Application of definite integrals

Prepared by :
Vaibhav Tandel(160330109xxx)
From : Electrical department
Subject : Calculus
Mahatma Gandhi Institute Of Technical
Education And Research Center

Area of a Plane Region Volume Arc Length and Surface Area
Calculus
Applications Of The Deﬁnite Integral (I)

Outline
1 Area of a Plane Region
Volume
Volume by Slicing
The Method of Disks
Volumes by Cylindrical Shells
Arc Length and Surface Area
Arc Length
Surface Area
Parametric Equations
2
3

Area of a Plane Region (I)
Let f (x) (f (x) > 0) be a continuous
function deﬁned on [a, b]. Consider
the partition
{xi |xi = x0 + i∆x, i = 0, 1, ·, n}
On each subinterval [xi−1 , xi ], we
construct a rectangle of height f (ci ),
for some ci [x∈ i−1 , xi ], as indicated in
Figure 5.1.
Figure: [5.1] Approximation
of area.
where x0 = a and x =
(b − a)
n
.

Area of a Plane Region (II)
We take the sum of the areas of the n
rectangles as an approximation of the
area A under the curve:
n
A≈ ∫
i=1
f (ci )∆x.
The exact area under the curve is
given by
n b
A = lim ∫n→∞
f (ci )∆x = ∫
i=1 a
f (x) dx. Figure: [5.1] Approximation
of area.

Area of a Plane Region (III)
In this section, we extend the use of
the deﬁnite integral to compute areas
ever so slightly.
First, suppose that we have two
continuous functions,
f and g and f (x) ≥ g(x)
for all x on [a, b]. We would like to ﬁnd
the area bounded between the two
curves y = f (x) and y = g(x) on the
interval [a, b] (see Figure 5.2). Figure: [5.2] Area between
two curves.

Area of a Plane Region (IV)
As before, we partition the interval
[a, b] into n subintervals of equal size,
b−a
∆x =, so that the points in the
n
partition are
xi = a + i ∆x, for i = 0, 1, 2, . . . , n.
Now, on each subinterval [xi−1 , xi ], we
construct a rectangle, stretching from
the lower curve y = g(x) to the upper
curve y = f (x), as shown in Figure
5.3a.
Figure: [5.3a] Approximate
area.

Area of a Plane Region (V)
Referring to Figure 5.3b, the ith
rectangle has height hi = f (ci ) − g(ci ),
for some ci [x∈ i−1 , xi ].
So, the area of the ith rectangle is
Area = length × width
= hi ∆x
= [f (ci ) − g(ci )] ∆x.
Figure: [5.3a] Approximate
area.

Area of a Plane Region (VI)
The total area is approximated by the
sum of the areas of the n rectangles,
n
A≈
i=1
[f (ci ) − g(ci )] ∆x.
If the limit as n → ∞ exists, we then
get the exact area. Finally, recognize
the limit of Riemann sums as a
deﬁnite integral:
n
A =
=
n→∞
b
lim [f (ci ) − g(ci )] ∆x
i=1 Figure: [5.3a] Approximate
area.
[f (x) − g(x)] dx.
a

Area of a Plane Region (VII)
REMARK 1.1
The formula
b
A=
a
f (x) − g(x) dx.
is valid only when f (x) ≥ g(x) on the interval [a, b].
In general, the area between y = f (x) and y = g(x) for a ≤ x ≤ b
is given by
b
|f (x) − g(x)| dx.
a

Area of a Plane Region (VIII)
Notice that to evaluate
b
|f (x) − g(x)| dx,
a
we must
d
evaluate
c
d
f (x) − g(x) dx on all subintervals where
f (x) ≥ g(x), then
evaluate
c
[g(x) − f (x)] dx on all subintervals where
g(x) ≥ f (x) and ﬁnally,
add the integrals together.

Finding the Area between Two Curves
Example (1.1)
Find the area bounded by the
graphs of y = 3 − x and
y = x2 − 9 (see Figure 5.4).
Figure: [5.4] y = 3 − x and
y = x2 − 9.

Finding the Area between Two Curves That Cross
Example (1.2)
graphs of y = x2 and y = 2 − x2
for 0 ≤ x ≤ 2.
Figure: [5.5] y = x2 and
y = 2 − x2 .

