6.2 1

Section 6.2: Volumes

• Objective– Use definite integrals to volumes of solids

• Slabs• Disks• Washers

6.2 2

Volume of a Solid (Slabs)

6.2 3

Typed: Volume of a Solid (Slabs)

Consider the solid below. Slice the solid in such a way that cross sections are perpendicular to the x axis and that for a fix x in the interval [a,b], A(x) is the area of the cross section.

A(x)

a xb

6.2 4

Volume of a SlabPartition [a,b] as: 0 1 2 na x x x x b= < < < < =L

Slice the solid into thin slabs by passing planes through these points perpendicular to the x axis. The volume of the slab is

a xb

1( ) ,i i i i iA x x where x x x−Δ < <

6.2 5

Volume of a Solid

0 1 2 na x x x x b= < < < < =L

The volume of the solid is approximated by:

As the norm of the partition approaches zero, we obtained a definite integral which is defined as the volume V of the solid and is defined as:

a xb

( )b

a

V A x dx= ∫

1( )

n

i ii

V A x x=

= Δ∑

6.2 6

Example: Right Circular Cone

Find the volume of a right circular cone of base radius r

and altitude h

xr

y

h

6.2 7

Right Circular Cone: Solution

Find the volume of a right circular cone of base radius r

and altitude h

We orient the cone on the xy-coordinate system such that x lies in the interval [0,h]. For a fixed x the cross section is a circle with area 2( )A x yπ=

x

xr

h

y

y

6.2 8

Right Circular Cone Solution

x

xr

h

y

y

6.2 9

Right Circular Cone Solution

x

xr

h

y

y

6.2 10

Right Circular Cone:Typed Solution

x

xr

h

y

y

y xr h=

ry xh

=

22 2

2( ) rA x y xh

π π= =

2 22 2

2 20 0

h hr rV x dx x dxh h

π π= =∫ ∫2 3

203

hr xh

π⎡ ⎤

= ⎢ ⎥⎣ ⎦

2 3

2 3r hh

π=

213

r hπ=

6.2 11

Conclusion

Since the area of the circular base of the cone is weconclude that the volume of a right circular cone is one-third the area of the base times the altitude.

2rπ

6.2 12

Solid of Revolution

Definition: If a region in the plane is revolved about a line, the resulting solid is called a solid of revolution and the line is called the axis of revolution.

Note: The region under investigation is the region under a curve y =f(x) bounded by x=a and x=b.

6.2 13

Example: Cone

h

r

Triangular region

Rotate region about the x axis. A cone is generated

x

xr

h

y

y

Line passing through (0,0) and (h,r)

6.2 14

Summary: Volume of a Solid

The volume of the solid is given by the formula:

where A(x) is the area of a cross-section at the point x.

( )b

a

V A x dx= ∫

x

A(x)y=A(x)

6.2 15

Example: Rectangle Rotation

Example: If a vertical rectangle in the first quadrant is rotated about the x axis, what is the resulting solid?

r

h

6.2 16

Solution: Rectangle Rotation

Example: If a vertical rectangle in the first quadrant is rotated about the x axis, what is the resulting solid?

r

h

Volume of disk =(area of disk) (width of disk)

2 2

0

h

V r dx r hπ π= =∫

6.2 17

The Disk Method

Suppose that and f is continuous on [a,b]. Consider the region bounded by the curve y=f(x), the x axis, and . Revolve this region about the x axisgenerating a solid. The volume of this solid of using the disk method, can be found by using one of the following:

[ ]2( )b

a

V f x dxπ= ∫ Horizontal Axis of Revolution

[ ]2( )d

c

V g y dyπ= ∫ Vertical Axis of Revolution

a x b≤ ≤

( ) 0f x ≥

6.2 18

The Disk Method

[ ]2( )b

a

V f x dxπ= ∫Horizontal Axis of Revolution

cross-sectional area is

a b

y=f(x)

[ ]2( )f xπ

6.2 19

The Disk Method

[ ]2( )d

c

V g y dyπ= ∫ Vertical Axis of Revolution

c

dx=g(y)

6.2 20

Example : Method of disks with y as the independent variable

Find the volume of the solid resulting from revolving the region bounded by the curves

24 1 0 3y x and y from x to x about the y axis= − = = =

6.2 21

Solution: Disk Method

Find the volume of the solid resulting from revolving the region bounded by the curves

24 1 0 3y x and y from x to x about the y axis= − = = =

2

2

44

4

y xx y

x y

= −

= −

= −

Solution: Express x as a function of y

( )44 4 22

1 1 1

94 (4 ) 42 2yV y dy y dy y ππ π π

⎡ ⎤= − = − = − =⎢ ⎥

⎣ ⎦∫ ∫

6.2 22

Example

Find the volume of the solid generated when the indicated region is revolved about the (a) x-axis and (b) y-axis

1 2 3

1

2

3

4

5

y = 4 - 2x

6.2 23

Solution: Revolve about x axis

Find the volume of the solid generated when the indicated region is revolved about the (a) x-axis and (b) y-axis

1 2 3

1

2

3

4

5

y = 4 - 2x

6.2 24

Typed Solution: Revolve about x axis

Solution:

( )24 2V x xπΔ ≈ − Δ

0 2x≤ ≤

( )2 2

04 2V x dxπ= −∫

Find the volume of the solid generated when the indicated region is revolved about the (a) x-axis and (b) y-axis

1 2 3

1

2

3

4

5

y = 4 - 2x

a. slice vertically

( )2

3

0

1 324 2 33.516 3

x ππ ⎡ ⎤= − − = ≈⎢ ⎥⎣ ⎦

6.2 25

Example: Revolve about y axis

Find the volume of the solid generated when the indicated region is revolved about the (a) x-axis and (b) y-axis

1 2 3

1

2

3

4

5

x = 2 – y/2

6.2 26

Typed Solution: Revolve about y axis

Soluton:2

22yV yπ ⎛ ⎞Δ ≈ − Δ⎜ ⎟

⎝ ⎠0 4y≤ ≤

24

02

2yV dyπ ⎛ ⎞= −⎜ ⎟

⎝ ⎠∫

Find the volume of the solid generated when the indicated region is revolved about the (a) x-axis and (b) y-axis

b. slice vertically

43

0

2 162 16.763 2 3

y ππ⎡ ⎤⎛ ⎞= − − = ≈⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥⎣ ⎦

1 2 3

1

2

3

4

5

x = 2 – y/2