4 Applications of the Definite Integral

In this chapter we will learn how a definite integral arise in many problems such as finding the
volume and surface area of a solid of revolution, length of a plane curve, calculating the work done
by a force, and force exerted by a fluid on a submerged object.
Although these problems are diverse, the required calculation can all be approximated by the
same procedure that we used to find area.

4.1 Area Between Two Curves

Let us review the basic principle that underlies the calculation of area as a definite integral.
1. Divide the interval [a, b] into n equal subintervals of width ∆x. Use those subintervals
to divide the region into n strips.
2. Approximate the area of the the k th strip by the area of the rectangle of width ∆x and
height f (x∗k ).
3. Add the approximate areas of the strips to approximate the entire area AR .
n
X
AR ≈ f (x∗k )∆x
k=1

4. Take the limite of the sum as the number of subintervals n approach infinitity. This
causes the error in the approximations to approach zero.
Xn Z b
∗
AR = lim f (xk )∆x = f (x) dx
n→∞ a
k=1

We now consider the region R in the interval [a, b] bounded above by the curve of y = f (x) and
bounded below by the graph of y = g(x) (Fig . 4.1a). The area of the approximating rectangle is

f (x∗k ) − g(x∗k ) ∆x
 

so that the area of the region is

n
X Z b
f (x∗k ) − g(x∗k ) ∆x =
 
AR = lim f (x) − g(x) dx
n→∞ a
k=1

It is possible that the two curves intersect at one or both endpoints, in which case the sides of
the region will be points rather than verical line segments (Fig .4.1b). When that occurs, you will
have to determine the points of intersection to obtain the limit/s of integration.
Example 4.1 Determine the area of the region in the interval [−1, 3] bounded by the curves y = x2 and
y = 2x + 4.
Z 3 i3
(2x + 4) − x2 dx = x2 + 4x − 13 x3
−1 −1
 
= (12) − − 83 = 44 3

-1 3

In the example, an accompanying figure allows an easy set up of the definite integral. In many
a problem, one needs to determine analytically which curve serves as the upper boundary of the
region.

41
 4 Applications of the Definite Integral

f (x∗
k) f (x∗
k)

a f (x∗ ) − g(x∗ ) a f (x∗ ) − g(x∗ )

k k b k k b

g(x∗
k)
g(x∗
k)

(a) (b)

Figure 4.1: Two plane regions bounded by curves and representative rectangular elements of each.

Example 4.2 Express as a definite integral the area of the region bounded by the curves p(x) = x3 + 2x2 − 11
and q(x) = 2x3 − 3x2 + 2x − 3.
Let h(x) denote the height function, where

h(x) = q(x) − p(x) = x3 − 5x2 + 2x + 8

Setting h(x) = 0 gives the points of intersection of the curves. The factor theorem and a synthetic
division yields the solutions x = −1, 2, 4. Hence, the curves intersect three times. Therefore, the
region is composed of two subregions.
At a point in [−1, 2], say x = 1,
h(1) = 1 − 5 + 2 + 8 = 6
which means the region is bounded above by q(x).
At a point in [2, 4] say x = 3,
h(3) = 27 − 45 + 6 + 8 = −4
so that the region is now bounded above by p(x) (Fig. 4.2a). The total area of the region is given
by Z 2 Z 4
x3 − 5x2 + 2x + 8 dx + −x3 + 5x2 − 2x − 8 dx
−1 2

Example 4.3 Find the area of the region bounded by the parabola x = y 2 and the line y = x − 6. Refer to
Fig. 4.2b.
The first rectangular element represent all the approximating rectangles in the interval [0, 4]. The
other rectangular element is representative of all the apaproximating rectangles in the interval [4, 9].
the area of each subregion is
Z 4 Z 9
√ √ √
AR 1 = x − (− x) dx AR2 = x − (x − 6) dx
0 4

The total area of the region is

Z 4 √
Z 9 √ 126
AR = 2 x dx + x − x + 6 dx = 6
0 4

√
y= x
6
−
x
=
x3 + 2x2 − 11 y
2x3 − 3x2 + 2x − 3 4

√
y=− x

(a) (b)

Figure 4.2: Plane regions for Examples 4.2 and 4.3.

