2. In this section, we derive methods for calculating the
volumes of solids obtained from rotation in space.
Solids of Revolution
3. In this section, we derive methods for calculating the
volumes of solids obtained from rotation in space.
Solids of Revolution
The key for the calculation is to find the areas of the
surfaces formed by revolving line segments in space.
4. In this section, we derive methods for calculating the
volumes of solids obtained from rotation in space.
Solids of Revolution
The key for the calculation is to find the areas of the
surfaces formed by revolving line segments in space.
We start with the rules for finding the horizontal and
vertical distances, i.e. the coordinate-distances
between the end points of line segments.
5. In this section, we derive methods for calculating the
volumes of solids obtained from rotation in space.
Solids of Revolution
The key for the calculation is to find the areas of the
surfaces formed by revolving line segments in space.
We start with the rules for finding the horizontal and
vertical distances, i.e. the coordinate-distances
between the end points of line segments.
We then find the areas of the surfaces these line
segments formed in space as they move,
and we “integrate” these surfaces into volumes.
6. Let’s review the following important formulas about
distance and area.
The Real Line
7. Let’s review the following important formulas about
distance and area.
The Real Line
RL
Let L and R be two points on the real line as shown
where R is the point to the right and L is the point to
the left, then the distance h between them is R – L,
i.e. right – left. x
8. Let’s review the following important formulas about
distance and area.
The Real Line
RL
Let L and R be two points on the real line as shown
where R is the point to the right and L is the point to
the left, then the distance h between them is R – L,
i.e. right – left.
Distance h = R – L
x
9. Let’s review the following important formulas about
distance and area.
The Real Line
RL
Let L and R be two points on the real line as shown
where R is the point to the right and L is the point to
the left, then the distance h between them is R – L,
i.e. right – left.
Distance h = R – L
For the vertical y-axis,
the distance v between two points is H – L,
i.e. high – low.
y
H
L
Distance
v = H – L
x
10. Let’s review the following important formulas about
distance and area.
The Real Line
RL
Let L and R be two points on the real line as shown
where R is the point to the right and L is the point to
the left, then the distance h between them is R – L,
i.e. right – left.
Distance h = R – L
For the vertical y-axis,
the distance v between two points is H – L,
i.e. high – low. So given the following points
y
H
L
Distance
v = H – L
x
P, Q and R in the xy-plane,
P(a, b) Q(c, b)
R(a, d)
11. Let’s review the following important formulas about
distance and area.
The Real Line
RL
Let L and R be two points on the real line as shown
where R is the point to the right and L is the point to
the left, then the distance h between them is R – L,
i.e. right – left.
Distance h = R – L
For the vertical y-axis,
the distance v between two points is H – L,
i.e. high – low. So given the following points
y
H
L
Distance
v = H – L
x
P, Q and R in the xy-plane,
the horizontal distance between
P and Q is h = c – a,
P(a, b) Q(c, b)
R(a, d)
h = c – a
12. Let’s review the following important formulas about
distance and area.
The Coordinate Distances
RL
Let L and R be two points on the real line as shown
where R is the point to the right and L is the point to
the left, then the distance h between them is R – L,
i.e. right – left.
Distance h = R – L
For the vertical y-axis,
the distance v between two points is H – L,
i.e. high – low. So given the following points
y
H
L
Distance
v = H – L
x
P, Q and R in the xy-plane,
the horizontal distance between
P and Q is h = c – a, and
the vertical distance between
R and P is v = d – b.
R(a, d)
v = d – b
P(a, b) Q(c, b)
h = c – a
14. Given a point (x, y) that’s on
the graph of an equation,
y
x
(x, y)
Equations and Coordinate-Distances
15. Given a point (x, y) that’s on
the graph of an equation,
y = f(x) is vertical distance
to the x-axis,
y
x
y = f(x)
x
(x, y)
Equations and Coordinate-Distances
16. Given a point (x, y) that’s on
the graph of an equation,
y = f(x) is vertical distance
to the x-axis, and that
x = g(y) is its horizontal
distance to the y-axis.
y
x
x = g(y)
y = f(x)
y
x
(x, y)
Equations and Coordinate-Distances
17. Given a point (x, y) that’s on
the graph of an equation,
y = f(x) is vertical distance
to the x-axis, and that
x = g(y) is its horizontal
distance to the y-axis.
y
x
x = g(y)
y = f(x)
y
x
(x, y)
2x – y2 = 4
(x, y)
Equations and Coordinate-Distances
Example A. a. Given 2x – y2 = –4 for 0 ≤ y,
express y as f(x), and x as g(y). Draw.
