b
x
y (^) y = f ( x )
y = g ( x ) a b
x
y (^) y = f ( x )
y = g ( x ) a b
A
x
y (^) y = f ( x )
y = g ( x )
= −
Area between f and g Area below f Area below g
Figure 6.1.
Example 3 Figure 6.1.8 shows velocity versus time curves for two race cars that move
T
t
v v = v 2 ( t )
v = v 1 ( t )
Car 2
Car 1
0
A
Figure 6.1.
along a straight track, starting from rest at the same time. Give a physical interpretation of
the area A between the curves over the interval 0 ≤ t ≤ T.
Solution. From (1)
A =
∫ T
0
[v 2 (t) − v 1 (t)] dt =
∫ T
0
v 2 (t) dt −
∫ T
0
v 1 (t) dt
Since v 1 and v 2 are nonnegative functions on [ 0 , T ], it follows from Formula (4) of Section
5.7 that the integral of v 1 over [ 0 , T ] is the distance traveled by car 1 during the time interval
0 ≤ t ≤ T , and the integral of v 2 over [ 0 , T ] is the distance traveled by car 2 during the
same time interval. Since v 1 (t) ≤ v 2 (t) on [ 0 , T ], car 2 travels farther than car 1 does over
the time interval 0 ≤ t ≤ T , and the area A represents the distance by which car 2 is ahead
of car 1 at time T.
Some regions may require careful thought to determine the integrand and limits of
integration in (1). Here is a systematic procedure that you can follow to set up this formula.
It is not necessary to make an extremely accurate sketch in Step 1; the only pur- pose of the sketch is to determine which curve is the upper boundary and which is the lower boundary.
Finding the Limits of Integration for the Area Between Two Curves
Step 1. Sketch the region and then draw a vertical line segment through the region at
an arbitrary point x on the x-axis, connecting the top and bottom boundaries
(Figure 6.1.9 a ).
Step 2. The y-coordinate of the top endpoint of the line segment sketched in Step 1
will be f(x), the bottom one g(x), and the length of the line segment will be
f(x) − g(x). This is the integrand in (1).
Step 3. To determine the limits of integration, imagine moving the line segment left and
then right. The leftmost position at which the line segment intersects the region
is x = a and the rightmost is x = b (Figures 6.1.9 b and 6.1.9 c ).
6.1 Area Between Two Curves 417
b
x
y
a
x
y
a x b
x
y f ( x )
g ( x )
( a ) ( b ) ( c )
f ( x ) − g ( x )
Figure 6.1.
There is a useful way of thinking about this procedure:
If you view the vertical line segment as the “cross section” of the region at the point x ,
then Formula (1) states that the area between the curves is obtained by integrating the
length of the cross section over the interval [a, b].
It is possible for the upper or lower boundary of a region to consist of two or more
different curves, in which case it will be convenient to subdivide the region into smaller
pieces in order to apply Formula (1). This is illustrated in the next example.
Example 4 Find the area of the region enclosed by x = y^2 and y = x − 2.
Solution. To determine the appropriate boundaries of the region, we need to know where
the curves x = y^2 and y = x − 2 intersect. In Example 2 we found intersections by equating
the expressions for y. Here it is easier to rewrite the latter equation as x = y + 2 and equate
the expressions for x, namely,
x = y^2 and x = y + 2 (3)
This yields
y^2 = y + 2 or y^2 − y − 2 = 0 or (y + 1 )(y − 2 ) = 0
from which we obtain y = − 1 , y = 2. Substituting these values in either equation in (3)
we see that the corresponding x-values are x = 1 and x = 4 , respectively, so the points of
intersection are ( 1 , − 1 ) and ( 4 , 2 ) (Figure 6.1.10 a ).
4 − 1
2
x
y
A
(4, 2)
(1, −1)
( a )
x = y^2 y = x − 2 ( x = y + 2)
4 − 1
2
x
y
A 2
(4, 2)
(1, −1)
A 1
( b )
y = x − 2
y = (^) √ x
y = −√ x
Figure 6.1.