42
 4 Applications of the Definite Integral

It is not always necessary to draw rectangular elements with a horizontal base. In certain
situations, a ‘vertical’ base will yield a simple integral.
For the region in Fig 4.2b, construct a ‘horizontal’ rectangle with height ∆y. The width of the
rectangle is the horizontal distance between the two curves,

(yk∗ + 6) − (yk∗ )2

so that the area of the representative element is

(yk∗ + 6) − (yk∗ )2 ∆y

The sum of the n approximating rectangles is

n
X
(yk∗ + 6) − (yk∗ )2 ∆y


AR ≈
k=1

and the area of the region is

n
X Z 3
(yk∗ + 6) − (yk∗ )2 ∆y = y + 6 − y 2 dy


AR = lim
n→∞ −2
k=1

∗
yk ∆y
∗ ∗ 2
(yk + 6) − (yk )

-2

Figure 4.3: ‘Horizontal’ rectangular element for Example 4.3.

Exercises
Sketch the region enclused by the given boundaries. De- 1.5. y = 12 − x2 , y = x2 − 6
cide whether to integrate with respect to x or y. Draw a
1.6. y = ex , y = xex , x = 0
typical approximating rectangle and label its height and
width. Then find the area of the region. 1.7. x = 2y 2 , x = 4 + y 2
√
1.1. y = ex , y = x2 − 1, x = −1, x = 1 1.8. y = x − 1, x − y = 1
1.2. y = sin x, y = x, x = π/2, x = π √
1.9. x = y 4 , y = 2 − x, y = 0
1.3. y = x2 − 2x, y = x + 4
Use calculus to find the area of the triangle with the
1.4. x = 1 − y 2 , x = y 2 − 1 given vertices.
Sketch the region enclosed by the given curves and find
its area. 1.10 (0, 0), (3, 1), (1, 2) 1.11 (2, 0) (0, 2) (−1, 1)

4.2 Volumes by Slicing: Discs and Washers

In this section we will see that the same basic principle to determine the area of a plane region,
can be used to find volumes of certain three-dimensional solids.
One of the simplest examples of a solid with congruent sections is a right cylinder. If the cross
section is a triangle, the cylinder is called a trianglular prism. A prism with a rectangular cross
section is commonly called a (rectangular) box. Of special interest in this section is the cylinder
where the cross section is a circle.
The volume of a right circular cylinder is given by

V = [area of a cross section] × [height] = πr2 h

In this discussion, the height of the circular cylinder is usually small, and we shall refer to it as
disc.

43
 4 Applications of the Definite Integral

Consider a plane region R in the interval [a, b] above the x-axis and bounded above by the curve
y = f (x). Let the region revolve about the x-axis. A rectangular element of width ∆x and height
f (x∗k ), generates a disc with a volume π[f (x∗k )]2 ∆x (Fig. 4.4a).
Summing the volumes of the n discs, the approximate volume of the solid of revolution is
n
X
VR ≈ π[f (x∗k )]2 ∆x
k=1

Taking the limit as n approach infinity, the volume of the solid is

n
X Z b
VR = lim π[f (x∗k )]2 ∆x = π[f (x)]2 dx
n→∞ a
k=1

(a) (b)

Figure 4.4: A disc generated by a rectangular element (a); and the solid of revolution generated by R (b).