18. Given a point (x, y) that’s on
the graph of an equation,
y = f(x) is vertical distance
to the x-axis, and that
x = g(y) is its horizontal
distance to the y-axis.
y
x
x = g(y)
y = f(x)
y
x
Solving for the vertical length y,
2x – y2 = –4 so y2 = 2x + 4, (x, y)
2x – y2 = 4
(x, y)
Equations and Coordinate-Distances
Example A. a. Given 2x – y2 = –4 for 0 ≤ y,
express y as f(x), and x as g(y). Draw.
19. Given a point (x, y) that’s on
the graph of an equation,
y = f(x) is vertical distance
to the x-axis, and that
x = g(y) is its horizontal
distance to the y-axis.
y
x
x = g(y)
y = f(x)
y
x
Solving for the vertical length y,
2x – y2 = –4 so y2 = 2x + 4,
since 0 ≤ y → y = f(x) = √2x + 4.
(x, y)
2x – y2 = 4
(x, y)
y = √2x + 4
Equations and Coordinate-Distances
Example A. a. Given 2x – y2 = –4 for 0 ≤ y,
express y as f(x), and x as g(y). Draw.
20. Given a point (x, y) that’s on
the graph of an equation,
y = f(x) is vertical distance
to the x-axis, and that
x = g(y) is its horizontal
distance to the y-axis.
y
x
x = g(y)
y = f(x)
y
Equations and Coordinate-Distances
x
Example A. a. Given 2x – y2 = –4 for 0 ≤ y,
express y as f(x), and x as g(y). Draw.
Solving for the vertical length y,
2x – y2 = –4 so y2 = 2x + 4,
since 0 ≤ y → y = f(x) = √2x + 4.
(x, y)
2x – y2 = 4
Solving for the horizontal length x,
2x – y2 = –4 so 2x = –4 + y2,
(x, y)
y = √2x + 4
x = (–4 + y2)/2
21. Given a point (x, y) that’s on
the graph of an equation,
y = f(x) is vertical distance
to the x-axis, and that
x = g(y) is its horizontal
distance to the y-axis.
y
x
x = g(y)
y = f(x)
y
Equations and Coordinate-Distances
x
Example A. a. Given 2x – y2 = –4 for 0 ≤ y,
express y as f(x), and x as g(y). Draw.
Solving for the vertical length y,
2x – y2 = –4 so y2 = 2x + 4,
since 0 ≤ y → y = f(x) = √2x + 4.
(x, y)
2x – y2 = 4
Solving for the horizontal length x,
2x – y2 = –4 so 2x = –4 + y2,
or that x = g(y) = (–4 + y2)/2.
(x, y)
y = √2x + 4
x = (–4 + y2)/2
23. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
h1
h2
2x – y2 = –4
3y = 2x
y
24. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
h1
h2
2x – y2 = –4
3y = 2x
(x=3y/2,y)((–4 + y2)/2,y)
y
25. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
we have that
h1 = 3y/2 – (y2 – 4)/2
h1
h2
2x – y2 = –4
3y = 2x
(x=3y/2,y)((–4 + y2)/2,y)
y
26. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
we have that
h1 = 3y/2 – (y2 – 4)/2
h1
h2
2x – y2 = –4
3y = 2x
(x=3y/2,y)
((–4 + y2)/2,y)
(3y/2,y)
((–4 + y2)/2,y)
y
27. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
we have that
h1 = 3y/2 – (y2 – 4)/2
h2 = (y2 – 4)/2 – 3y/2
h1
h2
2x – y2 = –4
3y = 2x
(x=3y/2,y)
((–4 + y2)/2,y)
(3y/2,y)
((–4 + y2)/2,y)
y
28. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
we have that
h1 = 3y/2 – (y2 – 4)/2
h2 = (y2 – 4)/2 – 3y/2
h1
h2
2x – y2 = –4
3y = 2x
v1
v2
2x – y2 = –4
3y = 2x
(x=3y/2,y)
((–4 + y2)/2,y)
(3y/2,y)
c. find the vertical lengths v1 and v2
as shown in terms of x’s.