To apply Formula (1), the equations of the boundaries must be written so that y is
expressed explicitly as a function of x. The upper boundary can be written as y =
x
(rewrite x = y^2 as y = ±
x and choose the + for the upper portion of the curve). The
lower boundary consists of two parts:
y = −
x for 0 ≤ x ≤ 1 and y = x − 2 for 1 ≤ x ≤ 4
(Figure 6.1.10 b ). Because of this change in the formula for the lower boundary, it is
necessary to divide the region into two parts and find the area of each part separately.
From (1) with f(x) =
x, g(x) = −
x, a = 0 , and b = 1 , we obtain
A 1 =
0
[
x − (−
x )] dx = 2
0
x dx = 2
[
x^3 /^2
] 1
0
From (1) with f(x) =
x, g(x) = x − 2 , a = 1 , and b = 4 , we obtain
A 2 =
1
[
x − (x − 2 )] dx =
1
x − x + 2 ) dx
[
x^3 /^2 −
x^2 + 2 x
] 4
1
6.1 Area Between Two Curves 419
✔QUICK CHECK EXERCISES 6.1 ( See page 421 for answers. )
1. An integral expression for the area of the region between the curves y = 20 − 3 x^2 and y = e x^ and bounded on the sides by x = 0 and x = 2 is. 2. An integral expression for the area of the parallelogram bounded by y = 2 x + 8 , y = 2 x − 3 , x = −1, and x = 5 is. The value of this integral is. 3. (a) The points of intersection for the circle x^2 + y^2 = 4 and the line y = x + 2 are and.
(b) Expressed as a definite integral with respect to x, gives the area of the region inside the circle x^2 + y^2 = 4 and above the line y = x + 2. (c) Expressed as a definite integral with respect to y, gives the area of the region described in part (b).
4. The area of the region enclosed by the curves y = x^2 and y = 3
x is.
EXERCISE SET 6.1 Graphing Utility C^ CAS
1–4 Find the area of the shaded region. ■
1.
y = x
y = x^2 + 1
− 1 2
5
x
y (^) 2. y = (^) √ x
y = − 14 x
4
3
x
y
x = 1 / y^2
x = y
2
2
x
y (^) 4.
x = 2 − y^2
x = − y
− 2 2
2 x
y
5–6 Find the area of the shaded region by (a) integrating with respect to x and (b) integrating with respect to y. ■
5.
2
4
y = x^2
y = 2 x
x
y
(2, 4)
5
5
x
y
y = 2 x − 4
y^2 = 4 x (4, 4)
(1, −2)
7–18 Sketch the region enclosed by the curves and find its area. ■
7. y = x^2 , y =
x, x = 14 , x = 1
8. y = x^3 − 4 x, y = 0 , x = 0 , x = 2 9. y = cos 2x, y = 0 , x = π/ 4 , x = π/ 2 10. y = sec^2 x, y = 2 , x = −π/ 4 , x = π/ 4 11. x = sin y, x = 0 , y = π/ 4 , y = 3 π/ 4 12. x^2 = y, x = y − 2 13. y = e x^ , y = e^2 x^ , x = 0 , x = ln 2 14. x = 1 /y, x = 0 , y = 1 , y = e 15. y =
1 + x^2
, y = |x| 16. y =
1 − x^2
, y = 2
17. y = 2 + |x − 1 |, y = − 15 x + 7 18. y = x, y = 4 x, y = −x + 2
19–26 Use a graphing utility, where helpful, to find the area of the region enclosed by the curves. ■
19. y = x^3 − 4 x^2 + 3 x, y = 0 20. y = x^3 − 2 x^2 , y = 2 x^2 − 3 x 21. y = sin x, y = cos x, x = 0 , x = 2 π 22. y = x^3 − 4 x, y = 0 23. x = y^3 − y, x = 0 24. x = y^3 − 4 y^2 + 3 y, x = y^2 − y 25. y = xe x
2 , y = 2 |x|
26. y =
x
1 − (ln x)^2
, y =
x
27–30 True–False Determine whether the statement is true or false. Explain your answer. [In each exercise, assume that f and g are distinct continuous functions on [a, b] and that A de- notes the area of the region bounded by the graphs of y = f(x), y = g(x), x = a, and x = b.] ■
27. If f and g differ by a positive constant c, then A = c(b − a). 28. If ∫^ b
a
[f(x) − g(x)] dx = − 3
then A = 3.
