Section 6.3 Lecture Slides

1. Chapter 6: Applications of Integration
Section 6.3: Volumes by Cylindrical Shells
Alea Wittig
SUNY Albany

2. Outline
Cylindrical Shells
Volume of a Cylindrical Shell
Examples
Cylindrical Shells Vs Washers
Example

3. Cylindrical Shells

4. ▶ Consider the problem of finding the volume of the solid
obtained by rotating about the y−axis the region bounded by
y = 2x2 − x3 and y = 0.
▶ To use the washer/disk
method with a vertical axis
of rotation, we would need
to find the inner and outer
radii by determining the left
and right curves.
▶ But how would we
determine xR and xL?
▶ xR and xL in this case correspond to the same function,
y = 2x2 − x3, which additionally is very difficult to write as a
function of y (g(y) = x).
▶ Too hard!

5. ▶ Idea: Form n rectangles "parallel" to the axis of rotation,
which in this example is a vertical line, specifically, the y-axis.
▶ We take "vertical" rectangles in the sense that the rectangle
has width △x and height y = f (x).
▶ Rotating about the y−axis we get a new type of solid called a
cylindrical shell.

6. ▶ The shape of a shell versus washer is different in the following
sense:
▶ A cylindrical shell looks like
a paper towel roll.
▶ The thickness of the wall of
the roll corresponds to the
width of the rectangle
being rotated (△x or △y).
▶ A washer looks more like,
well, a washer.
▶ The thickness of the washer
corresponds to the width of
the rectangle being rotated
(△x or △y).

7. Volume of a Cylindrical Shell

8. ▶ The volume of a cylindrical
shell is obtained by
subtracting the inner volume
V1 from the outer volume V2
V1 = πr2
1 h
V2 = πr2
2 h
Vshell = V2 − V1
= π(r2
2 − r2
1 )h
= π(r2 + r1)(r2 − r1)h
= 2π
(r2 + r1)
2
(r2 − r1)h
▶ Let △r = r2 − r1
▶ Let r denote the average value of r1 and r2, ie,
r =
(r2 + r1)
2

9. Vshell = 2π
(r2 + r1)
2
(r2 − r1)h
= 2πr · △r · h
▶ When we take n → ∞, the
thickness △r will become
arbitrarily close to 0.
▶ So the wall of the shell is
going to become arbitrarily
thin like a thin sheet of
paper rolled up.
▶ So we can think of r as just
the radius and 2πr as the
circumference of the shell.
Vshell = 2πr
|{z}
circumference
· h
|{z}
height
· △r
|{z}
thickness

10. ▶ Notice that if we unfurl the cylindrical shell, we have
something like a rectangular box with dimensions
length = 2πx
width = △x
height = f (x)
So the volume is
V = length · height · width
= 2πx
|{z}
circumference
· f (x)
|{z}
height
· △x
|{z}
thickness

11. ▶ Now we want to sum up the volumes of the shells to
approximate the volume of the solid so
V ≈
n
X
i=1
Vi =
n
X
i=1
2πri hi △r where n = number of shells
▶ We intuitively expect that as we take n → ∞ we have
V = lim
n→∞
n
X
i=1
2πri hi △r
If, as in the first example, we have a solid obtained by rotating
about the y-axis the region under the curve y = f (x) from a
to b, then ri = xi , hi = f (xi ), and △r = △x so that the
volume of the solid is
V =
Z b
a
2πxf (x)dx where 0 ≤ a < b

12. Examples

13. Example 1
Now let’s use cylindrical shells to solve that first problem:
Find the volume of the solid obtained by rotating the region
between y = 2x2 − x3 and y = 0 about the y−axis.
▶ From the sketch we see
that the height of a shell is
f (x) = 2x2 − x3, the radius
is x, and the curve
intersects y = 0 (the
x−axis) at x = 0 and
x = 2.
V =
Z 2
0
2π x
|{z}
radius
(2x2
− x3
)
| {z }
height
dx

14. Example 1
V =
Z 2
0
2πx(2x2
− x3
)dx
= 2π
Z 2
0
x(2x2
− x3
)dx
= 2π
Z 2
0
2x3
− x4
dx
= 2π
2x4
4
−
x5
5

17. 2
0
= 2π
x4
2
−
x5
5

20. 2
0
= 2π
24
2
−
25
5
−
*0
04
2
−
05
5
= 2π
8 −
32
5
= 2π
40
5
−
32
5
= 2π
8
5
=
16π
5

21. Remark
▶ In general, to use cylindrical shells we shouldn’t rely on the
formula
V =
Z b
a
2πxf (x)dx
because we won’t always have that type of geometry.
▶ Rather, we should remember the more general formula
V =
Z
2πr
|{z}
circumference
h
|{z}
height
dr
|{z}
thickness
where dr is
▶ dx if axis of rotation is a vertical line
▶ dy if axis of rotation is a horizontal line
▶ And we must find the radius r, the height h, and the limits of
integration by sketching the graph.

22. Example 2
Find the volume of the solid obtained by rotating about the y−axis
the region between y = x and y = x2.

23. Example 2
2/3
▶ Not shown in the image but the curves intersect when
x2 = x =⇒ x2 − x = 0 =⇒ x(x − 1) = 0 =⇒ x = 0, 1.
▶ To get cylindrical shells we form rectangles parallel to the
axis of rotation. In this case the axis of rotation is vertical (the
y-axis). So we form rectangles of width △x.
▶ A sample rectangle has width △x and height yt − yB = x − x2.
▶ Rotating the rectangle we get a cylindrical shell of
thickness=△x, height =x − x2 and radius= x.
▶ Note that △x will become dx in the integrand.