((–4 + y2)/2,y)
y
x
29. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
we have that
h1 = 3y/2 – (y2 – 4)/2
h2 = (y2 – 4)/2 – 3y/2
h1
h2
2x – y2 = –4
3y = 2x
v1
v2
2x – y2 = –4
3y = 2x
(x=3y/2,y)
((–4 + y2)/2,y)
(3y/2,y)
(x,y = 2x/3)
(x,y =√2x+4)
c. find the vertical lengths v1 and v2
as shown in terms of x’s.
Labeling the y-coordinates,
in terms of x’s,
((–4 + y2)/2,y)
y
x
30. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
we have that
h1 = 3y/2 – (y2 – 4)/2
h2 = (y2 – 4)/2 – 3y/2
h1
h2
2x – y2 = –4
3y = 2x
v1
v2
2x – y2 = –4
3y = 2x
(x=3y/2,y)
((–4 + y2)/2,y)
(3y/2,y)
(x,y = 2x/3)
(x,y =√2x+4)
c. find the vertical lengths v1 and v2
as shown in terms of x’s.
Labeling the y-coordinates,
in terms of x’s, so
v1 = √2x + 4 – 2x/3
((–4 + y2)/2,y)
y
x
31. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
we have that
h1 = 3y/2 – (y2 – 4)/2
h2 = (y2 – 4)/2 – 3y/2
h1
h2
2x – y2 = –4
3y = 2x
v1
v2
2x – y2 = –4
3y = 2x
(x=3y/2,y)
((–4 + y2)/2,y)
(3y/2,y)
(x,y = 2x/3)
(x,y =√2x+4)
(x,y =√2x+4)
(x,y = 2x/3)c. find the vertical lengths v1 and v2
as shown in terms of x’s.
Labeling the y-coordinates,
in terms of x’s, so
v1 = √2x + 4 – 2x/3
((–4 + y2)/2,y)
y
x
32. Equations and Coordinate-Distances
Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
b. find the horizontal lengths h1 and h2
as shown in terms of y.
Labeling the x-coordinates,
in terms of y’s as shown,
we have that
h1 = 3y/2 – (y2 – 4)/2
h2 = (y2 – 4)/2 – 3y/2
h1
h2
2x – y2 = –4
3y = 2x
v1
v2
2x – y2 = –4
3y = 2x
(x=3y/2,y)
((–4 + y2)/2,y)
(3y/2,y)
(x,y = 2x/3)
(x,y =√2x+4)
(x,y =√2x+4)
(x,y = 2x/3)c. find the vertical lengths v1 and v2
as shown in terms of x’s.
Labeling the y-coordinates,
in terms of x’s, so
v1 = √2x + 4 – 2x/3
v2 = 2x/3 –√2x + 4
((–4 + y2)/2,y)
y
x
33. If we follow a line segment in space as it orbits around
an axis that's in the same plane, it forms the following
surfaces: rings/circles, cylinders, or rostrums.
Surfaces of Orbiting Lines
34. If we follow a line segment in space as it orbits around
an axis that's in the same plane, it forms the following
surfaces: rings/circles, cylinders, or rostrums.
We are interested in the following cases.
Surfaces of Orbiting Lines
35. If we follow a line segment in space as it orbits around
an axis that's in the same plane, it forms the following
surfaces: rings/circles, cylinders, or rostrums.
We are interested in the following cases.
Surfaces of Orbiting Lines
I. If the line segment L is
perpendicular to the axis of
rotation, it generates a ring.
L
36. If we follow a line segment in space as it orbits around
an axis that's in the same plane, it forms the following
surfaces: rings/circles, cylinders, or rostrums.
We are interested in the following cases.
Surfaces of Orbiting Lines
I. If the line segment L is
perpendicular to the axis of
rotation, it generates a ring.
L
L
37. If we follow a line segment in space as it orbits around
an axis that's in the same plane, it forms the following
surfaces: rings/circles, cylinders, or rostrums.
We are interested in the following cases.
Surfaces of Orbiting Lines
I. If the line segment L is
perpendicular to the axis of
rotation, it generates a ring.
Let R be the outer radius
and r be the inner radius of the rotation,
then the surface area formed
is π(R2 – r2).
L
L
38. If we follow a line segment in space as it orbits around
an axis that's in the same plane, it forms the following
surfaces: rings/circles, cylinders, or rostrums.
We are interested in the following cases.
Surfaces of Orbiting Lines
I. If the line segment L is
perpendicular to the axis of
rotation, it generates a ring.