29. If ∫^ b
a
[f(x) − g(x)] dx = 0
then the graphs of y = f(x) and y = g(x) cross at least once on [a, b].
30. If A =
∫ (^) b
a
[f(x) − g(x)] dx
420 Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering
then the graphs of y = f(x) and y = g(x) don’t cross on [a, b].
31. Estimate the value of k ( 0 < k < 1 ) so that the region en- closed by y = 1 /
1 − x^2 , y = x, x = 0 , and x = k has an area of 1 square unit.
32. Estimate the area of the region in the first quadrant enclosed by y = sin 2x and y = sin−^1 x.
C (^) 33. Use a CAS to find the area enclosed by y = 3 − 2 x and
y = x^6 + 2 x^5 − 3 x^4 + x^2.
C (^) 34. Use a CAS to find the exact area enclosed by the curves
y = x^5 − 2 x^3 − 3 x and y = x^3.
35. Find a horizontal line y = k that divides the area between y = x^2 and y = 9 into two equal parts. 36. Find a vertical line x = k that divides the area enclosed by x =
y, x = 2 , and y = 0 into two equal parts.
37. (a) Find the area of the region enclosed by the parabola y = 2 x − x^2 and the x-axis. (b) Find the value of m so that the line y = mx divides the region in part (a) into two regions of equal area. 38. Find the area between the curve y = sin x and the line seg- ment joining the points ( 0 , 0 ) and ( 5 π/ 6 , 1 / 2 ) on the curve.
39–43 Use Newton’s Method (Section 4.7), where needed, to approximate the x-coordinates of the intersections of the curves to at least four decimal places, and then use those approximations to approximate the area of the region. ■
39. The region that lies below the curve y = sin x and above the line y = 0. 2 x, where x ≥ 0. 40. The region enclosed by the graphs of y = x^2 and y = cos x. 41. The region enclosed by the graphs of y = (ln x)/x and y = x − 2. 42. The region enclosed by the graphs of y = 3 − 2 cos x and y = 2 /( 1 + x^2 ). 43. The region enclosed by the graphs of y = x^2 − 1 and y = 2 sin x.
C (^) 44. Referring to the accompanying figure, use a CAS to esti- mate the value of k so that the areas of the shaded regions are equal. Source: This exercise is based on Problem A1 that was posed in the Fifty-Fourth Annual William Lowell Putnam Mathematical Competition.
c
1
y = sin x y = k
x
y
Figure Ex-
F O C U S O N C O N C E P TS
45. Two racers in adjacent lanes move with velocity func- tions v 1 (t) m/s and v 2 (t) m/s, respectively. Suppose that the racers are even at time t = 60 s. Interpret the
value of the integral ∫ (^60)
0
[v 2 (t) − v 1 (t)] dt
in this context.
46. The accompanying figure shows acceleration versus time curves for two cars that move along a straight track, accelerating from rest at the starting line. What does the area A between the curves over the interval 0 ≤ t ≤ T represent? Justify your answer.
t
a a = a 2 ( t )
a = a 1 ( t )
Car 2
Car 1
T (^) Figure Ex-
47. Suppose that f and g are integrable on [a, b], but neither f(x) ≥ g(x) nor g(x) ≥ f(x) holds for all x in [a, b] [i.e., the curves y = f(x) and y = g(x) are intertwined]. (a) What is the geometric significance of the integral ∫ (^) b
a
[f(x) − g(x)] dx?
(b) What is the geometric significance of the integral ∫ (^) b
a
|f(x) − g(x)| dx?
48. Let A(n) be the area in the first quadrant enclosed by the curves y = n
x and y = x. (a) By considering how the graph of y = n
x changes as n increases, make a conjecture about the limit of A(n) as n → +. (b) Confirm your conjecture by calculating the limit.
