APPLICATIONS OF THE DEFINITE INTEGRAL
1. Volume: Slicing, disks and washers
1.1. Volumes by Slicing. Suppose a solid object has boundaries extending from x = a, to
x = b, and that its cross-section in a plane passing through (x, 0, 0) and parallel to the yz-plane
has area A(x). To a first order approximation, the volume of the slice of the object on the right
to the plane of P
thickness ∆x is then A(x)∆x so that the volume of the solid is the limit of the
elemental sum
A(x)∆x.
This being a Riemann sum, the volume is given by the formula
ˆ b
A(x) dx.
V =
a
Example 1.1.
(1) The Pyramid Arena in Memphis has a square base of side approximately
600 feet and a height of approximately 320 feet. Find the volume of the pyramid with
these measurements.
(2) Find the volume of a sphere of radius R.
1.2. The Method of Disks. Suppose that f (x) ≥ 0 and f is continuous on the interval
[a, b]. Take the region bounded by the curve y = f (x) and the x-axis, for a ≤ x ≤ b, and
revolve it about the x-axis, generating a solid. We can find the volume of this solid by slicing it
perpendicular to the x-axis and recognizing that each cross section is a circular disk of radius
r = f (x),
we then have that the volume of the solid is
ˆ b
V =
π[f (x)]2 dx.
a
1
2
APPLICATIONS OF THE DEFINITE INTEGRAL
Since the cross sections of such a solid of revolution are all disks, we refer to this method of
finding volume as the method of disks.
√
Example 1.2.
(1) Revolve the region under the curve y = x on the interval [0, 4] about
the x-axis and find the volume of the resulting solid of revolution.
(2) Find the volume of a sphere of radius R.
(3) Find the volume of the solid generated by revolving the region bounded by y = x2 + 1,
y = 0, x = 1, x = 0 about the x-axis.
In a similar way, suppose that g(y) ≥ 0 and g is continuous on the interval [c, d]. Then,
revolving the region bounded by the curve x = g(y) and the y-axis, for c ≤ y ≤ d, about the
y-axis generates a solid. Notice that the cross sections of the resulting solid of revolution are
circular disks of radius r = g(y). All that has changed here is that we have interchanged the
roles of the variables x and y.
The volume of the solid is then given by
ˆ
d
π[g(y)]2 dy.
V =
c
Example 1.3.
(1) Find the volume of the solid resulting from
√ revolving the region bounded
2
by the curves y = 4 − x and y = 1 from x = 0 to x = 3 about the y-axis.
(2) Find the volume of the solid obtained by revolving the region bounded by the curves
y = x3 and y-axis and y = 8 about the y-axis.
(3) Find the volume of the solid generated by revolving the region bounded by y = 4 − x2 ,
the positive x-axis, and the positive y-axis, about the y-axis.
1.3. The Method of Washers. One complication that occurs in computing volumes is that
the solid may have a cavity or “hole” in it. Another occurs when a region is revolved about a
line other than the x-axis or the y-axis.
For example, consider the solid obtained by revolving the region bounded by the graphs of
y = 14 x2 , x = 0 and y = 1 about the x-axis, and the line y = 2, respectively:
APPLICATIONS OF THE DEFINITE INTEGRAL
3
Example 1.4.
(1) Let R be the region bounded by the graphs of y = 14 x2 , x = 0 and y = 1.
Compute the volume of the solid formed by revolving R about
(a) the y-axis,
(b) the x-axis, and
(c) the line y = 2.
(2) Let R be the region bounded by y = 4 − x2 and y = 0. Find the volume of the solids
obtained by revolving R about each of the following:
(a) the y-axis,
(b) the line y = −3,
(c) the line y = 7, and
(d) the line x = 3.
(3) Find the volume of the solid obtained by revolving the region R bounded by the curves
(a) y = x and y = x2 about the x-axis,
(b) y = x2 and y 2 = 8x about the x-axis,
(c) y = 6p− x2 , y = 2 about the y = 1,
(d) x = 4 − y 2 and y-axis about the x = −1,
The technique used to solve the problems above is a slight generalization of the method of disks
and is referred to as the method of washers, since the cross sections of the solids look like
washers.
