5 volumes and solids of revolution i x

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 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.

Let’s review the following important formulas about
distance and area.
The Real Line

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

distance and area.
The Real Line
RL
i.e. right – left.
Distance h = R – L
x

distance and area.
The Real Line
RL
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

distance and area.
The Real Line
RL
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)

distance and area.
The Real Line
RL
y
H
L
Distance
v = H – L
x
the horizontal distance between
P and Q is h = c – a,
P(a, b) Q(c, b)
R(a, d)
h = c – a

distance and area.
The Coordinate Distances
RL
y
H
L
Distance
v = H – L
x
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

Equations and Coordinate-Distances

Given a point (x, y) that’s on
the graph of an equation,
y
x
(x, y)

y = f(x) is vertical distance
to the x-axis,
y
x
y = f(x)
x
(x, y)

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)

y
x
x = g(y)
y = f(x)
y
x
(x, y)
2x – y2 = 4
(x, y)
Example A. a. Given 2x – y2 = –4 for 0 ≤ y,
express y as f(x), and x as g(y). Draw.

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)

y
x
x = g(y)
y = f(x)
y
x
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

y
x
x = g(y)
y = f(x)
y
x
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

y
x
x = g(y)
y = f(x)
y
x
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

Example A. Given 2x – y2 = –4 and 3y = 2x for 0 ≤ y,
2x – y2 = –4
3y = 2x
y

b. find the horizontal lengths h1 and h2
as shown in terms of y.
h1
h2
2x – y2 = –4
3y = 2x
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

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

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

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

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

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)
Labeling the y-coordinates,
in terms of x’s,
((–4 + y2)/2,y)
y
x

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)
in terms of x’s, so
v1 = √2x + 4 – 2x/3
((–4 + y2)/2,y)
y
x

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
v1 = √2x + 4 – 2x/3
((–4 + y2)/2,y)
y
x

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
v1 = √2x + 4 – 2x/3
v2 = 2x/3 –√2x + 4
((–4 + y2)/2,y)
y
x

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

We are interested in the following cases.

I. If the line segment L is
perpendicular to the axis of
rotation, it generates a ring.
L

L
L

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

is π(R2 – r2).
L
L
If L is connecter to the
rotational-axis so r = 0,
L
L

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

II. If a line segment L is
parallel to the axis of rotation,
it forms a cylinder.
L
R

L LR
R

L
Let R be radius of the rotation
then the surface area generates is 2πLR.
LR
R

L
LR
R
Finally if the line segment L is
tilted with respect to the axis of
rotation, it generates a rostrum.
L

L
LR
R
L
L

L
LR
R
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

L
LR
R
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.

Volume and Solids of Revolution
Let V be a solid and we want to find its volume.
V

Set a ruler x in a chosen direction and measure its
span, say from x = a to x = b.
V

span, say from x = a to x = b.
ba
V

Let S be a solid and we want to find its volume.
span, say from x = a to x = b. Let A(x) be its cross–
sectional area at x.
A(x)
ba
V
x

sectional area at x. Partition [a, b] into n equal size
subintervals, {x0=a, x1, x2, ..xi, .. xn=b}
ba
A(x)
V
x

subintervals, {x0=a, x1, x2, ..xi, .. xn=b}
a=x0 b=xn
x1 x2 xi–1 xi
A(x)
V
x

a=x0 b=xn
x1 x2 xi–1 xi
subintervals, {x0=a, x1, x2, ..xi, .. xn=b}, and select
an arbitrary point xi
* in each subinterval [xi–1, xi].
A(x)
V
x

a=x0 b=xn
xi–1 xi
xi
*x1 x2
V

Let Δx be the length of each subinterval.
a=x0 b=xn
xi–1 xi
xi
*x1 x2
V

Let Δx be the length of each subinterval.
a=x0 b=xn
xi–1 xi
xi
*x1 x2
V
Δx

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
xi
*
A(xi)*
Δx
x1 x2
Ci
*
V
Δx
A(xi)*

The volume vi of Ci is A(xi)Δx.
*
A(xi)*
Δx
Ci

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,
*
A(xi)*
Δx
Ci

approximates the volume of V, i.e.
*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi

*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
The volume of V is defined as Δx0 or n ∞.