49. Find the area of the region enclosed between the curve x^1 /^2 + y^1 /^2 = a^1 /^2 and the coordinate axes. 50. Show that the area of the ellipse in the accompanying figure is πab. [ Hint: Use a formula from geometry.]
x
y y^2 b^2
x^2 a^2
a
Figure Ex-
51. Writing Suppose that f and g are continuous on [a, b] but that the graphs of y = f(x) and y = g(x) cross sev- eral times. Describe a step-by-step procedure for determin- ing the area bounded by the graphs of y = f(x), y = g(x), x = a, and x = b.
422 Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering
This is a special case of a more general volume formula that applies to solids called right
cylinders. A right cylinder is a solid that is generated when a plane region is translated
along a line or axis that is perpendicular to the region (Figure 6.2.3).
Translated square
Some Right Cylinders
Translated disk Translated annulus Translated triangle
Figure 6.2.
If a right cylinder is generated by translating a region of area A through a distance h,
then h is called the height (or sometimes the width ) of the cylinder, and the volume V of
the cylinder is defined to be
V = A · h = [area of a cross section] × [height] (2)
(Figure 6.2.4). Note that this is consistent with Formula (1) for the volume of a right circular
cylinder.
Volume = A.^ h
Area A
h
Figure 6.2.4 We now have all of the tools required to solve the following problem.
6.2.1 problem Let S be a solid that extends along the x-axis and is bounded on the
left and right, respectively, by the planes that are perpendicular to the x-axis at x = a and
x = b (Figure 6.2.5). Find the volume V of the solid, assuming that its cross-sectional
area A(x) is known at each x in the interval [a, b].
To solve this problem we begin by dividing the interval [a, b] into n subintervals, thereby
a x b
Cross section S
Cross section area = A ( x )
Figure 6.2.5 dividing the solid into n slabs as shown in the left part of Figure 6.2.6. If we assume that
the width of the kth subinterval is x k , then the volume of the kth slab can be approximated
by the volume A(x k∗ )x k of a right cylinder of width (height) x k and cross-sectional area
A(x k∗ ), where x∗ k is a point in the kth subinterval (see the right part of Figure 6.2.6).
Figure 6.2.
x x (^) k Δ x (^) k
S (^) k a x 1 x 2 xn − 1 b
S
S 1 S^2
S (^) n
...
The cross section here has area A ( x * k ).
Adding these approximations yields the following Riemann sum that approximates the
volume V :
V ≈
∑^ n
k= 1
A(x k∗ )x k
6.2 Volumes by Slicing; Disks and Washers 423
Taking the limit as n increases and the widths of all the subintervals approach zero yields
the definite integral
V = lim
max xk → 0
∑^ n
k= 1
A(x k∗ )x k =
∫ b
a
A(x) dx
In summary, we have the following result.
6.2.2 volume formula Let S be a solid bounded by two parallel planes perpen-
dicular to the x-axis at x = a and x = b. If, for each x in [a, b], the cross-sectional area
of S perpendicular to the x-axis is A(x), then the volume of the solid is
V =
∫ b
a
A(x) dx (3)
provided A(x) is integrable.
It is understood in our calculations of volume that the units of volume are the cubed units of length [e.g., cubic inches (in^3 ) or cubic meters (m 3 )].
There is a similar result for cross sections perpendicular to the y-axis.
6.2.3 volume formula Let S be a solid bounded by two parallel planes perpen-
dicular to the y-axis at y = c and y = d. If, for each y in [c, d], the cross-sectional area
of S perpendicular to the y-axis is A(y), then the volume of the solid is
V =
∫ d
c
A(y) dy (4)
provided A(y) is integrable.
In words, these formulas state:
The volume of a solid can be obtained by integrating the cross-sectional area from one
end of the solid to the other.
Example 1 Derive the formula for the volume of a right pyramid whose altitude is h
and whose base is a square with sides of length a.