Example 4.4 A sphere is a solid formed by revolving the plane region bounded by a semicircle and a diameter.
Obtain the volume of the sphere of radius r using calculus.
The equation of the semicircle on the upper half of the plane is
p
y = r 2 − x2

Therefore, the volume of the sphere generated by the region bounded by the semicircle and the x-axis
is
Z r p Z r

2
V = π r2 − x2 dx = π r2 − x2 dx
9r 9r

ir
= π r2 x − 31 x3
−r
3 1 3
= π(r − 3r ) − π(−r3 + 13 r3 )
= 43 πr3

To determine the volume of a solid of revolution by the disc method, it is important to draw the
rectangular element perpendicular to the axis of revolution.
Example 4.5 Obtain the volume of the solid of revolution generated by the plane region bounded by the
y-axis, the line y = 3 and the parabola y = x2 − 1, when revolved about the y-axis.
The region is shown on the right.
Z 3 p Z 3

2
V = π y + 1 dy = π y + 1 dy 3
91 91 x∗
∗ k
1 2

i 3 yk
=π 2y +y
91

= π( 15
2 ) − π(− 12 ) = 8π -1

Not all solids of revolution have solid interiors; some have holes that create interior surfaces. In
Fig. 4.5, the plane region is revolved about an axis, and a cross section of the solid is a disc with
a hole inside.

44
 4 Applications of the Definite Integral

Figure 4.5: Annular cross section.

The outer radius of the annulus is R = f (x) and the inner radius is r = g(x). The volume of
the washer is
Vk = π[f (x∗k )]2 ∆x − π[g(x∗k )]2 ∆x
Thus, the volume of the solid of revolution is

n
X Z b
V = lim Vk = π [f (x)]2 − [g(x)]2 dx
n→∞ a
k=1

Example 4.6 Find the volume of the solid generated by revolving the region bounded by y = 2x and y = x2
about

(a) the x-axis; (b) the y-axis; (c) the line y = −1.

(a) about the x-axis.

Z b
V =π R2 − r2 dx
a
Z 2 R
=π (2x)2 − (x2 )2 dx
0 r

4 3

i2
=π 3x − 15 x5 = 64
15 π x∗
k
0

(b) about the y-axis.

Z b
V =π R2 − r2 dy
a
4
√
Z  y 2
=π ( y)2 − dy ∗
yk
0 2
r

i4
= π 12 y 2 − 12
1 3
y = 38 π R
0

(c) about y = −1.

Z b
V =π R2 − r2 dx ∗
a yk
R
Z 2
2 2 2 r
=π (2x + 1) − (x + 1) dx
0

i2 y = −1
= π − 51 x5 + 23 x4 + 2x2 = 104
15 π
0

Exercises
Find the volume of the solid that results when the re- 2.1. y = ex , y = 0 (x-axis), x = 0 (y-axis), x = ln 3
gion enclosed by the given curves is revolved about the
x-axis. 1
2.2. y = √ , x = −2, x = 2, y = 0
4 + x2

45
 4 Applications of the Definite Integral

2.3. y = 9 − x2 ,y = 0 2 y = 2 − x2
2 x
y=
2.4. y = cos x, y = sin x, x = 0, x = π/4

Find the volume of the solid that results when the re-
gion enclosed by the given curves is revolved about the −1 3 1
2.4 2.5
y-axis.
2.6. Find the volume of the solid that results when the
2.5. x = csc y, y = π/4, y = 3π/4, y = 0 √
region enclosed by y = x, y = 0, and x = 9 is revolved
about the line x = 9.
2.6. y = x2 , x = y2
2.7. Find the volume of the solid that results when the
2.7. x = y2 , x=y+2 region enclosed by x = y 2 and x = y is revolved about
the line y = −1.
2.8. x = 1 − y 2 , x = 2 + y 2 , y = −1, y = 1
2.8. Find the volume of the solid that results when the
Find the volume of the solid that results when the indi- region enclosed by y = x2 and y = x3 is revolved about
cated region is revolved about the indicated axis. the line x = 1.