2. VOLUMES BY CYLINDRICAL SHELLS
Let R denote the region bounded by the graph of y = f (x) and the x-axis on the interval [a, b],
where 0 < a < b and f (x) ≥ 0 on [a, b]. If we revolve this region about the y-axis, we get the
solid shown below:
If instead of taking a cross section perpendicular to the y-axis, we take a cross section perpendicular to the x-axis, and revolve it about the y-axis, we get a cylinder. Recall that the area of
a cylinder is given by:
A(x) = 2πrh,
where r is the radius of the cylinder and h is the height of the cylinder. We can see that the
radius is the x coordinate of the point on the curve, and the height is the y coordinate of the
curve. Hence
A(x) = 2πxy = 2πxf (x),
Therefore the volume is given by
ˆ
b
2πxf (x) dx.
a
4
APPLICATIONS OF THE DEFINITE INTEGRAL
Remark 2.1. Note that for a given solid, the variable of integration in the method of shells
is exactly opposite that of the method of washers. So, your choice of integration variable will
determine which method you use.
Example 2.1.
(1) Revolve the region bounded by the graphs of y = x and y = x2 in the
first quadrant about the y-axis.
(2) Find the volume of the solid formed by revolving the region bounded by the graph of
y = 4 − x2 and the x-axis about the line x = 3.
(3) Let R be the region bounded by the graphs of y = x, y = 2 − x and y = 0. Compute
the volume of the solid formed by revolving R about the lines
(a) y = 2,
(b) y = −1,
(c) x = 3.
(4) Let R be the region bounded by the graphs of y = x(x − 1)2 and the x-axis. Compute
the volume of the solid formed by revolving R about the y-axis.
4
(5) Let R be the region bounded by the graphs of y = , x = 1, x = 4, y = 0. Compute
x
the volume of the solid formed by revolving R about the y-axis.
(6) Let R be the region bounded by the graphs of x = y 2 , y = 2, x = 0. Compute the
volume of the solid formed by revolving R about the x-axis.
(7) Let R be the region bounded by the graphs of y = x2 and y = 2 − x2 in the first
quadrant. Compute the volume of the solid formed by revolving R about the y-axis.
(8) Set up an integral for the volume of the solid that results when the region bounded by
the curve y = 3 + 2x − x2 , the x-axis, and the y-axis, is revolved about
(a) the x-axis,
(b) the y-axis,
(c) the line y = −1.
2.1. Summary. We close this section with a summary of strategies for computing volumes of
solids of revolution.
• Sketch the region to be revolved.
• Determine the variable of integration (x if the region has a well-defined top and bottom,
y if the region has well-defined left and right boundaries).
• Based on the axis of revolution and the variable of integration, determine the method
(disks or washers for x-integration about a horizontal axis or y-integration about a vertical axis, shells for x-integration about a vertical axis or y-integration about a horizontal
axis).
• Label your picture with the inner and outer radii for disks or washers; label the radius
and height for cylindrical shells.
• Set up the integral(s) and evaluate.
3. Arc length and surface area
3.1. Arc Length. Let f (x) be continuous on [a, b] and differentiable on (a, b). Our aim is
to find the length of the curve y = f (x), a ≤ x ≤ b. We begin by partitioning the interval
b−a
[a, b] into n equal pieces: a = x0 < x1 < · · · < xn = b, where xi − xi−1 = ∆x =
for
n
each i = 1, 2, · · · , n. Between each pair of adjacent points on the curve, (xi−1 , f (xi−1 )) and
APPLICATIONS OF THE DEFINITE INTEGRAL
5
(xi , f (xi )), we approximate the arc length si by the straight-line distance between the two
points. From the usual distance formula, we have
si ≈ d((xi−1 , f (xi−1) ), (xi , f (xi )) =
p
(xi − xi−1 )2 + [f (xi ) − f (xi−1 )]2 .
Since f is continuous on all of [a, b] and differentiable on (a, b), f is also continuous on the
subinterval [xi−1 , xi ] and is differentiable on (xi−1 , xi ). By the Mean Value Theorem, we then
have f (xi ) − f (xi−1 ) = f 0 (ci )(xi − xi−1 ), for some number ci ∈ (xi−1 , xi ). This gives us the
approximation
p
si ≈ (xi − xi−1 )2 + [f (xi ) − f (xi−1 )]2
p
= (xi − xi−1 )2 + [f 0 (ci )(xi − xi−1 )]2
p
p
= 1 + [f 0 (ci )]2 (xi − xi−1 ) = 1 + [f (ci )]2 ∆x.