Solution. As illustrated in Figure 6.2.7 a , we introduce a rectangular coordinate system
O C
B
y
h − y
h (^12) s
(^12) a
( b )
( a )
x -axis
y -axis
y
B (0, h )
O (^) C (^) 12 a , 0
Figure 6.2.
in which the y-axis passes through the apex and is perpendicular to the base, and the x-axis
passes through the base and is parallel to a side of the base.
At any y in the interval [ 0 , h] on the y-axis, the cross section perpendicular to the y-
axis is a square. If s denotes the length of a side of this square, then by similar triangles
(Figure 6.2.7 b ) 1
2 s
1
2 a^
h − y
h
or s =
a
h
(h − y)
Thus, the area A(y) of the cross section at y is
A(y) = s^2 =
a^2
h^2
(h − y)^2
6.2 Volumes by Slicing; Disks and Washers 425
We can solve this problem by slicing. For this purpose, observe that the cross section
of the solid taken perpendicular to the x-axis at the point x is a circular disk of radius f(x)
(Figure 6.2.9 b ). The area of this region is
A(x) = π[f(x)]^2
Thus, from (3) the volume of the solid is
V =
∫ b
a
π[f(x)]^2 dx (5)
Because the cross sections are disk shaped, the application of this formula is called the
method of disks.
Example 2 Find the volume of the solid that is obtained when the region under the
curve y =
x over the interval [ 1 , 4 ] is revolved about the x-axis (Figure 6.2.10).
x
y
1 4
y = (^) √ x
Figure 6.2.10 (^) Solution. From (5), the volume is
V =
∫ b
a
π[f(x)]^2 dx =
1
πx dx =
πx^2
] 4
1
Example 3 Derive the formula for the volume of a sphere of radius r.
Solution. As indicated in Figure 6.2.11, a sphere of radius r can be generated by revolving
x
y
− r r
x^2 + y^2 = r^2
Figure 6.2.
the upper semicircular disk enclosed between the x-axis and
x^2 + y^2 = r^2
about the x-axis. Since the upper half of this circle is the graph of y = f(x) =
r^2 − x^2 ,
it follows from (5) that the volume of the sphere is
V =
∫ b
a
π[f(x)]^2 dx =
∫ r
−r
π(r^2 − x^2 ) dx = π
[
r^2 x −
x^3
]r
−r
πr^3
VOLUMES BY WASHERS PERPENDICULAR TO THE x -AXIS
Not all solids of revolution have solid interiors; some have holes or channels that create
interior surfaces, as in Figure 6.2.8 d. So we will also be interested in problems of the
following type.
6.2.5 problem Let f and g be continuous and nonnegative on [a, b], and suppose
that f(x) ≥ g(x) for all x in the interval [a, b]. Let R be the region that is bounded
above by y = f(x), below by y = g(x), and on the sides by the lines x = a and x = b
(Figure 6.2.12 a ). Find the volume of the solid of revolution that is generated by revolving
the region R about the x-axis (Figure 6.2.12 b ).
x
y
a b
y = f ( x )
y = g ( x )
x
y
( a )
( b )
x
R
f ( x )
g ( x ) a x b
Figure 6.2.
We can solve this problem by slicing. For this purpose, observe that the cross section of
the solid taken perpendicular to the x-axis at the point x is the annular or “washer-shaped”
426 Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering
region with inner radius g(x) and outer radius f(x) (Figure 6.2.12 b ); its area is
A(x) = π[f(x)]^2 − π[g(x)]^2 = π([f(x)]^2 − [g(x)]^2 )
Thus, from (3) the volume of the solid is
V =
∫ b
a
π([f(x)]^2 − [g(x)]^2 ) dx (6)
Because the cross sections are washer shaped, the application of this formula is called the
method of washers.
Example 4 Find the volume of the solid generated when the region between the graphs
of the equations f(x) = 12 + x^2 and g(x) = x over the interval [ 0 , 2 ] is revolved about the
x-axis.
