Volume: Slicing, Disks, and WashersMATH 211, Calculus II

J. Robert Buchanan

Department of Mathematics

Spring 2018

Background

Today’s discussion will center on using the definite integral tofind the volume of a three-dimensional object.

Recall: for a rectangular box

V = l w h = (length× width)︸ ︷︷ ︸cross-sectional area

×height

and for a cylinder

V = πr2h = π(radius)2︸ ︷︷ ︸cross-sectional area

×height

General IdeaSuppose a three-dimensional object lies along the x-axiscovering the interval [a,b] and the cross-sectional area is acontinuous function of x , call it A(x).

Riemann Sum Approach

Let ∆x =b − a

nfor n ∈ N and define xi = a + i∆x for

i = 1,2, . . . ,n.

V ≈n∑

i=1

A(xi)∆x

V = limn→∞

n∑i=1

A(xi)∆x

=

∫ b

aA(x) dx

ExampleFind the volume of a right circular cone whose base radius is 3and whose height is 7.

Solution

The cross sectional area of the cone at 0 ≤ x ≤ 7 isA(x) = π(3x/7)2.

V =

∫ 7

0π(3x/7)2 dx

=9π49

∫ 7

0x2 dx

=

[3π49

x3]x=7

x=0= 21π.

ExampleA solid object has as its base the circular region defined byx2 + y2 = 1. Every cross section of the object perpendicular tothe x-axis is a triangle whose base vertices are on the circleand whose height equals the length of the base. Find thevolume of this object.

Solution

The cross sectional area of the solid at −1 ≤ x ≤ 1 is

A(x) =12

(2√

1− x2)2.

V =

∫ 1

−1

12

(2√

1− x2)2 dx

= 2∫ 1

−1(1− x2) dx

= 4∫ 1

0(1− x2) dx

=

[4(

x − 13

x3)]x=1

x=0

=83.

ExampleA solid object has as its base the ellipse defined byx2

25+

y2

36= 1. Every cross section of the object perpendicular

to the x-axis is a square whose base vertices are on the ellipse.Find the volume of this object.

Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.

Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.

Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.

Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.

Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.

Solid of RevolutionIf a region in the plane is revolved around a line in the plane theresulting three-dimensional object is called a solid ofrevolution.

Method of Disks

Suppose f (x) ≥ 0 and continuous on [a,b] and the regionbounded beneath the graph of y = f (x), the x-axis, and thelines x = a and x = b is revolved around the x-axis.

Every cross section is a circular disk of radius r = f (x), thus thevolume of the solid of revolution is

V =

∫ b

aπ[f (x)]2︸ ︷︷ ︸

cross-sectional area

dx .

ExampleFind the volume of the solid of revolution generated wheny = 2− 1

2x is revolved around the x-axis while 0 ≤ x ≤ 4.

Method of Disks

Suppose f (x) ≥ 0 and continuous on [a,b] and the regionbounded beneath the graph of y = f (x), the x-axis, and thelines x = a and x = b is revolved around the x-axis.

Every cross section is a circular disk of radius r = f (x), thus thevolume of the solid of revolution is

V =

∫ b

aπ[f (x)]2︸ ︷︷ ︸

cross-sectional area

dx .

ExampleFind the volume of the solid of revolution generated wheny = 2− 1

2x is revolved around the x-axis while 0 ≤ x ≤ 4.

Solution

V = π

∫ 4

0

(2− 1

2x)2

dx

= π

∫ 4

0

(4− 2x +

14

x2)

dx

=

[π

(4x − x2 +

112

x3)]x=4

x=0

=16π

3

y as Independent VariableIf x = g(y) and g(y) ≥ 0 for c ≤ y ≤ d then the volume of thesolid of revolution generated by revolving the region boundedbetween the graph of x = g(y), the y -axis, and the lines y = cand y = d is given by

V =

∫ d

cπ[g(y)]2︸ ︷︷ ︸

cross-sectional area

dy .

x

c

d

y

x=g(y)

ExampleFind the volume of the solid of revolution generated wheny =√

x for 1 ≤ y ≤ 4 is revolved around the y -axis.