A Case Where the Intersection Points Are Known Only
Approximately
Example (1.3)
graphs of y = cos x and y = x2 .
Figure: [5.6] y = cos x and y = x2 .

The Area of a Region Determined by Three Curves
Example (1.4)
Find the area bounded by the graphs of y = x2 , y = 2 − x and
y = 0.
Figure: [5.7a] y = x2 and
y = 2 − x.
Figure: [5.7b] y = x2 and
y = 2 − x.

Compute Area by Integrating with Respect to y (I)
Observe that if we turn the Figure 5.7
sideways, it will look like a region with
a single curve determining each of the
upper and lower boundaries.
This observation,
treating the left and right
boundaries of the region as
functions of y,
leads to an alternative way to compute
the area of this type of region . Figure: [5.7a] y = x2 and
y = 2 − x.

Compute Area by Integrating with Respect to y (II)
Consider two continuous functions, f
and g of y, where f (y) ≥ g(y) for all y
on the interval c ≤ y ≤ d.
To ﬁnd the area bounded between
x = f (y) and x = g(y), we partition the
interval, [c, d] into n subintervals of
d−c
equal width ∆y =(see Figure
n
5.8a). The points in the partition is
denoted by
yi = c + i ∆y, for i = 0, 1, 2, . . . , n.
Figure: [5.8a] Area between
x = g(y) and x = f (y).

Compute Area by Integrating with Respect to y (III)
On each subinterval [yi−1 , yi ] (for
i = 1, 2, . . . , n), we construct a
rectangle, of width wi = [f (ci ) − g(ci )],
for some ci [y∈ i−1 , yi ]. The area of the
ith rectangle is then given by
Area = length × width
= [f (ci ) − g(ci )] ∆y.
The total area between the two curves
is then given approximately by the
Riemann sum
n
A≈
i=1
[f (ci ) − g(ci )] ∆y. Figure: [5.8b] Area of ith
rectangle.

Compute Area by Integrating with Respect to y (IV)
As we have done several times now,
we get the exact area by taking the
limit as n → ∞ and recognizing the
limit of Riemann sums as a deﬁnite
integral. We get
n
A =
=
n→∞
d
lim [f (ci ) − g(ci )] ∆y
i=1
[f (y) − g(y)] dy.
c
Figure: [5.8a] Area between
x = g(y) and x = f (y).

An Area Computed by Integrating with Respect to y
Example (1.5)
Repeat example 1.4, but
integrate with respect to y
instead.
Figure: [5.9] y = x2 and y = 2 − x.

The Area of a Region Bounded by Functions of y
Example (1.6)
graphs of x = y2 and
x = 2 − y2 .
Figure: [5.10] x = y2 and
x = 2 − y2 .

Estimating the Energy Lost by a Tennis Ball (I)
In collisions between a tennis racket and ball, the ball changes
shape, ﬁrst compressing and then expanding.
Let
x represent how far the ball is compressed, where
0 ≤ x ≤ m and
f (x) represent the force exerted on the ball by the racket.
Then, the energy transferred is proportional to the area under
the curve y = f (x).

Estimating the Energy Lost by a Tennis Ball (II)
Suppose that
fc (x) is the force during compression of the ball and
fe (x) is the force during expansion of the ball.
Energy is transferred to the ball during compression and
transferred away from the ball during expansion, so that the
energy lost by the ball in the collision (due to friction) is
proportional to
m
[fc (x) − fe (x)] dx.
0
The percentage of energy lost in the collision is then given by
100
m
0 [fc (x) − fe (x)] dx
.
m
0 fc (x) dx

Estimating the Energy Lost by a Tennis Ball (III)
Example (1.7)
Suppose that test measurements provide the following data on
the collision of a tennis ball with a racket. Estimate the
percentage of energy lost in the collision.
x (in.)
fc (x) (lb)
fe (x) (lb)
0.0
0
0
0.1
25
23
0.2
50
46
0.3
90
78
0.4
160
160

Estimating the Energy Lost by a Tennis Ball (IV)
Figure: [5.11] Force exerted on a tennis ball.