Let R be the outer radius
and r be the inner radius of the rotation,
then the surface area formed
is π(R2 – r2).
L
L
If L is connecter to the
rotational-axis so r = 0,
L
L
39. If we follow a line segment in space as it orbits around
an axis that's in the same plane, it forms the following
surfaces: rings/circles, cylinders, or rostrums.
We are interested in the following cases.
Surfaces of Orbiting Lines
I. If the line segment L is
perpendicular to the axis of
rotation, it generates a ring.
Let R be the outer radius
and r be the inner radius of the rotation,
then the surface area formed
is π(R2 – r2).
L
L
L
If L is connecter to the
rotational-axis so r = 0, then
we have a circle with area πL2.
L
L
40. Surfaces of Orbiting Lines
II. If a line segment L is
parallel to the axis of rotation,
it forms a cylinder.
L
R
41. Surfaces of Orbiting Lines
II. If a line segment L is
parallel to the axis of rotation,
it forms a cylinder.
L LR
R
42. Surfaces of Orbiting Lines
II. If a line segment L is
parallel to the axis of rotation,
it forms a cylinder.
L
Let R be radius of the rotation
then the surface area generates is 2πLR.
LR
R
43. Surfaces of Orbiting Lines
II. If a line segment L is
parallel to the axis of rotation,
it forms a cylinder.
L
Let R be radius of the rotation
then the surface area generates is 2πLR.
LR
R
Finally if the line segment L is
tilted with respect to the axis of
rotation, it generates a rostrum.
L
44. Surfaces of Orbiting Lines
II. If a line segment L is
parallel to the axis of rotation,
it forms a cylinder.
L
Let R be radius of the rotation
then the surface area generates is 2πLR.
LR
R
Finally if the line segment L is
tilted with respect to the axis of
rotation, it generates a rostrum.
L
L
45. Surfaces of Orbiting Lines
II. If a line segment L is
parallel to the axis of rotation,
it forms a cylinder.
L
Let R be radius of the rotation
then the surface area generates is 2πLR.
LR
R
Finally if the line segment L is
tilted with respect to the axis of
rotation, it generates a rostrum.
Let r and R be the top and bottom
radius of the rostrum respectively,
then the surface area formed by
the rotation is π(r + R)L.
L
L
r
R
L
46. Surfaces of Orbiting Lines
II. If a line segment L is
parallel to the axis of rotation,
it forms a cylinder.
L
Let R be radius of the rotation
then the surface area generates is 2πLR.
LR
R
Finally if the line segment L is
tilted with respect to the axis of
rotation, it generates a rostrum.
Let r and R be the top and bottom
radius of the rostrum respectively,
then the surface area formed by
the rotation is π(r + R)L.
L
L
r
R
L
Let’s see how these surfaces enable us to find the
volumes of solids of revolution.
47. Volume and Solids of Revolution
Let V be a solid and we want to find its volume.
V
48. Volume and Solids of Revolution
Let V be a solid and we want to find its volume.
Set a ruler x in a chosen direction and measure its
span, say from x = a to x = b.
V
49. Volume and Solids of Revolution
Let V be a solid and we want to find its volume.
Set a ruler x in a chosen direction and measure its
span, say from x = a to x = b.
ba
V
50. Volume and Solids of Revolution
Let S be a solid and we want to find its volume.
Set a ruler x in a chosen direction and measure its
span, say from x = a to x = b. Let A(x) be its cross–
sectional area at x.
A(x)
ba
V
x
51. Volume and Solids of Revolution
Let S be a solid and we want to find its volume.
Set a ruler x in a chosen direction and measure its
span, say from x = a to x = b. Let A(x) be its cross–
sectional area at x. Partition [a, b] into n equal size
subintervals, {x0=a, x1, x2, ..xi, .. xn=b}
ba
A(x)
V
x
52. Volume and Solids of Revolution
Let S be a solid and we want to find its volume.
Set a ruler x in a chosen direction and measure its
span, say from x = a to x = b. Let A(x) be its cross–
sectional area at x. Partition [a, b] into n equal size
subintervals, {x0=a, x1, x2, ..xi, .. xn=b}
a=x0 b=xn
x1 x2 xi–1 xi
A(x)
V
x
53. Volume and Solids of Revolution
Let S be a solid and we want to find its volume.
a=x0 b=xn
x1 x2 xi–1 xi
Set a ruler x in a chosen direction and measure its
span, say from x = a to x = b. Let A(x) be its cross–
sectional area at x. Partition [a, b] into n equal size
subintervals, {x0=a, x1, x2, ..xi, .. xn=b}, and select
an arbitrary point xi
* in each subinterval [xi–1, xi].
