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Math1014 Calculus II
Arc Lengths and Areas of Surfaces of Revolution

• Divide the interval [a, b] into n subintervals of equal length by the subdivision points a = x0 < x1 < x2 < · · · < xn = b,
b−a
with ∆x = xi − xi−1 = n . The distance between the two points (xi−1 , f (xi−1 )) and (xi , f (xi )) on the graph of a
differentiable function y = f (x) over the interval [a, b] is given by
y
(xi , f (xi ))
! y = f (x)
(xi − xi−1 )2 + ( f (xi ) − f (xi−1 ))2 (xi−1 , f (xi−1 ))
!
= (xi − xi−1 )2 + [ f ′ (ci )(xi − xi−1)]2
!
= 1 + [ f ′ (ci )]2 ∆x x
a x1 xi−1 xi b

for some ci in [xi−1 , xi ] by the Mean Value Theorem.1 Passing to Riemann sums, and taking limit as n → ∞, the arc
length L of the graph of the function y = f (x) is defined as
" b! n
!
L= 1 + [ f ′ (x)]2 dx = lim ∑ 1 + [ f ′(ci )]2 ∆x
a n→∞
i=1

• Similarly, the arc length of the graph of a differentiable function x = g(y), where c ≤ y ≤ d, is given by
" d!
L= 1 + [g′(y)]2 dy .
c

• Note that by rotating the line segment joining the points (xi−1 , f (xi−1 )) and (xi , f (xi )), we have a surface of revolution
which is part of a cone. By cutting and unrolling the surface into the part of a circular sector bounded between two
circle arc with length ri−1 θ = 2π f (xi−1 ) and ri θ = 2π f (xi ) respectively, its area can be found as
ri
1 2 1 2 1
r θ − ri−1 θ = (ri−1 θ + ri θ )(ri − ri−1 )
2 i 2 2 ri−1
2π f (xi )
2π f (xi−1 )
f (xi−1 ) + f (xi )
! θ
2π · (xi − xi−1)2 + ( f (xi ) − f (xi−1 )2
2
!
≈ 2π f (ci ) 1 + [ f ′ (ci )]2 ∆x

for some ci in the short interval [xi−1 , xi ], where f (xi−1 ) ≈ f (ci ) ≈ f (xi )
by continuity.
• If the graph of y = f (x) is rotated about the x-axis to generate a surface of revolution, the area A of the surface is
defined by the integration process as
n ! " b !
A = lim ∑ 2π f (x) 1 + [ f ′ (ci )]2 ∆x = 2π f (x) 1 + [ f ′ (x))]2 dx
n→∞ a
i=1
#
Using the notation ds = 1 + [y′]2 dx, the area formula above is sometimes written as
" b
A= 2π f (x)ds .
a

#
• If the graph of x = g(y) over the y interval [c, d] is rotated about the x-axis, using the notation ds = 1 + [x′]2 dy
accordingly, the area of the surface of revolution thus obtained is given by
" d ! " d
A= 2π g(y) 1 + [g′(y)]2 dy = 2π g(y)ds .
c c
1 Recall that the Mean Value Theorem says that for a function f continuous on [a,b], differentiable on (a,b), there is a number c in (a,b) such that
f (b) − f (a) = f ′ (c)(b − a).
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Example-Exercise Find the arc length of the graph of y = x 2 , 0 ≤ x ≤ 4.
!
3 12
Since dy
# 8 y

dx = 2 x , and 1 + (y′ )2 = 1 + 94 x, the arc length of the graph is 7

9 3/2 (4
" 4$ %4 2 & '
9 4
L= 1 + x dx = · 1+ x
0 4 9 3 4 0 3

2
8 ) 3
*
= 10 2 − 1 1

27 x
−3 −2 −1 1 2 3 4 5 6
−1

2
We can also consider the curve as the graph of a function of y; i.e., x = y . 3
dx 2 −1/3 ! ! #
Then = 3y , and 1 + [ dx
dy ]2 = 1 + 9y42/3 = 3y11/3 9y2/3 + 4.
dy
Hence " 8 " 4 √ (4
1 1 %1 8 ) 3
! 2/3 *
L= 9y 2/3 + 4 dy u=y
= 9u + 4du = (9u + 4)3/2
= 10 2 −1
0 3y1/3 0 2 27 0 27

√ √
Example-Exercise Find the arc length of the graph y = ln(x − x2 − 1), for 1 ≤ x ≤ 2.
√
Note that ey = x − x2 − 1 =⇒ (ey − x)2 = x2 − 1, or e2y − 2xey = −1. Hence

e2y + 1 1 y √
x= = (e + e−y ), ln( 2 − 1) ≤ y ≤ 0
2ey 2
√ dx
Note that ln( 2 − 1) < 0, and dy = ey − ey . The arc length is

" 0 $ " 0
1 1 # 2y
L= √ 1 + (ey − e−y )2 dy = √ e + 2 + e−2y dy
ln( 2−1) 4 ln( 2−1) 2
" 0 0
1 1 y
! "
= √ (e + e−y ) dy
(ey + e−y )2 dy = √
ln( 2−1) 2 2 ln( 2−1)

1% (0 1% √ 1 (
= ey − e−y √ = − ( 2 − 1) + √ =1
2 ln( 2−1) 2 2−1
√
Example-Exercise Find the area of the surface obtained by rotating the graph of y = 4 − x2, 0 ≤ x ≤ 2, about the x-axis ;
i.e., a hemi-sphere. 2

y′ = −x(4 − x2)−1/2 ,
x 2 4 0
and 1 + (y′)2 = 1 + 4−x2 = 4−x2
" 2 " 2
# 2 -2
2 2

A= 2π 4 − x2 · √ dx = 4π dx = 8π
0 4 − x2 0 0

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Example-Exercise Find the surface area of a torus obtained by rotating the circle (x − R)2 + y2 = r2 , 0 < r < R, about the
y-axis. # dx y
Note x = R ± r2 − y2 , with = ∓# , Then
dy r − y2
2
) dx *2 r2
1+ = 2 . y

dy r − y2
The area of the torus is given by
" r # r x
A= 2π (R − r 2 − y2 ) · # dy
−r r − y2
2

" r
# r
+ r 2 − y2 ) · #
2π (R + dy
−r r − y2
2
" r
1 % y (r
= 4π Rr # dy = 4π Rr sin−1 = 4π 2Rr = (2π R)(2π r)
−r 2
r −y 2 r −r
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Exercise

1. Find the arc length of the graph of y = 2x3/2 , 0 ≤ x ≤ 4

x2
2. Express the arc length of the ellipse + y2 = 1 as an integral. Do not evaluate the integral.
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3. Express the area of the surface of revolution by an definite integral:

(a) rotating y = cos 2x over the interval [0, π /6]about the x-axis;
(b) rotating y = 1 − x2 over the interval [0, 1] about the y-axis.
(c) How about rotating the curve y = 1 − x2 over the interval [0, 1] about the line x = 3?