33

3

Find the volume of the pyramid:

Consider a horizontal slice through the pyramid.

sdh

The volume of the slice is s2dh.

If we put zero at the top of the pyramid and make down the positive direction, then s=h.

0

3

h

2slice

V h dh=

32

0V h dh= ò

3

3

0

13h= 9=

This correlates with the formula:13

V Bh=1

9 33

= × × 9=

Method of Slicing:

1

Find a formula for A(x).

Sketch the solid and a typical cross section.

2

3 Find the limits of integration.

4 Integrate V(x) to find volume.

®

Volume formula :: Let S be a solid bounded by two parallel planes perpendicular to the x-axis at x=a and x=b. If, for each x in [a,b], the cross-sectional area of S perpendicular to the x-axis is A(x), then the volume of the solid is

( )b

a

V A x dx=ò

provided A(x) is integrable.

Volume formula :: Let S be a solid bounded by two parallel planes perpendicular to the y-axis at y=c and y=d. If, for each y in [c,d], the cross-sectional area of S perpendicular to the y-axis is A(y), then the volume of the solid is

( )d

c

V A y dy= ò

provided A(y) is integrable.

xy

A 45o wedge is cut from a cylinder of radius 3 as shown.

Find the volume of the wedge.

You could slice this wedge shape several ways, but the simplest cross section is a rectangle.

If we let h equal the height of the slice then the volume of the slice is: ( ) 2= ××V x y h dx

Since the wedge is cut at a 45o angle:x

h45o h x=

Since 2 2 9x y+ = 29y x= -®

xy

( ) 2V x y h dx= × ×h x=

29y x= -

( ) 22 9V x x x dx= - × ×

3 2

02 9V x x dx= -ò

29u x= -2 du x dx= -

( )0 9u =( )3 0u =

10

29

V u du=-ò9

32

0

23u=

227

3= × 18=

Even though we started with a cylinder, does not enter the calculation!

®

Cavalieri’s Theorem:

Two solids with equal altitudes and identical parallel cross sections have the same volume.

Identical Cross Sections

y x= Suppose we start with this curve.

If we want to build a nose cone in this shape.

So we put a piece of wood in a lathe and turn it to a shape to match the curve.

®

y x=How could we find the volume of the cone?

One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes.

The volume of each flat cylinder (disk) is:

2 the thicknessrp ×

In this case:

r= the y value of the function

thickness = a small change

in x = dx

( )2

x dx

®

y x=

The volume of each flat cylinder (disk) is:

2 the thicknessrp ×

If we add the volumes, we get:

( )24

0x dxpò

4

0 x dxp= ò

4

2

02x

p= 8p=

( )2

x dx

®

This application of the method of slicing is called the disk method. The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk.

If the shape is rotated about the x-axis, then the formula is:

A shape rotated about the y-axis would be:

®

2 2( )

b b

a a

V r dx f x dxp pé ù é ù= =ë û ë ûò ò

2( )

d

c

V gy dypé ù= ê úë ûò

Find the volume of the solid generated by revolving the regionsabout the x-axis.2y x x and y 0= - =bounded by

12

0

r dxpò ( )1

22

0

dx xx= -pò

Find the volume of the solid generated by revolving the regionsabout the x-axis.2y 3x x and y 0= - - =bounded by

02

3

dxr-

pò ( )20

2

3

dx3x x-

- -= pò

Find the volume of the solid generated by revolving the regionsabout the y-axis.1

2y x, x 0, y 2= = =bounded by

22

0

r dypò ( )2

2

0

d2y y= pò

Find the volume of the solid generated by revolving the regionsabout the x-axis.2

y 2sin2x, 0 x p= £ £bounded by

/ 22

0

xr dp

pò ( )/ 2

2

0

dx2sin2xp

= pò

Find the volume of the solid generated by revolving the regionsabout the line y = -1.

2y 3 x , y 1= - = -bounded by

22

2

dxr-

pò ( ) ( )( )2 2

2

2

3 x 1 dx-

-= - -pò

The region between the curve , and the

y-axis is revolved about the y-axis. Find the volume.

1x

y= 1 4y£ £

y x

1 1

2

3

4

1.707

2=

1.577

3=

1

2

We use a horizontal disk.

dy

The thickness is dy.

The radius is the x value of the function .1

y=

24

1

1 V dy

yp

æ ö÷ç ÷ç= ÷ç ÷ç ÷çè øò

volume of disk

4

1

1 dy

yp= ò

4

1lnyp= ( )ln4 ln1p= -

02ln2p= 2 ln2p=

®

The natural draft cooling tower shown at left is about 500 feet high and its shape can be approximated by the graph of this equation revolved about the y-axis:

2.000574 .439 185x y y= - +x

y

500 ft

( )500 2

2

0.000574 .439 185 y y dyp - +ò

The volume can be calculated using the disk method with a horizontal disk.

324,700,000 ft»

The region bounded by and is revolved about the y-axis.Find the volume.

2y x= 2y x=

The “disk” now has a hole in it, making it a “washer”.

If we use a horizontal slice:

The volume of the washer is: ( )2 2 thicknessR rp p- ×

( )2 2R r dyp -

outerradius

innerradius

2y x=

2y

x=

2y x=

y x=

2y x=

2y x=

( )2

24

0 2y

V y dypæ öæö÷ç ÷÷çç ÷÷= - çç ÷÷çç ÷ç ÷è øç ÷çè ø

ò

42

0

14

V y y dypæ ö÷ç ÷= -ç ÷ç ÷çè øò

42

0

1

4V y y dyp= -ò

4

2 3

0

1 12 12y yp

é ùê ú= -ê úë û

168

3p

é ùê ú= -ê úë û

83p

=®

This application of the method of slicing is called the washer method. The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle.

The washer method formula is:

®

( )2 2

2 2( ) ( )

b

ab

a

V R r dx

f x g x dx

p

p

= -

æ öé ù é ù= -ç ÷ë û ë ûè ø

ò

ò

2y x=If the same region is rotated about the line x=2:

2y x=

The outer radius is:

22y

R = -

R

The inner radius is:

2r y= -

r

2y x=

2y

x=

2y x=

y x=

42 2

0V R r dyp= -ò

( )2

24

02 2

2y

y dypæ ö÷ç ÷= - - -ç ÷ç ÷çè øò

( )24

04 2 4 4

4y

y y y dypæ ö÷ç ÷= - + - - +ç ÷ç ÷çè ø

ò

24

04 2 4 4

4y

y y y dyp= - + - + -ò14

2 2

0

13 4

4y y y dyp= - + +ò

43

2 3 2

0

3 1 82 12 3y y yp

é ùê ú= ×- + +ê úê úë û

16 6424

3 3p

é ùê ú= ×- + +ê úë û

83p

=