A(x)
V
x
55. Let Δx be the length of each subinterval.
a=x0 b=xn
xi–1 xi
Volume and Solids of Revolution
xi
*x1 x2
V
56. Let Δx be the length of each subinterval.
a=x0 b=xn
xi–1 xi
Volume and Solids of Revolution
xi
*x1 x2
V
Δx
57. Let Δx be the length of each subinterval. The
cylinder Ci with cross–sectional area A(xi) and height
Δx approximates the slice between xi–1 and xi.
a=x0 b=xn
xi–1 xi
Volume and Solids of Revolution
xi
*
A(xi)*
Δx
x1 x2
Ci
*
V
Δx
A(xi)*
58. The volume vi of Ci is A(xi)Δx.
Volume and Solids of Revolution
*
A(xi)*
Δx
Ci
59. The volume vi of Ci is A(xi)Δx. The sum of all the
volumes vi of all such cylinders, for i = 1 to i = n,
approximates the volume of V,
Volume and Solids of Revolution
*
A(xi)*
Δx
Ci
60. The volume vi of Ci is A(xi)Δx. The sum of all the
volumes vi of all such cylinders, for i = 1 to i = n,
approximates the volume of V, i.e.
Volume and Solids of Revolution
*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
61. The volume vi of Ci is A(xi)Δx. The sum of all the
volumes vi of all such cylinders, for i = 1 to i = n,
approximates the volume of V, i.e.
Volume and Solids of Revolution
*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
The volume of V is defined as Δx0 or n ∞.
62. The volume vi of Ci is A(xi)Δx. The sum of all the
volumes vi of all such cylinders, for i = 1 to i = n,
approximates the volume of V, i.e.
Volume and Solids of Revolution
*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
Hence the volume of V = lim
i=1
n
∑ A(xi)Δx.*
n ∞
The volume of V is defined as Δx0 or n ∞.
63. The volume vi of Ci is A(xi)Δx. The sum of all the
volumes vi of all such cylinders, for i = 1 to i = n,
approximates the volume of V, i.e.
Volume and Solids of Revolution
*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
Hence the volume of V = lim
i=1
n
∑ A(xi)Δx.*
n ∞
By the FTC, the volume of V = ∫x=a
b
A(x) dx
The volume of V is defined as Δx0 or n ∞.
64. The volume vi of Ci is A(xi)Δx. The sum of all the
volumes vi of all such cylinders, for i = 1 to i = n,
approximates the volume of V, i.e.
Volume and Solids of Revolution
*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
Hence the volume of V = lim
i=1
n
∑ A(xi)Δx.*
n ∞
By the FTC, the volume of V = ∫x=a
b
A(x) dx
The volume of V is defined as Δx0 or n ∞.
the cross–sectional
area function
65. The volume vi of Ci is A(xi)Δx. The sum of all the
volumes vi of all such cylinders, for i = 1 to i = n,
approximates the volume of V, i.e.
Volume and Solids of Revolution
*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
Hence the volume of V = lim
i=1
n
∑ A(xi)Δx.*
n ∞
By the FTC, the volume of V = ∫x=a
b
A(x) dx
the cross–sectional
area function
This formula and the one for area from last section
is called Cavalieri’s Principle.
The volume of V is defined as Δx0 or n ∞.
66. Volume and Solids of Revolution
The above volume formula requires that we know
the cross–sectional area function A(x).
V = ∫x=a
b
A(x) dx
67. Volume and Solids of Revolution
The above volume formula requires that we know
the cross–sectional area function A(x). One type of
solids whose cross–sectional areas are known are
the solids of revolution.
V = ∫x=a
b
A(x) dx
68. Volume and Solids of Revolution
The above volume formula requires that we know
the cross–sectional area function A(x). One type of
solids whose cross–sectional areas are known are
the solids of revolution. These are solids obtained
by revolving a template in space around an axis.
V = ∫x=a
b
A(x) dx
69. Volume and Solids of Revolution
The above volume formula requires that we know
the cross–sectional area function A(x). One type of
solids whose cross–sectional areas are known are
the solids of revolution. These are solids obtained
by revolving a template in space around an axis.