Solution

Since y =√

x then x = y2 and the volume is calculated asfollows.

V = π

∫ 4

1(y2)2 dy

= π

∫ 4

1y4 dy

=[π

5y5]y=4

y=1

=π(4)5

5− π(1)5

5

=1024π

5− π

5=

1023π5

Method of WashersSometimes a solid region has a “hole” inside it.

x

y

Question: how can we find the volume of the solid of revolutionwith a cavity?

Method of Washers Formula

If f (x) and g(x) are continuous on [a,b] and 0 ≤ f (x) ≤ g(x) on[a,b] the volume of the solid of revolution generated byrevolving the region bounded between the graphs of y = f (x),y = g(x), and the lines x = a and x = b is given by

V =

∫ b

aπ( [g(x)]2 − [f (x)]2︸ ︷︷ ︸

(outer radius)2−(inner radius)2

) dx

Example

Find the volume of the solid of revolution generated byrevolving the region bounded by y = 2− x2/2, y = 1− x2/4,x = −1, and x = 1 around the x-axis.

SolutionSince 2− x2/2 ≥ 1− x2/4 for −1 ≤ x ≤ 1 then

V = π

∫ 1

−1

[(2− x2

2

)2

−(

1− x2

4

)2]dx

= π

∫ 1

−1

[(4− 2x2 +

x4

4

)−(

1− x2

2+

x4

16

)]dx

= π

∫ 1

−1

[3− 3

2x2 +

316

x4]

dx

= 2π∫ 1

0

[3− 3

2x2 +

316

x4]

dx

=

[2π(

3x − 12

x3 +380

x5)]x=1

x=0

V = 2π(

3− 12

+3

80

)=

203π40

.

Example

Find the volume of the solid of revolution generated when theregion bounded by the curves y =

√x and y = x2 is revolved

around the y -axis.

x

y

Solution

Since y =√

x ⇐⇒ y2 = x and y = x2 ⇐⇒ √y = x and√

y ≥ y2 for 0 ≤ y ≤ 1 then

V = π

∫ 1

0

[(√

y)2 − (y2)2]

dy

= π

∫ 1

0

[y − y4

]dy

=

[π

(y2

2− y5

5

)]y=1

y=0

V = π

(12− 1

5

)=

3π10.

Revolving Around Different LinesWe can modify our previous formulas to handle the cases inwhich we revolve a region around horizontal or vertical line notintersecting the region.

The key is to think about the radius of revolution in each case.

x

y

axis of revolution

inner radius

outer radius

Example

Find the volume of the solid of revolution generated when theregion bounded between y =

√x , y = 0 and x = 4 is revolved

around the line y = 2.

Solution

The outer radius is ro = 2− 0 = 2 while the inner radius isri = 2−

√x .

V = π

∫ 4

0

[(2)2 − (2−

√x)2]

dx

= π

∫ 4

0

[4− (4− 4

√x + x)

]dx

= π

∫ 4

0

[4x1/2 − x

]dx

=

[π

(83

x3/2 − 12

x2)]x=4

x=0

V = π

(643− 8)

=40π

3.

Example

Find the volume of the solid of revolution generated when theregion bounded between y =

√x and y = x2 is revolved

around the line x = 2.

0.5 1.0 1.5 2.0 2.5x

0.2

0.4

0.6

0.8

1.0

y

axis

Solution

The curve furthest from the axis is y =√

x ⇐⇒ y2 = x . Thecurve closest is y = x2 ⇐⇒ √

y = x .

V = π

∫ 1

0

[(2− y2)2 − (2−

√y)2]

dy

= π

∫ 1

0

[(4− 4y2 + y4)− (4− 4

√y + y)

]dy

= π

∫ 1

0

[4y1/2 − y − 4y2 + y4

]dy

=

[π

(83

y3/2 − 12

y2 − 43

y3 +15

y5)]y=1

y=0

V = π

(83− 1

2− 4

3+

15

)=

31π30

.

Homework

I Read Section 6.2I Exercises: see D2L/WebAssign