Solution. First sketch the region (Figure 6.2.13 a ); then imagine revolving it about the
x-axis (Figure 6.2.13 b ). From (6) the volume is
V =
∫ b
a
π([f(x)]^2 − [g(x)]^2 ) dx =
0
([ 1
2 +^ x
2 ]^2 − x 2 )^ dx
0
+ x^4
dx = π
[
x
x^5
] 2
0
Figure 6.2.
y = x
y = 12 + x^2
x
y
1 2
1
2
3
4
5
x 2
Unequal scales on axes
y
Region defined by f and g
( a )
The resulting solid of revolution
( b )
VOLUMES BY DISKS AND WASHERS PERPENDICULAR TO THE y -AXIS
The methods of disks and washers have analogs for regions that are revolved about the y-
axis (Figures 6.2.14 and 6.2.15). Using the method of slicing and Formula (4), you should
be able to deduce the following formulas for the volumes of the solids in the figures.
V =
∫ d
c
π[u(y)]^2 dy
Disks
V =
∫ d
c
π([w(y)]^2 − [v(y)]^2 ) dy
Washers
428 Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering
it follows by (3) that the volume of the solid is
V =
0
A(x) dx =
0
x^4 + 2 x^2
dx = π
[
x^5 +
x^3
] 2
0
Figure 6.2.
0
x^2
4
y = − 1
x
y
R
x 2
✔QUICK CHECK EXERCISES 6.2 ( See page 431 for answers. )
1. A solid S extends along the x -axis from x = 1 to x = 3. For x between 1 and 3, the cross-sectional area of S per- pendicular to the x-axis is 3x^2. An integral expression for the volume of S is. The value of this integral is . 2. A solid S is generated by revolving the region between the x-axis and the curve y =
sin x ( 0 ≤ x ≤ π) about the x- axis. (a) For x between 0 and π, the cross-sectional area of S perpendicular to the x-axis at x is A(x) =. (b) An integral expression for the volume of S is. (c) The value of the integral in part (b) is.
3. A solid S is generated by revolving the region enclosed by the line y = 2 x + 1 and the curve y = x^2 + 1 about the x-axis.
(a) For x between and , the cross- sectional area of S perpendicular to the x-axis at x is A(x) =. (b) An integral expression for the volume of S is.
4. A solid S is generated by revolving the region enclosed by the line y = x + 1 and the curve y = x^2 + 1 about the y- axis. (a) For y between and , the cross- sectional area of S perpendicular to the y-axis at y is A(y) =. (b) An integral expression for the volume of S is.
EXERCISE SET 6.2 C^ CAS
1–8 Find the volume of the solid that results when the shaded region is revolved about the indicated axis. ■
1.
− 1 3
2
x
y y = √ 3 − x
1
2
x
y (^) y = x
y = 2 − x^2
2
2
x
y
y = 3 − 2 x
2
2
x
y
y = (^1) / x
6.2 Volumes by Slicing; Disks and Washers 429
3 6
1
x
y
y = (^) √cos x
1
1
x
y
y = x^3
y = x^2
(1, 1)
2
3
x
y
x = (^) √ 1 + y
3
2
x
y
y = x^2 − 1
(2, 3)
9. Find the volume of the solid whose base is the region bounded between the curve y = x^2 and the x-axis from x = 0 to x = 2 and whose cross sections taken perpendic- ular to the x-axis are squares. 10. Find the volume of the solid whose base is the region bounded between the curve y = sec x and the x-axis from x = π/4 to x = π/3 and whose cross sections taken per- pendicular to the x-axis are squares.
11–18 Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis. ■
11. y =
25 − x^2 , y = 3
12. y = 9 − x^2 , y = 0 13. x =
y, x = y/ 4
14. y = sin x, y = cos x, x = 0 , x = π/ 4 [ Hint: Use the identity cos 2x = cos^2 x − sin 2 x.] 15. y = e x^ , y = 0 , x = 0 , x = ln 3 16. y = e−^2 x^ , y = 0 , x = 0 , x = 1 17. y =
4 + x^2
, x = − 2 , x = 2 , y = 0
18. y =
e^3 x √ 1 + e^6 x^
, x = 0 , x = 1 , y = 0
19. Find the volume of the solid whose base is the region bounded between the curve y = x^3 and the y-axis from y = 0 to y = 1 and whose cross sections taken perpendic- ular to the y-axis are squares. 20. Find the volume of the solid whose base is the region en- closed between the curve x = 1 − y^2 and the y-axis and whose cross sections taken perpendicular to the y-axis are squares.