Volumes of Solids
As we shall see throughout this chapter, the integral is an
amazingly versatile tool.
In section 5.1, we saw how to use deﬁnite integrals to compute
area, a principal measure of the size of a two-dimensional
region.
In this section, we see how to use integrals to compute volume,
a principal measure of the size of a three-dimensional solid.

Cylinder (I)
You already know how to compute
some volumes. For instance, the
building shown in Figure 5.12a is
essentially a rectangular box, whose
volume is given by
lwh,
where
l is the length,
w is the width and
h is the height.
Figure: [5.12a]

Cylinder (II)
The right circular cylinders seen in the
buildings in Figure 5.12b have volume
given by
πr2 h,
where
h is the height and
r is the radius of the circular
cross section.
Figure: [5.12b]

Volume (I)
Notice in each case that the building has a familiar cross
section (a rectangle in Figure 5.12a and a circle in Figure
5.12b) that is extended vertically.
We call any such solid a cylinder (any solid whose cross
sections perpendicular to some axis running through the solid
are all the same).

Volume (II)
Now, notice the connection between the volume formulas for
these two cylinders. The volume of a right circular cylinder is
V= (πr )
cross-sectional area
× h
height
.
Likewise, in the case of a box, we have
V = (length × width) × height.
cross-sectional area
In general, the volume of any cylinder is found by
V = (cross-sectional area) × (height).
2

Volume (III)
To compute the volume of a pyramid or dome, we need to
extend this formula to solids that do not have constant
cross-sectional areas.
Figure: [5.13a] Figure: [5.13b]

Volumes by Slicing (I)
For any solid that extends from x = a
to x = b, we partition the interval [a, b]
on the x-axis into n subintervals of
b − a
equal width ∆x = n . As usual, we
denote
xi = a + i∆x, for i = 0, 1, 2, . . . , n.
We then slice the solid perpendicular
to the x-axis at each of the (n − 1)
points x1 , x2 , . . . , xn−1 (see Figure
5.14a). Figure: [5.14a] Sliced solid.

Volumes by Slicing (II)
Notice that if n is large, then each
slice of the solid will be thin and the
cross-sectional area is nearly
constant throughout any given slice.
Let A(x) denote the area of the cross
section corresponding to a value of x.
Observe that the slice between
x = xi−1 and x = xi is nearly a cylinder
(see Figure 5.14b). Figure: [5.14b] ith slice of
solid.

Volumes by Slicing (III)
So, for any point ci in the interval
[xi−1 , xi ], the areas of the cross
sections on that interval are all
approximately A(ci ).
Hence, an approximation to the
volume Vi of the ith slice is given by
the
Vi ≈ A(ci ) ∆x ,
Figure: [5.14b] ith slice of
solid.
cross-sectional area width
where ∆x is the width of the slice.

Volumes by Slicing (IV)
we can approximate the total volume
V of the solid by
n
V≈
i=1
A(ci ) ∆x.
The exact volume is then given by
n
V =
=
n→∞
b
lim A(ci ) ∆x,
i=1
A(x) dx.
a
Figure: [5.14b] ith slice of
solid.

Computing Volume from Cross-Sectional Areas
Example (2.1)
The Pyramid Arena in
Memphis (pictured in the
margin) 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.
Figure: The Pyramid Arena in
Memphis.

Estimating Volume from Cross-Sectional Data
Example (2.2)
In medical imaging, such as CT (computerized tomography) and MRI
(magnetic resonance imaging) processes, numerous measurements
are taken and processed by a computer to produce a
three-dimensional image that resembles the tissue the physician
wishes to study. The process is similar to the slicing process we have
used to ﬁnd the volume of a solid. In this case, however,
mathematical representations of various slices of the tissue are
combined to produce a three-dimensional image that can be
displayed and sliced back apart for a physician to determine the
health of the tissue. Suppose that an MRI scan indicates that the
cross-sectional areas of adjacent slices of a tumor are given by the
values in the following table.x (cm)
A(x) (cm2 )
0
0.0
0.1
0.1
0.2
0.4
0.3
0.3
0.4
0.6
0.5
0.9
0.6
1.2
0.7
0.8
0.8
0.6
0.9
0.2
1.0
0.1
Estimate the volume of the tumor.