4.3 Volume by Cylindrical Shells

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.
In Example 4.5 of the previous section, the plane region bounded by the x = 0, y = 3 and the
parabola y = x2 − 1 is revolved about the y axis. Under the disc method, a rectangular element
is drawn perpendicular to the axis of p revolution. In this case, the height of the representative
rectangle is ∆y and its width is x∗k = yk∗ + 1, which is found by solving x in terms of y. The
integration is carried out in the variable y.
Solving x in terms of y works for a limited case but in general it is difficult if not impossible.
Clearly, we need a different approach when a region is revolved about the y axis and solving x in
terms of y is not an option.
A cylindrical shell is a solid enclosed by two concentric right circular cylinders (Fig. 4.6a) .
The volume of a cylindrical shell with inner radius r1 , outer radius r2 , and height h can be written
as

V = V2 − V1
= πr22 h − πr12 h = π(r22 − r12 )h
= π(r2 + r1 )(r2 − r1 )h
r2 + r1
= 2π · (r2 − r1 )h
2
If we let ∆r = r2 − r1 (the thickness of the shell) and r = 21 (r2 + r1 ) (the average radius of the
shell), then this formula for the volume of a cylindrical shell becomes

V = 2πrh∆r

and it can be remembered as

V = [circumference][height][thickness]

Now let S be the solid obtained by rotating about the y-axis the region bounded by x = a,
x = b, f (x) and g(x) where f (x) ≥ g(x). Let x∗k = x̄k be the midpoint of the k th subinterval. If

r2 2πr
r1

h h

∆r

(a) (b)

Figure 4.6: A cylindrical shell (a), cut and flattened cylidrical shell (b).

46
 4 Applications of the Definite Integral

the rectangle with base ∆x and height f (x̄k ) − g(x̄k ) is rotated about the y-axis, the result is a
cylindrical shell with an average radius x̄k , height f (x̄k ) − g(x̄k ) and thickness ∆x. By the formula
above,
Vk = 2πx̄k [f (x̄k ) − g(x̄k )]∆x
Therefore an approximation to the volume V of S is given by the sum of the volumes of these
shells:
Xn Xn
V ≈ Vk = 2πx̄k [f (x̄k ) − g(x̄k )]∆x
k=1 k=1

This approximation appears to become better as n → ∞. But from the defition of an integral, we
have
Xn Z b
V = lim 2πx̄k [f (x̄k ) − g(x̄k )]∆x = 2πx[f (x) − g(x)] dx
n→∞ a
k=1

The best way to remember the formula is to think of a typical shell, cut and flattened, with
radius x, circumference 2πx, height f (x) − g(x), and thickness ∆x.
Example 4.7 Obtain the volume of the solid of revolution generated by the region bounded by the y-axis,
y = 3 and y = x2 − 1, about the y-axis, by the cylindrical shell method (Example 4.5).
We have the limits of integration x = 0 and x = 2 and the curves f (x) = 3 and g(x) = x2 − 1 as
upper and lower boundaries respectively.
Z 2 Z 2
2
V = 2π x[3 − (x − 1)] dx = 2π 4x − x3 dx
0 0
2 1 4

i 2
= 2π 2x − 4x x̄k
0
= 2π(8 − 4) = 8π x̄k

Example 4.8 Use cylindrical shell method to find the volume of the solid generated by revolving the region
bounded by y = 2x and y = x2 (Example 4.6) about
(a) the x-axis; (b) the y-axis; (c) the line y = −1.

(a) about the x-axis.

Z b
V = 2πyh dy
a
4
√
Z
ȳk
= 2π y( y − 12 y) dy
0 ȳk
2 5/2 1 3

i4 64
= 2π 5y − 6y = 15 π
0

(b) about the y-axis.

Z b
V = 2πxh dx
a
Z 2 x̄k
= 2π x(2x − x2 ) dx
0
2 3

i2
= 2π 3x − 14 x4 = 38 π x̄k
0

(c) about y = −1.