*
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 ∞

*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
i=1
n
∑ A(xi)Δx.*
n ∞
By the FTC, the volume of V = ∫x=a
b
A(x) dx

*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
i=1
n
∑ A(xi)Δx.*
n ∞
b
A(x) dx
the cross–sectional
area function

*
A(xi)*
Δx
Ci
Volume of V ≈
i=1
n
∑ A(xi)Δx*=
i=1
n
∑ vi
i=1
n
∑ A(xi)Δx.*
n ∞
b
A(x) dx
area function
This formula and the one for area from last section
is called Cavalieri’s Principle.

The above volume formula requires that we know
the cross–sectional area function A(x).
V = ∫x=a
b
A(x) dx

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

the solids of revolution. These are solids obtained
by revolving a template in space around an axis.
V = ∫x=a
b
A(x) dx

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.

Axis of rotationb. A holed solid
from the rotation.

Their cross–sections are simple.
from the rotation.

Their cross–sections are simple.
For example a, its cross– sections are (circular) discs.
from the rotation.

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)

around the x–axis,
x=a x=b x=a x=b
y = f(x)

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)

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)

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)
area function

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)
area function

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

r = √x
The cross–sections are discs
with radius r = √x.
0 2
y = √x

r = √x
Therefore the cross –sectional
area function is A(x) = π (√x)2.
0 2
y = √x

r = √x
∫x=0
2
dx
Hence the volume of V is
πx
0 2
y = √x

r = √x
∫x=0
2
dx
Hence the volume of V is
πx
0 2
= πx2
2
2
0
= 2π
y = √x

For example b, its cross–sections are washers
(rings).

(rings).
R
r

(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

(rings).
That is, the area of the ring =
R
r
π(R2 – r2).

(rings).
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.

y = x
3
y = –2
0
Example D. Find the volume of revolution around
the axis y = –2 formed by the template as shown.

y = x
30
y = –2

y = x
3
The cross–sections are washers.
0
y = –2

y = x
3
r=2
r
The inside radius r = 2, 0
y = –2

0
y = x
3
R = 2 + x
r=2
r
The inside radius r = 2, the
outside radius R = 2 + x.
R
y = –2

0
y = x
3
R = 2 + x
r=2
r
R
Therefore
A(x) = π [(2 + x)2 – 22]
y = –2

0
y = x
3
R = 2 + x
r=2
r
R
Therefore
A(x) = π [(2 + x)2 – 22]
= π (x2 + 4x)
y = –2

0
y = x
3
R = 2 + x
r=2
r
R
Therefore
A(x) = π [(2 + x)2 – 22]
= π (x2 + 4x)
∫x=0
3
dx
So the volume is
V = π (x2 + 4x)
y = –2

0
y = x
3
R = 2 + x
r=2
r
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
y = –2

0
y = x
3
R = 2 + x
r=2
r
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=
y = –2

x = f(y)
y=a
y=b
We may rotate the function x = f(y) around the
y–axis as shown.

y–axis as shown.
x = f(y)
y=a
y=b
x = f(y)
y=a
y=b
A(y)

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)

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)

y–axis as shown.
x = f(y)
y=a
y=b
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)

y–axis as shown.
x = f(y)
y=a
y=b
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)

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)

∫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)

∫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)
the axis x = –½ formed by the template as shown.
y =
1
x2y =1
y=4
x = –½

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

y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
The inside radius is r = ½.
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
x = y–1/2

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
x2 , we get that
x2 =
1
y or that x =
1
√y
= y –1/2
x = y–1/2

y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
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

y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
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

y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
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

y =1
y=4
x = –½
y =
1
x2 y =1
y=4
x = –½
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

5 volumes and solids of revolution i x

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