Computing the Volume of a Dome
Example (2.3)
Suppose that a dome has
circular cross sections, with
outline y = ± 45 (90 − x) for
2
0 ≤ x ≤ 90. (in units of feet,
this gives dimensions similar
to the Capitol Dome in Figure
5.13b. Agraph of this sideways
parabola is shown in Figure
5.15c). Find the volume of the
dome. Figure: [5.15c]

The Method of Disks (I)
Example 2.3 can be generalized to the method of disks used
to compute the volume of a solid formed by revolving a
two-dimensional region about a vertical or horizontal line. We
consider this general method next.

The Method of Disks (II)
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
(see Figures 5.16a and 5.16b).
Figure: [5.16a] y = f (x) ≥ 0.
Figure: [5.16b] Solid of
revolution.

Using the Method of Disks to Compute Volume
Example (2.4)
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.
√
Figure: [5.17a] y =
√
x. Figure: [5.17b] Solid of
revolution.

The Method of Disks (V)
The volume of the solid is then given by
d
V=
c
π[g(y)]2
cross-sectional area = πr2
dy.
Figure: [5.18a] Revolve about the Figure: [5.18b] Solid of
revolution.

Using the Method of Disks with y as the Independent
Variable (I)
Example (2.5)
Find the volume of the solid resulting from revolving the region
bounded by the curve y = 4 − x2 from x = 0 to x = 2 and the
x-axis about the y-axis.

Using the Method of Disks with y as the Independent
Variable (II)
Figure: [5.19a] y = 4 − x2 .
revolution.

The Method of Washers
There are two complications that can be found in the types of
volume calculations we have been studying.
The first of these is that you may need to compute the
volume of a solid that has a cavity or “hole” in it.
The second of these occurs when a region is revolved
about a line other than the x-axis or the y-axis.
In either case, if you look carefully at the figures, these will not
present you with any significant difficulties.
We illustrate these ideas in examples 2.6 and 2.7.

Computing Volumes of Solids with and without
Cavities (I)
Example (2.6)
Let R be the region bounded by the graphs of y = x2 /4, x = 0
and y = 1. Compute the volume of the solid formed by revolving
R about
1
2
3
the y-axis,
the x-axis and
the line y = 1.

Cavities (II)
Figure: [5.20a] x =
√
4y.
revolution.

Cavities (III)
Figure: [5.21a] x =
√
4y.
Figure: [5.21b] Solid with cavity.

Cavities (IV)
Figure: [5.22a] Revolve about
y = 2.
revolution.

The Method of Washers (II)
In parts (b) and (c) of example 2.6, the volume was computed
by subtracting an inner volume from an outer volume in order to
compensate for a cavity inside the solid.
This technique 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.

Revolving a Region about Different Lines (I)
Example (2.7)
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:
1
2
the line y = −3 and
the line x = 3.

Revolving a Region about Different Lines (II)
y = −3.
revolution.

Revolving a Region about Different Lines (III)
x = 3. Figure: [5.24b] Solid of
revolution.

Volumes by Cylindrical Shells (I)
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] (see Figure 5.25a). If we revolve this region about the
y-axis, we get the solid shown in Figure 5.25b.
y-axis.
revolution.

Volumes by Cylindrical Shells (II)
We ﬁrst partition the interval [a, b] into
n subintervals of equal
with On each subinterval∆x =b-a
. [xi−1 , xi ], pick a point ci and
construct
the rectangle of height f (ci ) as
indicated in Figure 5.26a.
Figure: [5.26a] ith rectangle.
n

Volumes by Cylindrical Shells (III)
Revolving this rectangle about the
y-axis forms a thin cylindrical shell
(i.e., a hollow cylinder, like a soda
can), as in Figure 5.26b.
Figure: [5.26b] Cylindrical
shell.

Volumes by Cylindrical Shells (IV)
To ﬁnd the volume of this thin cylindrical shell, imagine cutting
the cylinder from top to bottom and then ﬂattening out the shell.
After doing this, we should have a thin rectangular sheet, as
seen in Figure 5.26c
Figure: [5.26c] Flattened cylindrical shell.