Adding together the lengths of these n line segments, we get an approximation of the total arc
length,
n
X
p
1 + [f 0 (ci )]2 ∆x.
s≈
i=1
Notice that as n gets larger, this approximation should approach the exact arc length, that is,
s = lim
n
X
p
n→∞
1 + [f 0 (ci )]2 ∆x.
i=1
So, the arc length is given exactly by the definite integral:
ˆ bp
s=
1 + [f 0 (x)]2 dx,
a
whenever the limit exists.
Example 3.1.
(1) A cable is to be hung between two poles of equal height that are 20 feet
apart. It can be shown that such a hanging cable assumes the shape of a catenary, the
general form of which is y = a cosh xa = a2 (ex/a + e−x/a ). In this case, suppose that the
cable takes the shape of y = 5(ex/10 + e−x/10 ), for −10 ≤ x ≤ 10. How long is the cable?
3
(2) Find the arc length of the curve y = 31 (x2 + 2) 2 between 0 ≤ x ≤ 3.
2
3 43
(3) Find the length of the curve x = 16
y − 32 y 3 between 0 ≤ y ≤ 8.
3.2. Surface Area. One can easily show that the curved surface area of the right circular cone
of base radius r and slant height l is A = πrl,
6
APPLICATIONS OF THE DEFINITE INTEGRAL
and so the curved surface area of the frustum of the cone shown below is A = π(r1 + r2 )L.
Now, suppose that f is nonnegative and continuous on [a, b] and differentiable on (a, b). If
we revolve the graph of y = f (x) about the x-axis on the interval [a, b], we get the surface of
revolution seen below:
We partition [a, b] into n many pieces of equal size as we have done so many times. On each
subinterval, we can approximate the curve by the straight line segment joining the points
(xi−1 , f (xi−1 )) and (xi , f (xi )).
Notice that revolving this line segment around the x-axis generates the frustum of a cone. The
surface area of this frustum will give us an approximation to the actual surface area on the
interval [xi−1 , xi ]. First, observe that the slant height of this frustum is
p
Li = d((xi−1 , f (xi−1 )), (xi , f (xi ))) = (xi − xi−1 )2 + [f (xi ) − f (xi−1 )]2 ,
APPLICATIONS OF THE DEFINITE INTEGRAL
7
from the usual distance formula. Because of our assumptions on f , we can apply the Mean
Value Theorem, to obtain f (xi ) − f (xi−1 ) = f 0 (ci )(xi − xi−1 ), for some number ci ∈ (xi−1 , xi ).
This gives us
p
p
Li = (xi − xi−1 )2 + [f (xi ) − f (xi−1 )]2 = 1 + [f 0 (ci )]2 (xi − xi−1 ).
The surface area Si of that portion of the surface on the interval [xi−1 , xi ] is approximately the
surface area of the frustum of the cone,
p
Si ≈ π[f (xi ) + f (xi−1 )] 1 + [f 0 (ci )]2 ∆x
p
≈ 2πf (ci ) 1 + [f 0 (ci )]2 ∆x.
since if ∆x is small, f (xi ) + f (xi−1 ) ≈ 2f (ci ). Repeating this argument for each subinterval
[xi−1 , xi ], i = 1, 2, · · · , n, gives us an approximation to the total surface area S,
n
X
p
S≈
2πf (ci ) 1 + [f 0 (ci )]2 ∆x.
i=1
As n gets larger, this approximation approaches the actual surface area,
n
X
p
S = lim
2πf (ci ) 1 + [f 0 (ci )]2 ∆x.
n→∞
i=1
Recognizing this as the limit of a Riemann sum gives us the integral
ˆ b
p
S=
2πf (x) 1 + [f 0 (x)]2 dx.
a
whenever the integral exists.
Example 3.2.
(1) Find the area of the surface obtained by revolving the curve
√
y = 25 − x2 , −2 ≤ x ≤ 3 about the x-axis.
(2) Find the area of the surface obtained by revolving the curve y = x2 , 0 ≤ x ≤ 1 about
the y-axis.
(3) Find the area of the surface obtained by revolving the curve x = y 3 , 0 ≤ y ≤ 1 about
the y-axis.
(4) Find the area of the surface obtained by revolving the curve y = 6x, 0 ≤ x ≤ 1 about
the x-axis.
√
(5) Find the area of the surface obtained by revolving the curve y = 12 x2 − 1, 0 ≤ x ≤ 2 2
about the y-axis.