Z b
V = 2πrh dy
a ȳk
4
√ ȳk
Z
= 2π (y + 1)( y − 12 y) dy ȳk + 1
0
2 5/2

i4 y = −1
= 2π 5y − 16 y 3 + 23 y 3/2 − 14 y 2 = 104
15 π
0

47
 4 Applications of the Definite Integral

Exercises
Use the method of cylindrical shells to find the volume Use the method of cylindrical shells to find the volume
of the solid that results when the indicated region is generated by rotating the region bounded vy the given
revolved about the indicated axis. curves about the x-axis.
3.8. xy = 1, x = 0, y = 1, y = 3
2 y = 2 − x2 √
2 x 3.9. y = x, x = 0, y = 2
y=
3.10. x = 4y 2 − y 3 , x = 0
3.11. x = 1 + (y + 2)2 , x = 2
−1 3 1
3.1 3.2
3.12. x + y = 3, x = 4 − (y − 1)2
Use the method of cylindrical shells to find the volume Use the method of cylindrical shells to find the volume
generated by rotating the region bounded vy the given generated by rotating the region bounded vy the given
curves about the y-axis. curves about the specified axis.
√
3.3. y = 3 x, y = 0, x = 1 3.13. y = x4 , y = 0, x = 1; about x = 2
√
3.4. y = x3 , y = 0, x = 1, x = 2 3.14. y = x, y = 0, x = 1; about x = −1
2
3.5. y = e−x , y = 0, x = 0, x = 1 3.15. y = 4x − x2 , y = 3; about x = 1
3.6. y = 4x − x2 , y = x 3.16. x = y 2 + 1, x = 2; about x = 2
3.7. y = x2 , y = 6x − 2x2 3.17. y = x3 , y = 0, x = 1; about y = 1

4.4 Length of a Curve and Area of a Surface of Revolution

In this section, we learn to determine the length of a curve and the area of a surface of revolution
by applying the same principles to determine the area of a plane region.
To define the length of a curve, or arc length, we start by breaking the curve y = f (x) into
small segments. Then we approximate the curve segments by line segments and add the lengths of
the line segments to approximate the arc length. Finally, we increase the number of subdivisions
of the interval to infinity. If the limiting value of the sum of line segments exists, then we define
the limit as the length of the curve.

a b a b

(a) (b)

Figure 4.7: Arc length approximated using (a) 4 segments, and (b) 8 segments.

On the subinterval [xk91 , xk ], the line segment that approximates the length of the segment of
the curve is
p
Lk = (xk − xk91 )2 + (f (xk ) − f (xk91 ))2
r  ∆y 2
k
p
2 2
= (∆x) + (∆yk ) = 1 + · ∆x
∆x
If L is the length of the curve, then
n
r  ∆y 2
k
X
L≈ 1+ · ∆x
∆x
k=1

As n → ∞, the quantity ∆yk /∆x → f ′ (x∗k ) so that the length of the curve is

n
r  ∆y 2 Z b
r h dy i2
k
X
L = lim 1+ · ∆x = 1+ dx
n→∞ ∆x a dx
k=1

48
 4 Applications of the Definite Integral

Example 4.9 Find the arc length of the curve y = x3/2 from (1, 1) to (4, 8) in two ways:
a) Integrating with respect to x

b) Integrating with respect to y

(a) Differentiating y with respect to x, dy/dx = (3/2)x1/2 .

Z 4 q Z 4 √ Substitution
3 1/2 2 1


L= 1+ 2x dx = 2 4 + 9x dx
1 1 s = 4 + 9x =⇒ ds = 9 dx
Z 40 √ i40
1 3/2
1 x=1 s = 13
= s ds = 27
18s
13 13 x=4 s = 40
√ √
80 10 − 13 13
=
27

(b) Expressing in terms of y and differentiating, dx/dy = (2/3)y −1/3 .

Z 8 q Z 8 p Substitution
1 −1/3
L= 1 + 49 y −2/3 dy = 3y 9y 2/3 + 4 dy
1 1 t = 9y 2/3 + 4
Z 40 √ i40
= 11 3/2
t dt = 27 t dt = 6y −1/3 dy
18
13 13
√ √ y=1 t = 13
80 10 − 13 13 y=8 t = 40
=
27
If a section of the curve y = f (x) is revolved about the x-axis, it generates a surface S with an
area p
AS ≈ 2πf (x∗k ) (∆x)2 + (∆yk )2
The total surface area is then defined as the limiting value of the sum of the areas of the individual
surfaces generated by each section of the curve, and is given by

n
X p Z b
r  dy 2
AS = lim 2πf (x∗k ) (∆x)2 + (∆yk )2 = 2πy 1+ dx
n→∞ a dx
k=1

Figure 4.8: A surface generated by a curve revolved about an axis.