Volumes by Cylindrical Shells (VI)
The total volume V of the solid can then be approximated by
the sum of the volumes of the n cylindrical shells:
n
Vi ≈
i=1
2π ci f (ci ) ∆x .
radius height thickness
We can get the exact volume of the solid by taking the limit as
n → ∞ and recognizing the resulting deﬁnite integral. We have
n b
V = lim
n→∞
2πci f (ci ) ∆x =
i=1 a
2π x f (x) dx .
radius height thickness

A Volume Where Shells Are Simpler Than Washers (I)
Example (2.8)
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.

A Volume Where Shells Are Simpler Than Washers (II)
Figure: [5.27a] Typical rectangle
generating a cylindrical shell. Figure: [5.27b] Solid of
revolution.

Computing Volumes Using Shells and Washers (I)
Example (2.9)
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
1
y = 2,
y = −1 and
x=3
Figure: [5.28a] y = x and
y=2−x
2
3

Computing Volumes Using Shells and Washers (II)
Figure: [5.28b] Revolve about y = 2.

Computing Volumes Using Shells and Washers (III)
Figure: [5.28c] Revolve about y = −1.

Computing Volumes Using Shells and Washers (IV)
Figure: [5.28d] Revolve about x = 3.

Volume of A Solid of Revolution
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.

Arc Length (I)
In the ﬁrst part of this section, we develop the notion of ﬁnding
the arc length of the curve y = f (x) on the interval [a, b].
Here, we’ll assume that f is continuous on [a, b] and
differentiable on (a, b).
As usual, we partition the interval [a, b] into n equal pieces:
a = x0 < x1 < · · · < xn = b, where
xi − xi−1
for each i = 1, 2, . . . , n.
= ∆x = b-a
n

Arc Length (II)
Between each pair of adjacent points
on the curve, (xi−1 , f (xi−1 )) and
(xi , f (xi )), we approximate the arc
length si by the straight-line distance
between the two points (see Figure
5.30).
Figure: [5.30] Straight line
approximation of arc length.

Arc Length (III)
From the usual distance formula, we
have
si ≈ d{(xi−1 , f (xi−1 )), (xi , f (xi ))}
= (xi − xi−1 )2 + [f (xi ) − f (xi−1 )]2 .
Since f is continuous and
differentiable on [a, b], f is also
continuous and differentiable on the
subinterval [xi−1 , xi ]. Recall that by the
Mean Value Theorem, we have
f (xi ) − f (xi−1 ) = f (ci )(xi − xi−1 ),
for some number ci (x∈ i−1 , xi ).
Figure: [5.30] Straight line
approximation of arc length.

Arc Length (IV)
This gives us the approximation
si ≈
=
=
=
(xi − xi−1 )2 + [f (xi ) − f (xi−1 )]2
(xi − xi−1 )2 + [f (ci )(xi − xi−1 )]2
1 + [f (ci )]2 (xi − xi−1 )
∆x
1 + [f (ci )]2 ∆x.
Adding together the lengths of these n line segments, we get an
approximation of the total arc length,
n
s ≈
i=1
1 + [f (ci )]2 ∆x.

Arc Length (V)
Notice that as n gets larger, this approximation
n
s ≈
i=1
1 + [f (ci )]2 ∆x
should approach the exact arc length, that is,
n
s = lim
n→∞
1 + [f (ci )]2 ∆x.
i=1
We should recognize this as the limit of a Riemann sum for
1 + [f (x)]2 , so that the arc length is given exactly by the
deﬁnite integral:
b
s=
a
1 + [f (x)]2 dx,

Using the Arc Length Formula
Example (3.1)
Find the arc length of the portion of the curve y = sin x with
0 ≤ x ≤ π. (Recall that we estimated this as 3.79 in our
introductory example.)

Estimating an Arc Length
Example (3.2)
Find the arc length of the portion of the curve y = x2 with
0 ≤ x ≤ 1.

A Comparison of Arc Lengths of Power Functions
Example (3.3)
Find the arc length of the
portion of the curve y = x4 with
0 ≤ x ≤ 1 and compare to the
arc length of the portion of the
curve y = x2 on the same
interval.
Figure: [5.31] y = x2 and y = x4 .

