Areas of Surfaces of Revolution
P. Sam Johnson
April 10, 2017
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
1/29
Overview
When you jump rope, the rope sweeps out a surface in the space
around you called a surface of revolution.
The “area” of this surface depends on the length of the rope and
the distance of each of its segments from the axis of revolution.
In this lecture, we define areas of surfaces of revolution.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
2/29
Defining Surface Area
Rotating the semicircle
y=
p
a2 − x 2
of radius a with center at the origin generates a spherical surface
with area 4πa2 .
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
3/29
Defining Surface Area
We begin by rotating horizontal and slanted line segments about
the x-axis.
If we rotate the horizontal line segment AB having length ∆x
about the x-axis, we generate a cylinder with surface area 2πy ∆x.
This area is the same as that of a rectangle with side lengths ∆x
and 2πy .
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
4/29
Defining Surface Area
The length 2πy is the circumference of the circle of radius y
generated by rotating the points (x, y ) on the line AB about the
x-axis.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
5/29
Defining Surface Area
Suppose the line segment AB has length ∆s and is slanted rather
than horizontal.
Now when AB is rotated about the x-axis, it generates a frustum
of a cone.
From classical geometry, the surface area of this frustum is
2πy ∗ ∆s
where
y1 + y2
2
is the average height of the slanted segment AB above the x-axis.
y∗ =
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
6/29
Defining Surface Area
This surface area is the same as that of a rectangle with side
lengths ∆s and 2πy ∗ .
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
7/29
Defining Surface Area
Let’s build on these geometric principles to define the area of a
surface swept out by revolving more general curves about the
x-axis.
Suppose we want to find the area of the surface swept out by
revolving the graph of a nonnegative continuous function
y = f (x), a ≤ x ≤ b,
about the x-axis.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
8/29
Defining Surface Area
We partition the closed interval [a, b] in the usual way and use the
points in the partition to subdivide the graph into short arcs.
The following figure shows a typical arc PQ and the band it
sweeps out as part of the graph of f .
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
9/29
Defining Surface Area
As the arc PQ revolves about the x-axis, the line segment joining
P and Q sweeps out a frustum of a cone whose axis lies along the
x-axis.
The surface area of this frustum approximates the surface area of
the band swept out by the arc PQ.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
10/29
Defining Surface Area
The surface area of the frustum of the cone shown below is 2πy ∗ L,
where y ∗ is the average height of the line segment joining P and
Q, and L is its length.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
11/29
Defining Surface Area
Since f ≥ 0, we see that the average height of the line segment is
y∗ =
f (xk−1 ) + f (xk )
2
and the slant length is
L=
P. Sam Johnson (NITK)
q
(∆xk )2 + (∆yk )2 .
Areas of Surfaces of Revolution
April 10, 2017
12/29
Defining Surface Area
q
f (xk−1 ) + f (xk )
Frustum surface area = 2π.
. (∆xk )2 + (∆yk )2
2
q
= π.(f (xk−1 ) + f (xk )).
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
(∆xk )2 + (∆yk )2 .
April 10, 2017
13/29
Defining Surface Area
The area of the original surface, being the sum of the areas of the
bands swept out by arcs like arc PQ, is approximated by the
frustum area sum
n
X
q
π.(f (xk−1 + f (xk )). (∆xk )2 + (∆yk )2 .
(1)
k=1
We expect the approximation to improve as the partition of [a, b]
becomes finer.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
14/29
Defining Surface Area
Moreover, if the function f is differentiable, then by the Mean
Value Theorem, there is a point (ck , f (ck )) on the curve between
P and Q where the tangent is parallel to the segment PQ. At this
point,
∆yk
f 0 (ck ) =
∆xk
∆yk = f 0 (ck ) ∆xk .
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
15/29
Defining Surface Area
With this substitution for ∆yk , the sums in Equation (1) take the
form
n
X
q
π.(f (xk−1 + f (xk )). (∆xk )2 + (f 0 (ck ) ∆xk )2
k=1
=
n
X
π.(f (xk−1 + f (xk )).
q
1 + (f 0 (ck ))2 ∆xk .
(2)
k=1
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
16/29
Defining Surface Area
These sums are not the Riemann sums of any function because the
points xk−1 , xk , and ck are not the same.
However, a theorem from advanced calculus assures us that as the
norm of the partition of [a, b] goes to zero, the sums in Equation
(2) converges to the integral
Z
b
q
2πf (x) 1 + (f 0 (x))2 dx.
a
We therefore define this integral to be the area of the surface
swept out by the graph of f from a to b.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
17/29
Surface Area for Revolution About the x-Axis
Definition
If the function
f (x) ≥ 0
is continuously differentiable on [a, b], the area of the surface
generated by revolving the curve
y = f (x)
about the x-axis is
r
Z b
Z b
q
dy 2
dx =
2πf (x) 1 + (f 0 (x))2 dx. (3)
S=
2πy 1 +
dx
a
a
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
18/29
Surface Area for Revolution About the x-axis
Exercise
1. Find the area of the surface generated by revolving the curve
√
y = 2 x, 1 ≤ x ≤ 2,
about the x-axis.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
19/29
Surface Area for Revolution About the y -axis
For revolution about the y -axis, we interchange x and y in
Equation (3).