21–26 Find the volume of the solid that results when the region enclosed by the given curves is revolved about the y-axis. ■
21. x = csc y, y = π/ 4 , y = 3 π/ 4 , x = 0 22. y = x^2 , x = y^2 23. x = y^2 , x = y + 2 24. x = 1 − y^2 , x = 2 + y^2 , y = − 1 , y = 1 25. y = ln x, x = 0 , y = 0 , y = 1 26. y =
1 − x^2 x^2
(x > 0 ), x = 0 , y = 0 , y = 2
27–30 True–False Determine whether the statement is true or false. Explain your answer. [In these exercises, assume that a solid S of volume V is bounded by two parallel planes perpen- dicular to the x-axis at x = a and x = b and that for each x in [a, b], A(x) denotes the cross-sectional area of S perpendicular to the x-axis.] ■
27. If each cross section of S perpendicular to the x-axis is a square, then S is a rectangular parallelepiped (i.e., is box shaped). 28. If each cross section of S is a disk or a washer, then S is a solid of revolution. 29. If x is in centimeters (cm), then A(x) must be a quadratic function of x, since units of A(x) will be square centimeters (cm 2 ). 30. The average value of A(x) on the interval [a, b] is given by V /(b − a). 31. Find the volume of the solid that results when the region above the x-axis and below the ellipse
x^2 a^2
y^2 b^2
= 1 (a > 0 , b > 0 )
is revolved about the x-axis.
32. Let V be the volume of the solid that results when the region enclosed by y = 1 /x, y = 0, x = 2 , and x = b ( 0 < b < 2 ) is revolved about the x-axis. Find the value of b for which V = 3. 33. Find the volume of the solid generated when the region enclosed by y =
x + 1 , y =
2 x, and y = 0 is revolved about the x-axis. [ Hint: Split the solid into two parts.]
34. Find the volume of the solid generated when the region enclosed by y =
x, y = 6 − x, and y = 0 is revolved about the x-axis. [ Hint: Split the solid into two parts.]
F O C U S O N C O N C E P TS
35. Suppose that f is a continuous function on [a, b], and let R be the region between the curve y = f(x) and the line y = k from x = a to x = b. Using the method of disks, derive with explanation a formula for the vol- ume of a solid generated by revolving R about the line y = k. State and explain additional assumptions, if any, that you need about f for your formula. 36. Suppose that v and w are continuous functions on [c, d], and let R be the region between the curves x = v(y) and x = w(y) from y = c to y = d. Using the method of washers, derive with explanation a formula for the vol- ume of a solid generated by revolving R about the line
6.2 Volumes by Slicing; Disks and Washers 431
(b) Use the average of left and right endpoint approxima- tions to approximate the volume.
1
x
y
cm
cm
cm
cm
cm
cm
cm
cm
cm
cm
5 cm Figure Ex-
58. Use the result in Exercise 55 to find the volume of the solid that remains when a hole of radius r/2 is drilled through the center of a sphere of radius r, and then check your answer by integrating. 59. As shown in the accompanying figure, a cocktail glass with a bowl shaped like a hemisphere of diameter 8 cm contains a cherry with a diameter of 2 cm. If the glass is filled to a depth of h cm, what is the volume of liquid it contains? [ Hint: First consider the case where the cherry is partially submerged, then the case where it is totally submerged.]
