Lecture_06_05.nb
6.5 - Areas of Surfaces of Revolution and
the Theorems of Pappus
Introduction
Suppose we rotate some curve about a line to obtain a surface, we can use a definite integral to calculate the area of the
surface.
Defining Surface Area
The simplest case to consider is a cylinder, the surface area is S = 2 p r h.
Consider a frustrum, which is obtained by rotating a line segment (that is not parallel to the axis of rotation) about some
line.
1
Lecture_06_05.nb
2
r1 + r2
The surface area of a frustrum is S = 2 p ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅ h = pHr1 + r2 L h where r1 is inner radius, r2 is the outer radius, and h is
2
slant length.
Now, consider the solid obtained rotating a curve C about some line. For the sake of simplicity, consider the curve
y = f HxL between x = a and x = b that is rotated about the x-axis.
y
x
a
b
As before, partition the interval @a, bD to obtain x0 = a, x1 , x2 , . . . , xn = b. We will use a frustrum to approximate the
surface area of each segment. Consider the kth segment, between xk-1 and xk . The surface area is
f Hxk-1 L + f Hxk L
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
Frustrum surface area = 2 p ÿ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
2
"################
#################
HD xk L2 + HD yk L2
= p ÿ H f Hxk-1 L + f Hxk LL
"################
#################
HD xk L2 + HD yk L2
D yk
%%%%%%%%%%%%%%%%
HD xk L2 J1
= p ÿ H f Hxk-1 L + f Hxk LL $%%%%%%%%%%%%%%%%
+ I ÅÅÅÅ
ÅÅÅÅÅÅ%%%%%%%
M N
D xk
2
Since f is differentiable on @a, bD, it is differentiable on every subinterval of @a, bD, thus by the Mean Value Theorem, there
is a number ck in @xk-1 , xk D such that
f Hxk L - f Hxk-1 L
D yk
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅ
ÅÅÅÅÅÅ
f ' Hck L = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
xk - xk-1
D xk
Thus,
Frustrum surface area = p ÿ H f Hxk-1 L + f Hxk LL
"##########################
1 + H f ' Hck LL2 D xk
The approximation of the total surface area is given by
S º ⁄ p ÿ H f Hxk-1 L + f Hxk LL
n
"##########################
1 + H f ' Hck LL2 D xk
k=1
We can obtain a better and better approximation by letting nض.
Lecture_06_05.nb
3
DEFINITION
Surface Area for Revolution About the x-axis
If the function f HxL ¥ 0 is continuously differentiable on @a, bD, the area of the surface generated by revolving the curve
y = f HxL about the x-axis is
S=·
b
d y %%%%
1 + I ÅÅÅÅ
2 p y $%%%%%%%%%%%%%%%%
ÅÅÅÅ M „ x =
dx
2
Ÿa 2 p f HxL
b
"##########################
1 + H f ' HxLL2 „ x
a
ü Example
Find the area of the surface generated by revolving the curve y =
è!!!
15
x between ÅÅÅÅ34 § x § ÅÅÅÅ
ÅÅ about the x-axis.
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Revolution About the y-axis
DEFINITION
Surface Area for Revolution About the y-axis
If the function x = gHyL ¥ 0 is continuously differentiable on @c, dD, the area of the surface generated by revolving the
curve x = gHyL about the y-axis is
S=·
d
d x %%%%
1 + I ÅÅÅÅ
2 p x $%%%%%%%%%%%%%%%%
ÅÅÅÅ M „ y =
dy
2
Ÿc
d
2 p gHyL
"##########################
1 + Hg ' HyLL2 „ y
c
ü Example
Find the area of the surface generated by revolving the region curve x = y + 2
è!!!!
y between 1 § y § 2 about the y-axis.
Lecture_06_05.nb
5
Parameterized Surfaces
We can also consider the case where x and y are given by a set of parametric equations. If the curve is parameterized by the
set of equations x = f HtL and y = gHtL, a § t § b, where f and g are continuously differentiable on @a, bD, then we have the
following.
Surface Area for Revolution for Parameterized Curves
If a smooth curve x = f HtL, y = gHtL, 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)
S=·
b
d y%%%%%%
dx
H ÅÅÅÅ
2 p y $%%%%%%%%%%%%%%%%
ÅÅÅÅ L +%%%%%%%%
I ÅÅÅÅ
ÅÅÅÅ M „ t =
dt
dt
2
2
a
2.
Revolution about the y-axis (x ¥ 0)
S=·
b
d y%%%%%%
dx
H ÅÅÅÅ
2 p x $%%%%%%%%%%%%%%%%
ÅÅÅÅ L +%%%%%%%%
I ÅÅÅÅ
ÅÅÅÅ M „ t =
dt
dt
2
2
Ÿa 2 p gHtL
"################
#####################
H f ' HtLL2 + Hg ' HtLL2 „ t
Ÿa 2 p f HtL
#####################
"################
H f ' HtLL2 + Hg ' HtLL2 „ t
b
b
a
ü Example
Find the area of the surface generated by revolving the curve x = ÅÅÅÅ23 t3ê2 , y = 2
è!!
è!!!
t , 0 § t § 3 about the x-axis.
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6
The Differential Form
The equations
S=·
b
2
d y %%%%
1 + I ÅÅÅÅ
2 p y $%%%%%%%%%%%%%%%%
ÅÅÅÅ M „ x and S = ·
dx
a
d
d x %%%%
1 + I ÅÅÅÅ
2 p x $%%%%%%%%%%%%%%%%
ÅÅÅÅ M „ y
dy
2
c
are often written in terms of d s =
############
"################
H„ xL2 + H„ yL2 as
S = Ÿa 2 p y „ s and S = Ÿc 2 p x „ s
b
d
In the first integral, y is the distance from the x-axis to an element of arc length ds. Similarly, for the second integral, x is
the distance from the y-axis to an element of arc length ds. Thus, both integrals are of the form
S =
Ÿ 2 p HradiusL Hband widthL = Ÿ 2 p r „ s
where r is the radius from the axis of revolution to an element of arc length ds.
Cylindrical Versus Conical Bands
Why not use cylindrical instead of conical (frustrum) bands? If we revolve the curve y = f HxL, a § x § b about the x-axis,
we obtain the integral
S = Ÿ 2 p f HxL „ s
b
a
This formula fails to give results consistent with the surface area formulas from geometry, which is what we want
(consistency).
The Theorems of Pappus
THEOREM 1
Pappus's Theorem for Volumes
If a plane region is revolved once about a line in the plane that does not cut through the region's interior, then the
volume of the solid it generates is equal to the region's area times the distance traveled by the region's centroid during
the revolution. If r is the distance from the axis of revolution to the centroid, then
V = 2p rA
Lecture_06_05.nb
THEOREM 2
Pappus's Theorem for Surface Areas
If an arc of a smooth plane curve is revolved once about a line in the plane that does not cut through the arc's interior,
then the area of the surface generated by the arc equals the length of the arc times the distance traveled by the arc's
centroid during the revolution. If r is the distance from the axis of revolution to the centroid, then
S = 2p rL
ü
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