2. LENGTHS OF CURVES
To find the length of the arc of the curve y = f (x) between x =
a and x = b let ds be the length of a small element of arc so
that:
( ) ( ) ( )
2
222
1thus
+≅+≅
dx
dy
dx
ds
dydxds
3. LENGTHS OF CURVES
In the limit as the arc length ds approaches zero:
and so: 2
1
ds dy
dx dx
= + ÷
2
1
b
x a
b
x a
ds
s dx
dx
dy
dx
dx
=
=
=
= + ÷
∫
∫
⇒
4. LENGTHS OF CURVES – PARAMETRIC EQUATIONS
Instead of changing the variable of the integral as before when
the curve is defined in terms of parametric equations, a special
form of the result can be established which saves a deal of
working when it is used.
Let: ( ) ( ) ( ) ( ) ( )
dt
dt
dy
dt
dx
sand
dt
dy
dt
dx
dt
ds
0dtasso
dt
dy
dt
dx
dt
ds
so
dydxdsbeforeAs.tFxandtfy
t
tt 1
∫
=
+
=
+
=
→
+
=
+≅==
2 2222
222
222
5. LENGTH OF AN ARC
A. Rectangular Coordinates
f(x)y,1
2
=
+= dx
dx
dy
ds g(y)x,dy
dy
dx
1ds
2
=
+=
B. Parametric Form
dt
dt
dy
dt
dx
ds
22
+
= when x=x(t), y=y(t); where t is a parameter
1. 2.
C. Polar Coordinates
f(r),dr
dr
d
r1ds
2
2
=θ
θ
+= )g(r,d
d
dr
rds
2
2
θ=θ
θ
+=2.1.
6. EXAMPLE
Find the length of the arc of each of the following:
2
3
3
3
ty
ttx
=
−=1.
from t=o to t=1
2.
tsiney
tcosex
t
t
=
=
from t=o to t=4
5. Length of the arc of the semicircle
222
ayx =+
2xto1xfrom
e
e
lny x
x
==
+
−
=
1
1
3.
4.
),(to),(from
xy
2100
8 23
=
7. AREA OF SURFACE OF REVOLUTION
• DEFINITION:
Let y = f(x) have a continuous derivative on the interval [a, b].
The area S of the surface of revolution formed by revolving
the graph of f about a horizontal or vertical axis is
where r(x) is the distance between the graph of f and the axis
of revolution.
[ ] xoffunctionaisydx)x('f)x(rS
b
a
→+π= ∫
2
12
8. If x = g(y) on the interval [c, d], then the surface area is
where r(y) is the distance between the graph of g and the axis of
revolution.
[ ] yoffunctionaisxdy)y('g)y(rS
d
c
→+π= ∫
2
12
9. EXAMPLE
1. Find the area formed by revolving the graph of f(x) = x3
on
the interval [0,1] about the x-axis.
2. Find the area formed by revolving the graph of f(x) = x2
on
the interval [0, ] about the y – axis.
3. Find the area of the surface generated by revolving the
curve
, 1 ≤ x ≤ 2 about the x – axis.
4. The line segment x = 1 – y, 0 ≤ y ≤ 1, is revolved about the y
– axis to generate the cone. Find its lateral surface area.
2
xy 2=
10. EXAMPLE
1. Find the area formed by revolving the graph of f(x) = x3
on
the interval [0,1] about the x-axis.
2. Find the area formed by revolving the graph of f(x) = x2
on
the interval [0, ] about the y – axis.
3. Find the area of the surface generated by revolving the
curve
, 1 ≤ x ≤ 2 about the x – axis.
4. The line segment x = 1 – y, 0 ≤ y ≤ 1, is revolved about the y
– axis to generate the cone. Find its lateral surface area.
2
xy 2=