V = ∫x=a
b
A(x) dx
70. Volume and Solids of Revolution
The above volume formula requires that we know
the cross–sectional area function A(x). One type of
solids whose cross–sectional areas are known are
the solids of revolution. These are solids obtained
by revolving a template in space around an axis.
V = ∫x=a
b
A(x) dx
Example B.
Axis of rotation
a. A filled solid
from the rotation.
71. Volume and Solids of Revolution
The above volume formula requires that we know
the cross–sectional area function A(x). One type of
solids whose cross–sectional areas are known are
the solids of revolution. These are solids obtained
by revolving a template in space around an axis.
V = ∫x=a
b
A(x) dx
Example B.
Axis of rotation
a. A filled solid
from the rotation.
72. Volume and Solids of Revolution
Axis of rotationb. A holed solid
from the rotation.
73. Volume and Solids of Revolution
Axis of rotationb. A holed solid
from the rotation.
74. Volume and Solids of Revolution
Their cross–sections are simple.
Axis of rotationb. A holed solid
from the rotation.
75. Volume and Solids of Revolution
Their cross–sections are simple.
For example a, its cross– sections are (circular) discs.
Axis of rotationb. A holed solid
from the rotation.
76. Volume and Solids of Revolution
If the template is bounded by a function y = f(x) > 0,
the x–axis, from x = a to x = b, and we rotate it
around the x–axis,
x=a x=b
y = f(x)
77. Volume and Solids of Revolution
If the template is bounded by a function y = f(x) > 0,
the x–axis, from x = a to x = b, and we rotate it
around the x–axis,
x=a x=b x=a x=b
y = f(x)
78. Volume and Solids of Revolution
If the template is bounded by a function y = f(x) > 0,
the x–axis, from x = a to x = b, and we rotate it
around the x–axis, then the y = f(x) is the radius of
the cross–sectional disc at x.
x=a x=b x=a x=b
r = f(x)
r = f(x)
x x
y = f(x)
79. Volume and Solids of Revolution
If the template is bounded by a function y = f(x) > 0,
the x–axis, from x = a to x = b, and we rotate it
around the x–axis, then the y = f(x) is the radius of
the cross–sectional disc at x.
x=a x=b x=a x=b
r = f(x)
r = f(x)
x x
Then the cross–section–area A(x) = π [f (x)]2.
y = f(x)
80. Volume and Solids of Revolution
If the template is bounded by a function y = f(x) > 0,
the x–axis, from x = a to x = b, and we rotate it
around the x–axis, then the y = f(x) is the radius of
the cross–sectional disc at x.
x=a x=b x=a x=b
r = f(x)
r = f(x)
x x
∫x=a
b
dxHence the volume of V = π [f (x)]2
y = f(x)
Then the cross–section–area A(x) = π [f (x)]2.
the cross–sectional
area function
81. Volume and Solids of Revolution
If the template is bounded by a function y = f(x) > 0,
the x–axis, from x = a to x = b, and we rotate it
around the x–axis, then the y = f(x) is the radius of
the cross–sectional disc at x.
x=a x=b x=a x=b
r = f(x)
r = f(x)
x x
∫x=a
b
dxHence the volume of V = π [f (x)]2
We call this the disc method.
y = f(x)
Then the cross–section–area A(x) = π [f (x)]2.
the cross–sectional
area function
82. Volume and Solids of Revolution
Example C. Find the volume of revolution around the
x axis formed by the curve y = √x from x = 0 to x = 2.
83. Volume and Solids of Revolution
Example C. Find the volume of revolution around the
x axis formed by the curve y = √x from x = 0 to x = 2.
0 2
y = √x
84. Volume and Solids of Revolution
Example C. Find the volume of revolution around the
x axis formed by the curve y = √x from x = 0 to x = 2.
0 2
y = √x
85. Volume and Solids of Revolution
Example C. Find the volume of revolution around the
x axis formed by the curve y = √x from x = 0 to x = 2.
r = √x
The cross–sections are discs
with radius r = √x.
0 2
y = √x
86. Volume and Solids of Revolution
Example C. Find the volume of revolution around the
x axis formed by the curve y = √x from x = 0 to x = 2.
r = √x
The cross–sections are discs
with radius r = √x.
Therefore the cross –sectional
area function is A(x) = π (√x)2.
0 2
y = √x
87. Volume and Solids of Revolution
Example C. Find the volume of revolution around the
x axis formed by the curve y = √x from x = 0 to x = 2.
r = √x
The cross–sections are discs
with radius r = √x.