Example 4.10 Obtain the formula for the surface of a sphere of radius r.
√
The equation of the upper semicircle of radius r is y = r2 − x2 . The derivative dy/dx is

dy −x
=√
dx r − x2
2

so that r r r
 dy 2 x2 r2 r
1+ = 1+ 2 = =√
dx r − x2 r2 − x2 r 2 − x2
The surface area of the sphere of radius r is
Z r p Z r ir
r
AS = 2π r − x · √
2 2 dx = 2πr dx = 2πrx = 4πr2
9r r 2 − x2 9r 9r

49
 4 Applications of the Definite Integral

Exercises
√
Find the exact arc length of the curve over the interval. 4.11. x = 2 1 − y, −3 ≤ y ≤ 0
4.1. y = 3x3/2 − 1 from x = 0 to x = 1
4.2. x = 1
3
(y 2 + 2)3/2 from y = 0 to y = 1
4.12. The accompanying figure shows a spherical cap
x6 + 8 of height cut from a sphere of radius r. Show that the
4.3. y = from x = 2 to x = 3
16x2 surface area S of the cap is S = 2πrh.
4
4.4. 24xy = y + 48 from y = 2 to y = 4
h
4.5. y = ln sec x from x = 0 to x = π/4
Find the area of the surface generated by revolving the r
given curve about the x-axis.
4.6. y = 7x, 0 ≤ x ≤ 1
√
4.7. y = x, 1 ≤ x ≤ 4
√
3 y, 1 ≤ y ≤ 8
4.8. x =
Find the area of the surface generated by revolving the 4.13. The portion of a sphere that is cut by two parallel
given curve about the y-axis. planes is called a zone. Use the result of Exercise 4.12
to show that the surface area of a zone depends on the
4.9. x = 9y + 1, 0 ≤ y ≤ 2 radius of the sphere and the distance between the planes,
4.10. x = y 3 , 0 ≤ y ≤ 1 not not on the location of the zone.

4.5 Work
In this section we will be concerned with two related concepts, work and energy. A detailed study
of the relationship between work and energy is explained in a physics course, and our primary goal
here will be to explain the role of integration in the study of work.
If a constant force of magnitude F is applied in the direction of motion of an object, and if that
object moves a distance d, then the work W performed by the force on the object to be
W =F ·d
Many important problems are concerned with finding the work done by a variable force that
is applied in the direction of motion. For example, Fig. 4.9a shows a spring in its natural state
(neither compressed nor stretched). If we want to pull the block horizontally (Fig. 4.9b) then we
would have to apply more and more force to the block to overcome the resistive force of the already
stretched spring.
The basic idea for solving this problem is to break up the interval [a, b] into subintervals that
are sufficiently small that the force does not vary much on each subinterval. This will allow us to
treat the force as constant on each subinterval and to approximate the work on each subinterval.
On the subinterval [xk91 , xk ], the work done by a variable force F is approximated by
Wk ≈ F (x∗k )∆x
Adding these approximations and taking the limit of the sum as n increases to infinity, we find the
formula for the total work done.

n
X Z b
W = lim F (x∗k )∆x = F (x) dx
n→∞ a
k=1

Hooke’s law states that under appropriate conditions on a spring that is stretch x units beyond
its natural length pulls back with a force
F (x) = kx
where k is a constant called the spring constant or spring stiffness.

L L x
(a) (b)

Figure 4.9: A spring in its natural state (a), and a stretched spring (b).

50
 4 Applications of the Definite Integral

Example 4.11 A spring exerts a force of 5 N when stretched 1 m beyond its natural length.
(a) Find the spring constant k.
(b) How much work is required to stretch the spring 1.8 m beyond its natural length?

