MATH1120 Calculus II
Notes 3
September 17, 2013
1. Applications of Definite Integrals
1.1. The washer method for solids of revolution.
Exercise 1.1. Find the volume of an oblique cone of base radius r and height h. (Hint: you
may use Cavalieri’s principle to compare the oblique cone with another solid with known
volume. Alternatively, you may integrate the cross-sectional areas to get the volume.)
1.2. The shell method for solids of revolution. An alternative method of obtaining
the volume of solid of revolution is by means of cylindrical shells, i.e. decomposing the
solid into various thin layers of cylinders and integrating their (infinitesimal) volumes.
Theorem 1.1 (Shell method). The volume of the solid obtained by revolving the region
bounded by the graph of y = f (x), the x-axis, and the vertical lines x = a and x = b
(assuming that either 0 < a < b or a < b < 0, about the y-axis is
Z b
V = 2π
|x||f (x)|dx
a
In general, if the solid is obtained by revolving some general region bounded by x = a and
x = b about a vertical line x = L, then
Z b
(Shell radius)(Shell height)dx
V = 2π
a
Remark 1.2.
(1) If you are to use shell method to compute the volume of solid of
revolution whose axis of symmetry is x-axis(resp. y-axis), you should integrate
with respect to y(resp. x).
(2) There are quite a number of variants of the formula given in the theorem. For
example, suppose a solid is obtained by revolving the region bounded by the graphs
of y = f (x) and y = g(x), a ≤ x ≤ b around the vertical line x = c, where c is not
inside the interval [a, b]. Then the volume of the solid is
Z b
2π|x − c||f (x) − g(x)|dx
a
Example 1.3. Find the volume of the solid obtained by revolving the region bounded by
the graph of y = −x2 + 2x, the x-axis, x = 0 and x = 2 around the y-axis using (1)washer
method, and (2)shell method.
1
2
Solution:
(1)
y = −x2 + 2x =⇒ x = 1 ±
1
Z
V =π
0
p
p
1 − y =⇒ x2 = 2 − y ± 2 1 − y
1
p
3
2
8π
4 1 − ydy = 4π − (1 − y) 2 =
3
3
0
(2)
Z
V = 2π
0
2
2
4
x
2x3
8π
x · (−x2 + 2x)dx = 2π − +
=
4
3 0
3
Exercise 1.2. Find the volume of the jello as in Exercise 2.3 in Notes 2 using shell
method. Verify that the answer you obtained is the same as the one you found using
washer method. (Hint: With respect to which variable should we integrate? What are the
integration limits?
1.3. Some rules of thumb.
(1) Disk/Washer method If a solid is obtained by revolving a certain region around
a line parallel to x-axis (resp. y-axis), then one should integrate with respect to
x(resp. y) to get the volume.
(2) Shell method If a solid is obtained by revolving a certain region around a line
parallel to the x-axis(resp. y-axis), one should integrate with respect to y(resp. x)
to get the volume.
How can we decide which method is quicker? Let
(1) The region to be revolved be bounded by graphs of explicit functions of a(where a
can be x or y).
(2) The axis of rotation is parallel to the b-axis(where b can be x or y).
Then
(1) if a = b, then washer/disk method is better.
(2) if a 6= b, then shell method is better.
In any case, drawing a picture of the solid with a typical washer/shell always helps. Note
that
(1) Washers are always perpendicular to the axis of rotation.
(2) The heights of the thin cylindrical shells are always parallel to the axis of rotation.
Remark 1.4. If the graph is described by some implicit functions, where there is no
explicit independent variable, or the region concerned is bounded by graphs of functions
with different independent variables, then the above rule of thumb may not help you decide
which method is better.
3
Example 1.5. [Sect 6.1, Ex. 55] Solution:
Z a
p
p
V =π
((b + a2 − y 2 )2 − (b − a2 − y 2 )2 )dy
Z−a
a
p
=π
4b a2 − y 2 dy
−a
2
πa
= 4bπ
2
= 2π 2 a2 b
Example 1.6 (Sect. 6.1, Ex. 55). Redo it using the method of cylindrical shells.
Solution:
Since the axis of rotation is vertical, we should integrate with respect
√ to x to find the
volume. Note that the shell radius= b − x, while the shell height= 2 a2 − x2 .
Z a p
Volume = 2π
2 a2 − x2 (b − x)dx
Z−aa p
Z a p
2
2
x a2 − x2 dx
= 4πb
a − x dx − 4π
−a
−a
a
1 2
2 32
2 2
= 2π ba − 4π − (a − x )
3
−a
= 2π 2 ba2
Exercise 1.3. Suppose a solid is obtained by revolving the region bounded by the graphs
of y = −x2 + 6x − 8 and y = x − 2, 2 ≤ x ≤ 4, around the line y = −1. Write down
the integrals which compute the volume of the solid using (1)washer method, and (2)shell
method. Explain how you set up the integrals. DO NOT evaluate the integrals.
1.4. Arc Length.
Theorem 1.7. If f 0 is continuous on [a, b], then the arc length of the curve y = f (x) from
x = a to x = b is
s
2
Z b
dy
L=
1+
dx
dx
a
Z yp
sec4 t − 1dt,
Example 1.8 (Sect. 6.3, Ex. 9). Find the arc length of the curve x =
0
−π/4 ≤ y ≤ π/4.
Solution:
Z
π/4
s
1+
−π/4
dx
dy
2
Z
π/4
dy =
sec2 ydy
−π/4
π/4
= [tan y]−π/4
=2
4
1.5. Areas of surfaces of revolution.
Theorem 1.9. If f (x) ≥ 0 and f 0 is continuous on [a, b], the area of the surface obtained
by revolving the graph y = f (x), a ≤ x ≤ b, about the x-axis is
Z b
p
S = 2π
f (x) 1 + (f 0 (x))2 dx
a
Discussion: Use Theorem 1.9 to show that the surface area of a sphere of√radius a
is 4πa2 . (Hint: Such a sphere can be obtained by revolving the graph of y = a2 − x2 ,
−a ≤ x ≤ a around the x-axis)
Exercise 1.4 (Sect. 6.4, Ex. 21).
Exercise 1.5 (Sect. 6.4, Ex. 23).