Therefore the cross –sectional
area function is A(x) = π (√x)2.
∫x=0
2
dx
Hence the volume of V is
πx
0 2
y = √x
88. Volume and Solids of Revolution
Example C. Find the volume of revolution around the
x axis formed by the curve y = √x from x = 0 to x = 2.
r = √x
The cross–sections are discs
with radius r = √x.
Therefore the cross –sectional
area function is A(x) = π (√x)2.
∫x=0
2
dx
Hence the volume of V is
πx
0 2
= πx2
2
2
0
= 2π
y = √x
89. Volume and Solids of Revolution
For example b, its cross–sections are washers
(rings).
90. Volume and Solids of Revolution
For example b, its cross–sections are washers
(rings).
R
r
91. Volume and Solids of Revolution
For example b, its cross–sections are washers
(rings).
If R = radius of the outside circle and r = radius of
the inside circle, then the area of the ring is the
difference of the areas of the two discs.
R
r
92. Volume and Solids of Revolution
For example b, its cross–sections are washers
(rings).
If R = radius of the outside circle and r = radius of
the inside circle, then the area of the ring is the
difference of the areas of the two discs.
That is, the area of the ring =
R
r
π(R2 – r2).
93. Volume and Solids of Revolution
For example b, its cross–sections are washers
(rings).
If R = radius of the outside circle and r = radius of
the inside circle, then the area of the ring is the
difference of the areas of the two discs.
That is, the area of the ring =
R
r
π(R2 – r2).
We get washers if we revolve the template around
an axis that is away from the template.
94. Volume and Solids of Revolution
y = x
3
y = –2
0
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
95. Volume and Solids of Revolution
y = x
30
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
96. Volume and Solids of Revolution
y = x
3
The cross–sections are washers.
0
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
97. Volume and Solids of Revolution
y = x
3
The cross–sections are washers.
0
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
98. Volume and Solids of Revolution
y = x
3
The cross–sections are washers.
r=2
r
The inside radius r = 2, 0
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
99. Volume and Solids of Revolution
0
y = x
3
The cross–sections are washers.
R = 2 + x
r=2
r
The inside radius r = 2, the
outside radius R = 2 + x.
R
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
100. Volume and Solids of Revolution
0
y = x
3
The cross–sections are washers.
R = 2 + x
r=2
r
The inside radius r = 2, the
outside radius R = 2 + x.
R
Therefore
A(x) = π [(2 + x)2 – 22]
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
101. Volume and Solids of Revolution
0
y = x
3
The cross–sections are washers.
R = 2 + x
r=2
r
The inside radius r = 2, the
outside radius R = 2 + x.
R
Therefore
A(x) = π [(2 + x)2 – 22]
= π (x2 + 4x)
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
102. Volume and Solids of Revolution
0
y = x
3
The cross–sections are washers.
R = 2 + x
r=2
r
The inside radius r = 2, the
outside radius R = 2 + x.
R
Therefore
A(x) = π [(2 + x)2 – 22]
= π (x2 + 4x)
∫x=0
3
dx
So the volume is
V = π (x2 + 4x)
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
103. Volume and Solids of Revolution
0
y = x
3
The cross–sections are washers.
R = 2 + x
r=2
r
The inside radius r = 2, the
outside radius R = 2 + x.
R
Therefore
A(x) = π [(2 + x)2 – 22]
= π (x2 + 4x)
∫x=0
3
dx
So the volume is
V =
π
π (x2 + 4x)
=
x3
(
3 + 2x2)
3
0
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
104. Volume and Solids of Revolution
0
y = x
3
The cross–sections are washers.
R = 2 + x
r=2
r
The inside radius r = 2, the
outside radius R = 2 + x.
R
Therefore
A(x) = π [(2 + x)2 – 22]
= π (x2 + 4x)
∫x=0
3
dx
So the volume is
V =
π
π (x2 + 4x)
=
x3
(
3 + 2x2)
3
0
27π=
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.
y = –2
105. Volume and Solids of Revolution
x = f(y)
y=a
y=b
We may rotate the function x = f(y) around the
y–axis as shown.