F (x) 5N
(a) k = = = 5 N/m
x 1m
Z 1.8 i1.8
kx dx = 12 kx2 N
= 21 5 m (1.8 m)2 = 8.1 J


(b)
0 0

Some problems cannot be solved by mechanically substituting into formulas, and one must return
to basic principles to obtain solutions. This is illustrated in the next example.
Example 4.12 Figure 4.10 shows a conical container of radius 3 m and height 10 m. Suppose that this container
is filled with water to a depth of 5 m. How much work is required to pump all of the water out
through a hole in the top of the container?
Divide the water into n layers with ∆x denoting the thickness of each layer. The work Wk needed
to move the k th layer to the top of the container is approximately

Wk ≈ F (x∗k )x∗k

where F (x∗k ) is the force required to lift that layer. But the force required to lift the layer is the force
needed to overcome gravity.

F (x∗k ) = mk · g
= (vk ρ)g = (ρg)πrk2 ∆x
3 ∗ 2
= πρg( 10 xk ) ∆x

It follows that
Wk ≈ F (x∗k )x∗k = 0.09πρg(x∗k )3 ∆x
and hence the exact total work W required to move all of the water is
n
X
W = lim 0.09πρg(x∗k )3 ∆x
n→∞
k=1
Z 10
= 0.09πρg x3 dx
5
i10
= 0.0225πρgx4
5
= 0.0225πρg(9375 m4 )
kg m 4
= 0.0225π · 1000 m3 · 9.8 s2 · 9375 m

= 2067.1875π kJ

x∗
k x∗
k

5 10

∆x 5 rk
10 − x∗
k
3
10
3
x

Figure 4.10

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 4 Applications of the Definite Integral

Exercises
5.1. A variable force of 5x−2 pounds moves an object all the water to the top of the vat. Use 9810 N/m3 as
along a straight line when it is x feet from the ori- the weight density of water.
gin. Calculate the work done in moving the object from
x = 1 ft to x = 10 ft. 4m
6m
5.2. A force of 10 lb is required to hold a spring stretched
4 in beyond its natural length. How much work is done
in stretching it from its natural length to 6 in beyond its
natural length?
3m
5.3. A spring has a natural length of 20 cm. If a 25 N
force is required to keep it stretched to a length of 30 cm,
how much work is required to stretch it from 20 cm to
25 cm?
5.9. A 100 ft length of steel chain weighing 15 lb/ft is
5.4. A spring has natural length 20 cm. Compare the
dangling from a pulley. How much work is requred to
work W1 done in stretching the spring from 20 cm to
wind the onto the pulley?
30 cm with the work W2 done in stretching it from 30 cm
to 40 cm. How are W2 and W1 related? 5.10. A bucket that weighs 4 lb and a rope of negligi-
5.5. A cylindrical tank of radius 5 ft and height 9 ft is ble weight are used to draw water from a well that is
two-thirds filled with water. Find the work required to 80 ft deep. The bucket is filled with 40 lb of water and
pump all the water over the upper rim. is pulled up at a rate of 2 ft/s, but water leaks out of a
5.6. Solve Exercise 5.5 asuming that the tank is half- hole in the bucket at a rate of 0.2 lb/s. Find the work
filled with water. done in pulling the bucket to the top of the well.
5.7. A cone-shaped water reservoir is 20 ft in diameter 5.11. A leaky 10-kg bucket is lifted from the ground to
across the top and 15 ft deep. If the reservoir is filled to a height of 12 m at a constant speed with a rope that
a depth of 10 ft, how much work is required to pump all weighs 0.8 kg/m. Initially the bucket contains 36 kg of
the water to the top of the reservoir? water, but the water leaks at a constant rate and fin-
5.8. The vat shown in the accompanying figure contains ishes draining just as the bucket reaches the 12-m level.
water to a depth of 2 m. Find the work required to pump How much work is done?

52