8
APPLICATIONS OF THE DEFINITE INTEGRAL
ˆ
Answers
320 2
15
Answers 1.1.
(1)
600 −
dx = 38400000.
8
0
ˆ R √
4πR3
π( R2 − x2 )2 dx =
(2)
.
3
−R
ˆ 4
√
π( x)2 dx = 8π.
Answers 1.2.
(1)
0
ˆ R √
4πR3
π( R2 − x2 )2 dx =
(2)
.
3
−R
ˆ 1
28π
(3)
π(x2 + 1)2 dx =
.
15
0
ˆ 4 p
9π
.
Answers 1.3.
(1)
π( 4 − y)2 dy =
2
1
ˆ 8
1
96π
(2)
π(y 3 )2 dy =
.
5
ˆ0 4 p
(3)
π( 4 − y)2 dy = 8π.
0
ˆ
1
p
π( 4y)2 dy = 2π.
Answers 1.4.
(1) (a)
0 2 !
ˆ 2
1 2
8π
π 12 −
x
.
(b)
dx =
4
5
0
!
2
ˆ 2
1 2
56π
π
2− x
.
(c)
− 12 dx =
4
15
0
ˆ 4 p
(2) (a)
π( 4 − y)2 dy = 8π.
2 2 ˆ0 2 1472π
2
(b)
π
4 − x − (−3) − 0 − 3
dx =
.
15
−2
2 ˆ 2 576π
2
2
π 7 − 7 − (4 − x )
dx =
(c)
.
5
−2
2 2 ˆ 4 p
p
(d)
π
3 − (− 4 − y) − 3 − 4 − y
dy = 64π.
0
2 ˆ 1 2π
2
(3) (a)
π x − x2
dx =
.
15
0
ˆ 2 √ 2 2 48π
(b)
π
8x − x2
dx =
.
5
0
2 2 ˆ 2 832π
2
(c)
π
6−x −1 − 2−1
dx =
.
15
−2
2 2 ˆ 2 p
32π
2
(d)
π
4 − y − (−1) − 0 − 1
dy =
+ 4π 2 .
3
−2
APPLICATIONS OF THE DEFINITE INTEGRAL
ˆ
1
π
2πx(x − x2 ) dx = .
Answers 2.1.
(1)
6
0
ˆ 2
π
2π(3 − x)(4 − x2 ) dx = = 64π.
(2)
6
−2 ˆ
1
10π
(3) (a)
2π(2 − y)((2 − y) − y) dy =
.
3
ˆ0 1
8π
2π(y − (−1))((2 − y) − y) dy =
.
(b)
3
0
ˆ 2
ˆ 1
2π(3 − x)(2 − x) dx = 4π.
2π(3 − x)x dx +
(c)
1
ˆ 1 0
π
2πx(x(x − 1)2 ) dx = .
(4)
15
ˆ0 4
4
2πx
(5)
dx = 24π.
x
ˆ1 2
(6)
2πy(y 2 ) dy = 8π.
ˆ0 1
(7)
2πx((2 − x2 ) − x2 ) dx = π.
0
ˆ 3
153π
π(3 + 2x − x2 )2 dx =
(8) (a) Disk Method:
.
5
0
ˆ 3
45π
.
2πx(3 + 2x − x2 ) dx =
(b) Shell Method:
2
0
2 ˆ 3
2
243π
2
dx =
π
3 + 2x − x − (−1) − 0 − (−1)
(c) Washer Method:
.
5
0
s
2
ˆ 10 x/10
e
e−x/10
Answers 3.1.
(1)
−
+ 1 dx = 10 e1 − 10 e−1 .
2
2
−10
ˆ 3q
(2)
(1 + x2 )2 dx = 12.
0
2
ˆ 8 s √
3 y
1
−√
(3)
+ 1 dy = 9.
3 y
4
0
r
ˆ 3 √
x2
Answers 3.2.
(1)
2π 25 − x2
+ 1 dx = 50π.
2
25
−
x
−2
√
r
ˆ 1
1
π 5 5π
√
(2)
2π y 1 +
dy = − +
.
4y
6
6
0
ˆ 1
p
√
π
(3)
2πy 3 1 + 9y 4 dy = (10 10 − 1).
27
ˆ0 1
√
√
(4)
2π6x 37 dx = 6π 37.
0
s
ˆ 3 p
1
52π
(5)
2π 2(y + 1) 1 +
dy =
.
2(y + 1)
3
−1
9