Figure Ex-
60. Find the volume of the torus that results when the region en- closed by the circle of radius r with center at (h, 0 ), h > r, is revolved about the y-axis. [ Hint: Use an appropriate formula from plane geometry to help evaluate the definite integral.] 61. A wedge is cut from a right circular cylinder of radius r by two planes, one perpendicular to the axis of the cylinder and the other making an angle θ with the first. Find the volume of the wedge by slicing perpendicular to the y-axis as shown in the accompanying figure.
u
y
x
r
Figure Ex-
62. Find the volume of the wedge described in Exercise 61 by slicing perpendicular to the x-axis. 63. Two right circular cylinders of radius r have axes that inter- sect at right angles. Find the volume of the solid common to the two cylinders. [ Hint: One-eighth of the solid is sketched in the accompanying figure.] 64. In 1635 Bonaventura Cavalieri, a student of Galileo, stated the following result, called Cavalieri’s principle : If two solids have the same height , and if the areas of their cross sections taken parallel to and at equal distances from their bases are always equal , then the solids have the same vol- ume. Use this result to find the volume of the oblique cylin- der in the accompanying figure. (See Exercise 52 of Section 6.1 for a planar version of Cavalieri’s principle.)
Figure Ex-
h
r
r
Figure Ex-
65. Writing Use the results of this section to derive Cavalieri’s principle (Exercise 64). 66. Writing Write a short paragraph that explains how For- mulas (4)–(8) may all be viewed as consequences of For- mula (3).
✔QUICK CHECK ANSWERS 6.
1
3 x^2 dx; 26 2. (a) π sin x (b)
∫ (^) π
0
π sin x dx (c) 2π 3. (a) 0; 2; π[( 2 x + 1 )^2 − (x^2 + 1 )^2 ] = π[−x^4 + 2 x^2 + 4 x]
(b)
0
π[−x^4 + 2 x^2 + 4 x] dx 4. (a) 1; 2; π[(y − 1 ) − (y − 1 )^2 ] = π[−y^2 + 3 y − 2 ] (b)
1
π[−y^2 + 3 y − 2 ] dy
432 Chapter 6 / Applications of the Definite Integral in Geometry, Science, and Engineering
6.3 VOLUMES BY CYLINDRICAL SHELLS
The methods for computing volumes that have been discussed so far depend on our ability
to compute the cross-sectional area of the solid and to integrate that area across the solid.
In this section we will develop another method for finding volumes that may be applicable
when the cross-sectional area cannot be found or the integration is too difficult.
CYLINDRICAL SHELLS
In this section we will be interested in the following problem.
6.3.1 problem Let f be continuous and nonnegative on [a, b] ( 0 ≤ a < b), and let
R be the region that is bounded above by y = f(x), below by the x-axis, and on the
sides by the lines x = a and x = b. Find the volume V of the solid of revolution S that
is generated by revolving the region R about the y-axis (Figure 6.3.1).
Figure 6.3.
x
y
y = f ( x )
R
a b
x
S
y
Sometimes problems of the above type can be solved by the method of disks or washers
perpendicular to the y-axis, but when that method is not applicable or the resulting integral
is difficult, the method of cylindrical shells , which we will discuss here, will often work.
A cylindrical shell is a solid enclosed by two concentric right circular cylinders (Fig-
ure 6.3.2). The volume V of a cylindrical shell with inner radius r 1 , outer radius r 2 , and
h
r 1 r 2
Figure 6.3.
height h can be written as
V = [area of cross section] · [height]
= (πr 22 − πr 12 )h
= π(r 2 + r 1 )(r 2 − r 1 )h
[ 1
2 (r^1 +^ r^2 )
]
· h · (r 2 − r 1 )
But 12 (r 1 + r 2 ) is the average radius of the shell and r 2 − r 1 is its thickness, so
V = 2 π · [average radius] · [height] · [thickness] (1)
We will now show how this formula can be used to solve Problem 6.3.1. The underlying
idea is to divide the interval [a, b] into n subintervals, thereby subdividing the region R into
n strips, R 1 , R 2 ,... , Rn (Figure 6.3.3 a ). When the region R is revolved about the y-axis,
these strips generate “tube-like” solids S 1 , S 2 ,... , Sn that are nested one inside the other
and together comprise the entire solid S (Figure 6.3.3 b ). Thus, the volume V of the solid
can be obtained by adding together the volumes of the tubes; that is,
V = V (S 1 ) + V (S 2 ) + · · · + V (S n )
