This document discusses the surface area of surfaces of revolution. It begins by defining a surface of revolution as a surface formed by rotating a curve around an axis. It then presents the general surface area integral formula for computing the area of a surface of revolution. The document provides examples of applying the formula to compute the surface areas of rotating a semi-circular arc and a parabolic arc around different axes. It also contains worked examples integrating with respect to x and y.

1. Chapter 8: Further Applications of Integration
Section 8.2: Area of a Surface of Revolution
Alea Wittig
SUNY Albany

2. Outline
Surface Area Integral Formula
Examples

3. Surface Area Integral Formula

4. Surface of Revolution
▶ A surface of revolution is formed when a curve is rotated
about a line.
▶ A surface is ’hollow’ on the inside, unlike a solid of revolution.

5. Surface Area of a Circular Cylinder
eg, To compute the surface
area of a circular cylinder
with base radius r and
height h we could cut the
cylinder down the side to
obtain a rectangle of length
2πr and width h.
Surface Area = length · width
= 2πr · h

6. Surface Area of a Circular Cone
▶ Take circular cone with base radius r and slant height ℓ.
▶ Cut it down the slant to obtain a sector of the circle of radius l
and arc length 2πr.
▶ the area of the circle of radius ℓ is πℓ2
.
▶ the area of the sector is 2πr
2πℓ
times the area of the circle of
radius ℓ
πℓ2
·
2πr
2πℓ
= πrℓ

7. Area of a Surface of Revolution
▶ For a more complicated surface of revolution:
▶ Approximate the curve with
a polygonal path of n line
segments like in 8.1
▶ Rotate the approximated
path.
▶ The resulting approximated
surface is made up of
bands.
▶ Sum the areas of the bands
to approximate surface
area.
▶ Take the limit n → ∞ we
get the exact surface area.

8. Area of a Band
▶ Each band is a portion of a
circular cone.
▶ Below we find the formula
for the area of a band:
A = πr2(ℓ1 + ℓ)
| {z }
area of whole cone
− πr1ℓ1
| {z }
area of small cone
= πr2ℓ1 + πr2ℓ − πr1ℓ
= π(r2 − r1)ℓ1 + πr2ℓ
▶ Using similar triangle we have
ℓ1
r1
=
ℓ1 + ℓ
r2

9. Area of a Band
ℓ1
r1
=
ℓ1 + ℓ
r2
=⇒
r2ℓ1 = r1ℓ1 + r1ℓ =⇒
(r2 − r1)ℓ1 = r1ℓ =⇒
Area of band = π(r2 − r1)ℓ1 + πr2ℓ
= πr1ℓ + πr2ℓ
= π(r1 + r2)ℓ
A = 2πrl
where r = r1+r2
2 , the average radius of the band.

10. Area of a Surface of Revolution
▶ For the ith band has length
ℓi = |Pi−1Pi |,
ℓi =
q
(△x)2 + (△yi )2
and the average radius is
ri =
yi−1 + yi
2
▶ So the area of the ith band is
Ai = 2πri ℓi
= 2π
yi−1 + yi
2
·
q
(△x)2 + (△yi )2

11. Area of a Surface of Revolution
▶ As we take larger and larger n, we have yi ≈ yi−1 and so
Ai = 2π
yi−1 + yi
2
·
q
(△x)2 + (△yi )2
≈ 2πyi
q
(△x)2 + (△yi )2
= 2πyi
s
1 +
△yi
△xi
2
△x
= 2πf (x∗
i )
q
1 + [f ′(x∗
i )]2△x
for some x∗
i in (xi−1, xi ) by MVT.
▶ Now taking the limit as n → ∞
lim
n→∞
n
X
i=1
Ai = lim
n→∞
n
X
i=1
2πf (x∗
i )
q
1 + [f ′(x∗
i )]2△x
=
Z b
a
2πf (x)
q
1 + [f ′(x)]2dx

12. Surface Area Formulas (4) and (5)
▶ Now we have that the area of the surface obtained by rotating
a positive curve y = f (x), a ≤ x ≤ b with continuous
derivative about the x − axis is
S =
Z b
a
2πf (x)
q
1 + [f ′(x)]2dx (4)
▶ In Liebniz notation this is
S =
Z b
a
2πy
r
1 +
dy
dx
2
dx (5)
▶ Note: for equation (5) we plug in y = f (x) so the whole
integral is in terms of x.

