7.2: Volumes by Slicing

Greg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 2001

Little Rock Central High School,Little Rock, Arkansas

Bellwork:Find the area bound by and in the first quadrant.

Area =

The Method of Cross-Sections

Intersect S with a plane Px

perpendicular to the x-axis

Call the cross-sectional area A(x) A(x) will vary as x increases from a to b

Cross-Sections (cont’d) Divide S into “slabs” of equal width

∆xusing planes at x1, x2,…, xn Like slicing a loaf of bread! To add an infinite number of slices of

bread…..we must integrate

b

aV A x dx

The formula can be

applied to any solid for which the

cross-sectional area A(x) can be

found

This includes solids of revolution,

which we will cover today…

…but includes many other solids as

well

A Bigger Picture

b

aV A x dx

Method of Slicing:

1

Find a formula for A(x)dx (OR A(y)dy)

(Note that I used A(x)dx instead of dA(x).)

Sketch the solid and a typical cross section.

2

3 Find the limits of integration.

4 Integrate A(x)dx to find volume. ORIntegrate A(y)dy to find volume.

y x Suppose I start with this curve.

My boss at the ACME Rocket Company has assigned me to build a nose cone in this shape.

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

y xHow 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 thicknessr

In this case:

r= the y value of the function

thickness = a small change

in x = dx

2

x dx

𝐴 (𝑥 )𝑑𝑥=¿

y xThe volume of each flat cylinder (disk) is:

2 the thicknessr

If we add the volumes, we get:

24

0x dx

4

0 x dx4

2

02x

8

2

x dx

𝑉=𝑥1

𝑥2

𝐴 (𝑥 ) 𝑑𝑥

=

𝐴 (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:

2 b

aV y dx

2 b

aV x dy A shape rotated about the y-axis would be:

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

y

volume of disk

4

1

1 dy

y

4

1ln y ln 4 ln1

02ln 2 2 ln 2

𝐴 (𝑦 )𝑑𝑦=𝜋 (1/√ 𝑦 )  2

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 22

0.000574 .439 185 y y dy

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 r

2 2R r dy

outerradius

innerradius

2y x

2

yx

2y x

y x

2y x

2y x

2

24

0 2

yV y dy

4 2

0

1

4V y y dy

4 2

0

1

4V y y dy

42 3

0

1 1

2 12y y

168

3

8

3

𝐴 (𝑦 )𝑑𝑦=¿

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 b

aV R r dx

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

2y x

The outer radius is:

22

yR

R

The inner radius is:

2r y

r

2y x

2

yx

2y x

y x

4 2 2

0V R r dy

2

24

02 2

2

yy dy

24

04 2 4 4

4

yy y y dy

24

04 2 4 4

4

yy y y dy

14 2 2

0

13 4

4y y y dy

432 3 2

0

3 1 8

2 12 3y y y

16 64

243 3

8

3

p

Second Example of Washers

Problem Find the volume of the solid obtained by rotating about the x-axis the region enclosed by the curves y = x andy = x2

…but about the line y = 2 instead of thex-axis

The solid and a cross-section are illustrated on the next slide

Second Example of Washers (cont’d)

Second Example of Washers (cont’d)

Solution Here

So

2 22 4 22 2 5 4A x x x x x x

1 1 4 2

0 0

15 3 2

0

5 4

85 4

5 3 2 15

V A x dx x x x dx

x x x

The formula can be

applied to any solid for which the

cross-sectional area A(x) can be

found

This includes solids of revolution, as

shown above…

…but includes many other solids as

well

A Bigger Picture

b

aV A x dx

The Method of Cross-Sections

Intersect S with a plane Px

perpendicular to the x-axis

Call the cross-sectional area A(x) A(x) will vary as x increases from a to b

Cross-Sections (cont’d) Divide S into “slabs” of equal width

∆xusing planes at x1, x2,…, xn Like slicing a loaf of bread! To add an infinite number of slices of

bread…..we must integrate