Definition
If x = g (y ) ≥ 0 is continuously differentiable on [c, d], the area of
the surface generated by revolving the curve
x = g (y )
about the y -axis is
s
Z
d
S=
2πx
c
P. Sam Johnson (NITK)
1+
dx 2
dy
Z
dy =
d
q
2πg (y ) 1 + (g 0 (y ))2 dy .
c
Areas of Surfaces of Revolution
April 10, 2017
20/29
Surface Area for Revolution About the y -axis
Exercise
2. The line segment
x = 1 − y,
0≤y ≤1
is revolved about the y -axis to generate the cone in the figure.
Find its lateral surface area (which excludes the base area).
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
21/29
Parametrized Curves
If the curve is parametrized by the equations
x = f (t) and y = g (t),
a ≤ t ≤ b,
where f and g are continuously differentiable on [a, b], then the
corresponding square root appearing in the arc formula is
r q
dx 2 dy 2
[f 0 (t)]2 + [g 0 (t)]2 =
+
.
dt
dt
This observation leads to the following formulas for area of
surfaces of revolution for smooth parametrized curves.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
22/29
Surface Area of Revolution for Parametrized Curves
If a smooth curve x = f (t), y = g (t), a ≤ t ≤ b, is traversed
exactly once as t increases from a to b, then the areas of the
surfaces generated by revolving the curve about the coordinate
axes are as follows.
1. Revolution about the x-axis (y ≥ 0) :
Zb
S=
2πy
r dx 2
dt
+
dy 2
dt
dt
a
2. Revolution about the y -axis (x ≥ 0) :
Zb
S=
2πx
r dx 2
dt
+
dy 2
dt
dt.
a
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
23/29
The Differential Formula
The equations
Zb
r
2πy
S=
1+
dy 2
dx
dx
a
and
Zd
S=
r
πx
1+
dy 2
dx
dy
c
are often
p written in terms of the arc length differential
ds = dx 2 + dy 2 as
Zb
S=
Zd
2πy ds
and S =
a
P. Sam Johnson (NITK)
2πx ds.
c
Areas of Surfaces of Revolution
April 10, 2017
24/29
The Differential Formula
In the first of these, y is the distance from the x-axis to an
element of arc length ds. In the second, x is the distance from the
y -axis to an element of arc length ds.
Both integrals have the form
Z
Z
S = 2π(radius)(band width) = 2πρ ds
where ρ is the radius from the axis of revolution to an element of
arc length ds.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
25/29
The Differential Formula
We should express the radius function ρ and the arc length
differential ds in terms of a common variable and supply limits of
integration for that variable.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
26/29
Exercises
3. Find the surface area of the frustum of the cone generated by
revolving the line segment
y=
x +1
,
2
1≤x ≤3
about y -axis. Check your result with the geometry formula.
(Hint: LSA of a frustum of a cone = π(r1 + r2 )∗slant height)
4. Find the area of the surface generated by revolving the
portion of the astroid
2
2
x 3 + y 3 = 1,
(y ≥ 0)
about the x-axis.
(Also try the same problem by considering the corresponding
parametric equations: x = cos 3 t, y = sin3 t, 0 ≤ t ≤ π)
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
27/29
Exercises
5. Show that the area of the surface generated by revolving the
line segment
r x, 0 ≤ x ≤ h
y=
h
about the x-axis, agrees with the formula obtained using
geometry.
6. Find the surface area of the double cone generated by
revolving the line segment
y = x,
−1 ≤ x ≤ 2
about the x-axis.
7. Prove that the curved surface area of a sphere of radius r
intercepted between two parallel planes at a distance a and b
from the center of the sphere is 2πr (b − a) when b > a and
hence deduce the surface area of the sphere.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
28/29
References
1. M.D. Weir, J. Hass and F.R. Giordano, Thomas’ Calculus,
11th Edition, Pearson Publishers.
2. R. Courant and F.John, Introduction to calculus and analysis,
Volume II, Springer-Verlag.
3. N. Piskunov, Differential and Integral Calculus, Vol I & II
(Translated by George Yankovsky).
4. E. Kreyszig, Advanced Engineering Mathematics, Wiley
Publishers.
P. Sam Johnson (NITK)
Areas of Surfaces of Revolution
April 10, 2017
29/29