Volumes by Slicing: Disks and Washers

Objective: To find the volume of figures using the method of

disks/washers.

Volume

• Recall that the underlying principle for finding the area of a plane region is to divide the region into thin strips, approximate the area of each strip by the area of a rectangle, add the approximations to form a Riemann Sum, and take the limit of the Riemann Sums to produce an integral for the area.

Volume

• Under appropriate conditions, the same strategy can be used to find the volume of a solid. The idea is to divide the solid into thin slabs, approximate the volume of each slab, add the approximations to form a Riemann Sum, and take the limit of the Riemann Sums to produce an integral for the volume.

The Volume Problem

• 7.2.1 Let S be a solid that extends along the x-axis and is bounded on the left and right, respectively, by the planes that are perpendicular to the x-axis at x = a and x = b. Find the volume V of the solid, assuming that its cross-sectional area A(x) is known at each x in the interval [a, b].

The Volume Problem

• To solve this problem, we begin by dividing the interval [a, b] into n subintervals, thereby dividing the solid into n slabs. If we assume that the width of the kth subinterval is , then the volume of the kth slab can be approximated by the volume of a right cylinder of width (height) and cross sectional area , where is a point in the kth subinterval.

xkk xxA )( *

kx)( *kxA

*kx

The Volume Problem

• Adding these approximations yields the following Riemann Sum that approximates the volume V:

• Taking the limit as n increases and the widths of the subintervals approach zero yields the definite integral

n

kkk xxAV

1

* )(

b

a

n

kkk

xdxxAxxAV )()(lim

1

*

0max

The Volume Formula

• 7.2.2 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

provided A(x) is integrable.

b

a

dxxAV )(

The Volume Formula

• 7.2.2 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

provided A(y) is integrable.

d

c

dyyAV )(

Solids of Revolution

• A solid of revolution is a solid that is generated by revolving a plane region about a line that lies in the same plane as the region; the line is called the axis of revolution. Many familiar solids are of this type.

Volume Problem

• 7.2.4 Let f be continuous and nonnegative on [a, b], and let R be the region that is bounded above by y = f(x), below by the x-axis, and on the sides by the lines x = a and x = b. Find the volume of the solid of revolution that is generated by revolving the region R about the x-axis.

Volume Problem

• We can solve this problem by slicing. For this purpose, observe that the cross section of the solid taken perpendicular to the x-axis at the point x is a circular disk of radius f(x).

• The area of this region is

• The volume of the solid is

2)]([)( xfxA

dxxfVb

a 2)]([

Method of Disks

• Because the cross sections are disk shaped, the application of this formula is called the method of disks.

• The area of this region is

• The volume of the solid is

2)]([)( xfxA

dxxfVb

a 2)]([

Example 2

• Find the volume of the solid that is obtained when the region under the curve over the interval [1, 4] is revolved about the x-axis.

xy

Example 2

• Find the volume of the solid that is obtained when the region under the curve over the interval [1, 4] is revolved about the x-axis.

xy

2

15

2)]([

4

1

4

1

22

xxdxdxxfV

b

a

Example 3

• Derive the formula for the volume of a sphere of radius r.

Example 3

• Derive the formula for the volume of a sphere of radius r.

33

2222

3

4

3)()]([ r

xxrdxxrdxxfV

r

r

r

r

b

a

22 xry

Problem

• 7.2.5 Let f and g be continuous and nonnegative on [a, b], and suppose that f(x) > g(x) for all x in the interval [a, b]. Let R be the region that is bounded above by y = f(x), below by y = g(x), and on the sides by the lines x = a and x = b. Find the volume of the solid of revolution that is generated by revolving the region R about the x-axis.

Volumes By Washers

• We can solve this problem by slicing. For this purpose, observe that the cross section of the solid taken perpendicular to the x-axis at the point x is the annular or “washer-shaped” region with inner radius g(x) and outer radius f(x). Thus,

• The area is

• The volume is

b

a

dxxgxfV ))]([)](([ 22

22 )]([)]([)( xgxfxA

Example 4

• Find the volume of the solid generated when the region between the graphs of the equation and over the interval [0, 2] is revolved about the x-axis.

2

2

1)( xxf

xxg )(

Example 4

• Find the volume of the solid generated when the region between the graphs of the equation and over the interval [0, 2] is revolved about the x-axis.

2

2

1)( xxf

xxg )(

2

0

222

10

69)]5([.

dxxxV

Perpendicular to the y-axis

• The methods or disks and washers have analogs for regions that are revolved about the y-axis. Using the method of slicing, we should have no problem deducing the following formulas for the volumes of the solids.

Disks/Washers

• These formulas are:

d

c

dyyvywV ))]([)](([ 22d

c

dyyuV 2)]([

Example 5

• Find the volume of the solid generated when the region enclosed by , y = 2, and x = 0 is revolved about the y-axis.

xy

Example 5

• Find the volume of the solid generated when the region enclosed by , y = 2, and x = 0 is revolved about the y-axis.

5

32

5)(

2

0

2

0

522

ydyyV

xy

Example 6

• Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are:

a) squaresb) semi-circles

xy

Example 6

• Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are:

a) Squares• Remember, we will use the method of slicing to find

the area of a cross section and then integrate. The side of each square is the value of the function, so the area is the function squared.

2

27

2)(

3

0

23

0

2

x

dxxV

xy

Example 6

• Find the volume of the solid whose base is the region bounded between the curve and the x-axis from x = 0 to x = 3 whose cross sections taken perpendicular to the x-axis are:

b) semi-circles • We need to find the area of a semi-circle to use the

method of slicing. The diameter of the circle is the value of the function, so the radius is half of that.

16

9

1622

3

0

223

0

xdx

xV

xy

Example 7

• Now, we will try to rotate a figure around a line other than the x or y-axis. We will use the idea of the outer and inner radius to find the correct formula.

Example 7

• Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2.

2y x2y x

r

R

2

yx

y x

Example 7

• Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2.

• The outer radius is

• The inner radius is

• The volume is

2y x2y x

r

R

2

yx

y x

22

yR

2r y

4 2 2

0V R r dy

Example 7

• Find the volume of the figure generated by rotating the graphs of y = 2x and y = x2 from y = 0 to y = 4 around the line x = 2.

• The outer radius is

• The inner radius is

• The volume is

2y x2y x

r

R

2

yx

y x

22

yR

2r y

2

24

02 22

yy dy

8

3