Computing the Length of Cable Hanging between Two
Poles
Example (3.4)
A cable is to be hung between
two poles of equal height that
are 20 feet apart. In this case,
suppose that the cable takes
the shape of
y = 5(ex/10 + e−x/10 ),
−10 ≤ x ≤ 10, as seen in
Figure 5.32. How long is the
cable?
Figure: [5.32]
y = 5(ex/10 + e−x/10 ).

Surface Area (I)
In addition to computing the volume of
the solid, we often want to determine
the area of the surface that is
generated by the revolution.
For instance, when revolving the line
y = x + 1, for 0 ≤ x ≤ 1 about the
x-axis, the surface generated looks
like a megaphone with two open
ends, as shown in Figure 5.33.
Figure: [5.33] Surface of
revolution.

Surface Area (II)
The solid in Figure 5.33 is known as
the frustum of a cone (a right circular
cone whose top has been cut off by a
plane parallel to the base). For the
general frustum shown in Figure 5.34,
the curved surface area is given by
A = π(r1 + r2 )L,
where L is called the slant height.
Figure: [5.34] Frustum of a
cone.

Surface Area (III)
Returning to the original problem of
revolving the line y = x + 1 on the
interval [0, 1] about the x-axis (seen in
Figure 5.33), we have r1 = 1, r2 = √2
and L = 2 (from the Pythagorean
Theorem). The curved surface area is
then
A = π(1 + 2) √ 2 = 3π √2 ≈ 13.329.
revolution.

Surface Area (IV)
Since the average radius of the
circular cross sections is r = r1 + r2
the surface area can also be written
as
A = 2πrL.
revolution.
2

Surface Area (V)
For the general problem of ﬁnding the
curved surface area of a surface of
revolution, consider the case where
f (x) ≥ 0 and f is differentiable on the
interval (a, b).
Figure: [5.35a] Revolve
about x-axis.

Surface Area (VI)
If we revolve the graph of y = f (x) for a ≤ x ≤ b about the x-axis
(see Figure 5.35a), we get the surface of revolution shown in
Figure 5.35b.
x-axis.
Figure: [5.35b] Surface of
revolution.

Surface Area (VII)
To evaluate the surface area we proceed as follows:
1
Divide the surface into n small
pieces
Approximate the surface area of
each piece by the surface area of
the frustum of a cone.
Add together these approximate
surface areas and
Take the limit as n → ∞.
2
3
4
We then obtain the following formula.
b
S=
a
2πf (x) 1 + [f (x)]2 dx, Figure: [5.35b] Surface of
revolution.
whenever the integral exists.

Using the Surface Area Formula
Example (3.5)
Find the surface area of the surface generated by revolving
y = x4 , for 0 ≤ x ≤ 1, about the x-axis.

Arc Length of a Parametric Curve (I)
As we have for curves deﬁned by equations of the form
y = f (x), we can also derive formulas for arc length for the case
of curves described by parametric equations.
We present the formula for arc length here.
b
s=
a
[x (t)]2 + [y (t)]2 dt.

Arc Length of a Parametric Curve (II)
The formula for arc length:
b
s=
a
[x (t)]2 + [y (t)]2 dt.
Note: In the case where t represents time and the curve
represents the path of an object, the equation has a nice
interpretation.
[x (t)]2 + [y (t)]2 is the speed of the object
[x (t)]2 + [y (t)]2 dt is the product of the speed and the
time, which gives distance.

Arc Length of a Parametric Curve (III)
Example (3.6)
Find the arc length of the
curve deﬁned by the
parametric equations
x(t) = cos 5t
,
y(t) = sin 7t
for 0 ≤ t ≤ 2π.
Figure: [5.36] A Lissajous curve.

Arc Length for a Sliding Ladder
Example (3.7)
An 8-foot-tall ladder leans
vertically against a wall. The
bottom of the ladder then
slides along the ﬂoor, with the
top remaining in contact with
the wall, until the ladder rests
on the ﬂoor. Find the distance
traveled by the midpoint of the
ladder. Figure: [5.37] A sliding ladder.

Application of definite integrals

Application of definite integrals

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Application of definite integrals