24. Example 2
3/3
V =
Z 1
0
2πx
|{z}
circumference
(x − x2
)
| {z }
height
dx
|{z}
thickness
=
Z 1
0
2πx(x − x2
)dx
= 2π
Z 1
0
x2
− x3
dx
= 2π
x3
3
−
x4
4

27. 1
0
= 2π
1
3
−
1
4
−
0 − 0
!
= 2π
4
12
−
3
12
= 2π
1
12
=
π
6

28. Example 3
1/3
Use cylindrical shells to find the volume of the solid obtained by
rotating about the x−axis the region under the curve y =
√
x from
0 to 1.

29. Example 3
2/3
▶ The axis of rotation is horizontal so we take horizontal
rectangles, ie, rectangles of width △y.
▶ Since △y → dy in the integrand, this tells us we will be
integrating with respect to y.
▶ A sample rectangle has width △y and height = 1 − y2.
▶ Rotating the rectangle we get a cylindrical shell of thickness =
△y, height = 1 − y2 and radius = y.
▶ at x = 0, y =
√
0 = 0 and at x = 1, y =
√
1 = 1 so we are
integrating from y = 0 to y = 1.

30. Example 3
3/3
V =
Z 1
0
2πy
|{z}
circumference
(1 − y2
)
| {z }
height
dy
|{z}
thickness
= 2π
Z 1
0
y(1 − y2
)dy
= 2π
Z 1
0
y − y3
dy
= 2π
y2
2
−
y4
4

35. 1
0
= 2π
1
2
−
1
4
−
0 − 0
!
= 2π
1
4
=
π
2

36. Example 4
1/4
Find the volume of the solid obtained by rotating the region
bounded by y = x − x2 and y = 0 about the line x = 2.

37. Example 4
2/4
▶ First we must graph the curve y = x − x2.
▶ It’s graph is a transformation of the parabola y = x2.
▶ To see the transformation(s) we first complete the square.
y = x − x2
= −(x2
− x)
= − x2
− x +
−1
2(−1)
!2!
+
−1
2(−1)
!2
= −(x2
− x +
1
4
) +
1
4
= −(x −
1
2
)2
+
1
4
▶ So the transformation is:
▶ Shift right 1
2 unit.
▶ Shift up 1
4 unit.
▶ Flip over the x axis.

38. Example 4
3/3

39. Example 4
4/4
V =
Z 1
0
2π(2 − x)
| {z }
circumference
(x − x2
)
| {z }
height
dx
|{z}
thickness
= 2π
Z 1
0
(2 − x)(x − x2
)dx
= 2π
Z 1
0
2x − 2x2
− x2
+ x3
dx
= 2π
Z 1
0
2x − 3x2
+ x3
dx
= 2π
2x2
2
−
3x3
3
+
x4
4
!

44. 1
0
= 2π
1 − 1 +
1
4
−
0 − 0 + 0
=
π
2

45. Cylindrical Shells Vs Washers

46. How do we know when to use the cylindrical shell method or
the washer/disk method?
▶ Is the region more easily described by top and bottom
boundary curves of the form y = f (x), or by left and right
boundaries curves of the form x = g(y)?
▶ top/bottom - then integration will be wrt x and rectangles will
be drawn vertically.
▶ left/right - then integration will be wrt y and rectangles drawn
horizontally and int wrt y.
▶ Draw a sample rectangle in the region. Imagine the rectangle
revolving; it becomes either a disk/washer or a shell.

47. Example

48. Example 5
1/5
Find the volume of the solid
obtained by rotating the region
bounded by y = x2 and y = 2x
about the line x = −1 using
a. x as the variable of
integration and
b. y as the variable of
integration

49. Example 5
2/5
a. Integrating with respect to x we have got to take rectangles of
width △x because this will become dx in the integrand..
▶ This is shown in image (b).
▶ Rotating the vertical rectangle about a vertical line we get a
cylindrical shell.
b. Integrating with respect to y we have got to take rectangles of
width △y because this will become dy in the integrand.
▶ This is shown in image (c).
▶ Rotating the horizontal rectangle about a vertical line we
get a washer.

50. Example 5
3/5
▶ Now set up and compute the integrals on your own as an
exercise. You should get the same volume of course whether
you are using washers or shells.

51. Example 5
4/5
a. Integrating wrt x → rectangles of width △x → cylindrical
shells (since axis of rotation is vertical)
V =
Z b
a
[circumference] · [height] · [thickness]
=
Z 2
0
2π(x + 1)(2x − x2
)dx
= 2π
Z 2
0
2x2
− x3
+ 2x − x2
dx
= 2π
Z 2
0
x2
+ 2x − x3
dx
= 2π
x3
3
+
2x2
2
−
x4
4

56. 2
0
= 2π
8
3
+ 4 −
16
4
= 2π
8
3
+ 4 − 4
=
16π
3

57. Example 5
5/5
b. Integrating wrt y → rectangles of width △y → washers (since
axis of rotation is vertical)
V =
Z d
c
π[r2
outer − r2
inner]dy
=
Z 4
0
π[(1 +
√
y)2
− (1 +
y
2
)2
]dy
= π
Z 4
0
1 + 2
√
y + y − (1 + y +
y2
4
)dy
= π
Z 4
0
2
√
y −
y2
4
dy
= π
2y
3
2
3
2
−
y3
12

60. 4
0
= π
4y
3
2
3
−
y3
12

63. 4
0
= π
4
√
64
3
−
64
12
=
32π
3
−
16π
3
=
16π
3