106. Volume and Solids of Revolution
We may rotate the function x = f(y) around the
y–axis as shown.
x = f(y)
y=a
y=b
x = f(y)
y=a
y=b
A(y)
107. Volume and Solids of Revolution
We may rotate the function x = f(y) around the
y–axis as shown.
x = f(y)
y=a
y=b
Then A(y) = π [f(y)]2
r
x = f(y)
y=a
y=b
A(y)
108. Volume and Solids of Revolution
We may rotate the function x = f(y) around the
y–axis as shown.
x = f(y)
y=a
y=b
Then A(y) = π [f(y)]2 and the volume V = ∫y=a
b
dyπ [f (y)]2
r
x = f(y)
y=a
y=b
A(y)
109. Volume and Solids of Revolution
We may rotate the function x = f(y) around the
y–axis as shown.
x = f(y)
y=a
y=b
Then A(y) = π [f(y)]2 and the volume V = ∫y=a
b
dyπ [f (y)]2
If it orbits around a vertical axis away with left
boundary r = g(y) as shown,
R = f(y)
y=a
y=b
r = g(y)
r
x = f(y)
y=a
y=b
A(y)
110. Volume and Solids of Revolution
We may rotate the function x = f(y) around the
y–axis as shown.
x = f(y)
y=a
y=b
Then A(y) = π [f(y)]2 and the volume V = ∫y=a
b
dyπ [f (y)]2
If it orbits around a vertical axis away with left
boundary r = g(y) as shown,
R = f(y)
y=a
y=b
r = g(y) R = f(y)
y=a
y=b
r = g(y)
r
x = f(y)
y=a
y=b
A(y)
111. Volume and Solids of Revolution
then A(y) = π [R2 – r2], and
∫y=a
π [R2 – r2] dy
R = f(y)
y=a
y=b
R = f(y)
y=a
y=b
the volume V =
b
r = g(y) r = g(y)
112. Volume and Solids of Revolution
then A(y) = π [R2 – r2], and
∫y=a
π [R2 – r2] dy
R = f(y)
y=a
y=b
R = f(y)
y=a
y=b
the volume V =
b
= ∫y=a
π [f(y)2 – g(y)2] dy
b
r = g(y) r = g(y)
113. Volume and Solids of Revolution
then A(y) = π [R2 – r2], and
∫y=a
π [R2 – r2] dy
R = f(y)
y=a
y=b
R = f(y)
y=a
y=b
the volume V =
b
= ∫y=a
π [f(y)2 – g(y)2] dy
b
r = g(y) r = g(y)
Example D. Find the volume of revolution around
the axis x = –½ formed by the template as shown.
y =
1
x2y =1
y=4
x = –½
114. Volume and Solids of Revolution
y =1
y=4
x = –½
y =
1
x2
Solve for the x for the function y = 1
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
115. Volume and Solids of Revolution
y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
The inside radius is r = ½.
Solve for the x for the function y = 1
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
x = y–1/2
116. Volume and Solids of Revolution
y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
The inside radius is r = ½. The outside radius is
R = ½ + y –1/2. Hence the volume is
Solve for the x for the function y = 1
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
x = y–1/2
117. Volume and Solids of Revolution
y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
The inside radius is r = ½. The outside radius is
R = ½ + y –1/2. Hence the volume is
Solve for the x for the function y = 1
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
∫y=1
π [(1/2 + y–1/2)2 – 1/2 2] dy
4
V =
x = y–1/2
118. Volume and Solids of Revolution
y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
The inside radius is r = ½. The outside radius is
R = ½ + y –1/2. Hence the volume is
Solve for the x for the function y = 1
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
∫y=1
π [(1/2 + y–1/2)2 – 1/2 2] dy
4
V =
∫y=1
π(y –1/2 + y) dy
4
=
x = y–1/2
119. Volume and Solids of Revolution
y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
The inside radius is r = ½. The outside radius is
R = ½ + y –1/2. Hence the volume is
Solve for the x for the function y = 1
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
∫y=1
π [(1/2 + y–1/2)2 – 1/2 2] dy
4
V =
∫y=1
π(y –1/2 + y) dy
4
= = π(2y1/2 + y2)1
2 1
4
x = y–1/2
120. Volume and Solids of Revolution
y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
The inside radius is r = ½. The outside radius is
R = ½ + y –1/2. Hence the volume is
Solve for the x for the function y = 1
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
∫y=1
π [(1/2 + y–1/2)2 – 1/2 2] dy
4
V =
∫y=1
π(y –1/2 + y) dy
4
= = π(2y1/2 + y2)1
2 1
4
= π(12 – )5
2 2=19π
x = y–1/2