13. Surface Area Formulas (6) and (7)
▶ If the curve is described as x = g(y), c ≤ y ≤ d then
S =
Z d
c
2πy
s
1 +
dx
dy
2
dy (6)
▶ Note: for equation (6) we leave y as y so the whole integral is
in terms of y.
▶ Formulas (5) and (6) can be summarized as
S =
Z
2πyds (7)
where we can use either
ds =
r
1 +
dy
dx
2
dx
| {z }
if using (5)
or ds =
s
1 +
dx
dy
2
dy
| {z }
if using (6)

14. Surface Area Formula (8) for Rotation about y-axis
▶ For rotation about the y axis we use
S =
Z
2πxds (8)

15. Examples

16. Example 1
1/2
The curve y =
√
4 − x2, −1 ≤ x ≤ 1, is an arc of the circle
x2 + y2 = 4. Find the area of the surface obtained by rotating this
curve about the x-axis.

17. Example 1
2/2
▶ We use formula (5) A =
Z b
a
2πy
r
1 +
dy
dx
2
dx since
rotation is about the x-axis and the curve is given in the form
y = f (x).
y =
p
4 − x2,
dy
dx
=
−2x
2
√
4 − x2
=
−x
√
4 − x2
1 +
dy
dx
2
= 1 +
x2
(4 − x2)
=
4 − x2 + x2
4 − x2
=
4
4 − x2
A =
Z 1
−1
2π
p
4 − x2
r
4
4 − x2
dx
=
Z 1
−1
4πdx
= 8π

18. Example 2
1/2
The portion of the curve x = 2
3y
3
2 between y = 0 and y = 3 is
rotated about the x − axis. Find the area of the resulting surface.

19. Example 2
2/2
▶ We use formula (6) A =
Z d
c
2πy
s
1 +
dx
dy
2
dy since we
are rotating about the x-axis and the curve is given in the
form x = g(y).
x =
2
3
y
3
2 ,
dx
dy
=
√
y
Z 3
0
2πy
s
1 +
dx
dy
2
dy =
Z 3
0
2πy
p
1 + ydy
=
Z 4
1
2π(u − 1)
√
udu
= 2π
Z 4
1
u3/2
− u1/2
du
= 2π
2
5
u5/2
−
2
3
u3/2

22. 4
1
=
232π
15

23. Example 3
1/4
The arc of the parabola y = x2 from (1, 1) to (2, 4) is rotated
about the y − axis. Find the area of the resulting surface
a. integrating with respect to x
b. integrating with respect to y.

24. Example 3
2/4
▶ Rotation is about y-axis so use formula (8) A =
Z
2πxds
a. ds =
r
1 +
dy
dx
2
dx to integrate wrt x
b. ds =
r
1 +
dx
dy
2
dy to integrate wrt y.
y = x2
,
dy
dx
= 2x
A =
Z 2
1
2πx
p
1 + 4x2dx
| {z }
u=1+4x2, du=8xdx
u1=5, u2=17
=
2π
8
Z 17
5
√
udu
=
π
4
2u3/2
3

27. 17
5
=
π
6
(17
√
17 − 5
√
5)

28. Example 3
3/4
▶ Now to integrate wrt y we also need to write x as a function
of y.
x =
√
y,
dx
dy
=
1
2
√
y
A =
Z 4
1
2π
√
y
s
1 +
1
4y
dy
=
Z 4
1
2π
r
y +
1
4
dy
| {z }
u=y+ 1
4
, du=dy
u1=5
4
, u2=17
4

29. Example 3
4/4
= 2π
Z 17/4
5/4
√
udu
= 2π
2u3/2
3

32. 17/4
5/4
=
4π
3
17
√
17
8
−
5
√
5
8
=
π
6
(17
√
17 − 5
√
5) ✓