Volume: The Disc Method Section 6.2. If a region in the plane is revolved about a line, the...

Volume: The Disc Method

Section 6.2

If a region in the plane is revolved about a line, the resulting solid is a solid of revolution, and the line is called the axis of revolution. The simplest such solid is a right circular cylinder or disc, which is formed by revolving a rectangle about an axis .

Revolving a Function

• Consider a function f(x) on the interval [a, b]

• Now consider revolvingthat segment of curve about the x axis

• What kind of functions generated these solids of revolution?

f(x)

a b

Disks

• We seek ways of usingintegrals to determine thevolume of these solids

• Consider a disk which is a slice of the solid– What is the radius– What is the thickness– What then, is its volume?

dx

f(x)

2Volume of slice = ( )f x dx

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

The simplest is a right circular cylinder or disk, a rectangle revolved around the x-axis.

The volume is equal to the area of the

disk times the

width of the disk, V = πr2w.

r

w

The Disc Method

Volume of disc =

(area of disc)(width of disc)

xrV 2

Disks

• To find the volume of the whole solid we sum thevolumes of the disks

• Shown as a definite integral

f(x)

a b

2( )

b

a

V f x dx

R

w

Axis of Revolution

w

R

Revolve this rectangle about the x-axis.

Revolve this function about the x-axis.

f x

A B

Revolve this rectangle about the x-axis.

x

f x

A B

Forms a Disk.

4

2

1i

i

V r x

disk circleV A x

x

2r x

f x

A B

Revolve this function about the x-axis.

f x

A B

2b

a

V f x dx

Revolving About y-Axis

• Also possible to revolve a function about the y-axis– Make a disk or a washer to be horizontal

• Consider revolving a parabola about the y-axis– How to represent the

radius?– What is the thickness

of the disk?

Revolving About y-Axis

• Must consider curve asx = f(y)– Radius = f(y)– Slice is dy thick

• Volume of the solid rotatedabout y-axis

2( )

b

a

V f y dy

Horizontal Axis of Revolution

Volume = V =

Vertical Axis of Revolution

Volume = V =

dxxRb

a

2

)(

dyyRd

c

2

)(

Disk Method (to find the volume of a solid of revolution) f x

A B

Horizontal Axis of revolution:

f yVertical Axis of revolution:

Disk Method (to find the volume of a solid of revolution) f x

A B 2b

a

V f x dx Horizontal Axis of revolution:

dx

f y

dy 2b

a

V f y dy

Vertical Axis of revolution:

radius width

1. Find the volume of the solid formed by revolving f(x) over about the x-axis.

sinf x x

0,

1. Find the volume of the solid formed by revolving f(x) over about the x-axis.

2

0

sinV x dx

sinf x x

0,

2

1

-1

2 0 dx 0

sin x dx

0

cos x

cos cos0

1 1

2

2. Find the volume of the solid formed by revolving the region formed by f(x) and g(x) about y = 1.

22 and 1f x x g x

2. Find the volume of the solid formed by revolving the region formed by f(x) and g(x) about y = 1.

1

22

1

1V x dx

22 and 1f x x g x

11

dx

22 1x

2

1

-2 2

Length of Rectangle:

2

1

-2 2

22 x1

21 x

1

2 4

1

1 2x x dx

13 52 1

3 5 1x x x

2 1 2 13 5 3 51 1

4 23 52

16

15

3. Find the volume of the solid formed by revolving the region formed by over about the y-axis.

2y x 0,4y

3. Find the volume of the solid formed by revolving the region formed by over about the y-axis.

4 2

0

V y dy

2y x

4

0

dy4

0

y dy

0,4y

4212 0y

8

4

3

2

1

-1

-2 2

y x

The disc method can be extended to cover solids of revolution with holes by replacing the representative disc with a representative washer.

r

w

R

Washers• Consider the area

between two functions rotated about the axis

• Now we have a hollow solid

• We will sum the volumes of washers

• As an integral

f(x)

a b

g(x)

2 2( ) ( )

b

a

V f x g x dx

Volume of washer = dxrR )( 22

dxxrxRVb

a 22 )()(

How do you find the volume of the figure formed by revolving the shaded area about the x-axis?

f x

A B

g x

How do you find the volume of the figure formed by revolving the shaded area about the x-axis?

f x

A B

g x

A B

f x

A B

g x

Revolving this function creates a solid whose volume is larger than we want.

2b

a

V f x dx 2b

a

V g x dx Revolving this function carves out the part we don’t want.

2 2b

a

V f x g x dx Outside radius

Inside radius

The volume we want is the difference between the two.

Washer Method (for finding the volume of a solid of revolution)

f x

A B

g x

Washer Method (for finding the volume of a solid of revolution)

f x

A B

g x

2 2b

figure

a

V f x g x dx Outside radius

Inside radius

2 2outside radius inside radiuswasherA

1. Find the volume of the solid generated by revolving the area enclosed by the two functions about the x-axis.

2,f x x g x x

1. Find the volume of the solid generated by revolving the area enclosed by the two functions about the x-axis.

2,f x x g x x f x g x

Find intersection points first.2x x4x x40 x x 30 1x x

0,1x

12

0

22

V dx xx

1

4

0

x x dx 12 51 1

2 5 0x x

1 12 5 0

3

10

1

0.5

1

Washer Method:

2. Find the volume of the solid formed by revolving the area enclosed by the given functions about the y-axis.

2 1, 0, 0, 1y x y x x

2. Find the volume of the solid formed by revolving the area enclosed by the given functions about the y-axis.

2 1, 0, 0, 1y x y x x This region is not always created by the same two functions.

21y x

1

2

0

1V dy

2

1.5

1

0.5

1

1y x

dy

dy

The change occurs at y = 1.

We need to use two integrals to find the volume.

2 22

1

11 yy d

Since we’re revolving about the y-axis, each radius must be in terms of y.

2 1y x (distance from the parabola to the x-axis)

(distance from the parabola to the y-axis)

For y in [0,1] we use disk method.

For y in [1,2] we use washer method.

1

2

0

1V dy 2 22

1

11 yy d

21 2120 1

2y y y

1 2

2

0 1

1 2dy y dy

121 0 4 2 2

2

3

2

Volume of Revolution - XFind the volume of revolution about the x-axis of f(x) = sin(x) + 2 from x = 0 to x = 2.

Volume of Revolution - XFind the volume of revolution about the x-axis of f(x) = sin(x) + 2 from x = 0 to x = 2

Did you get 88.826 cubic units?

2 2

0(sin( ) 2)A x dx

Use the TI to integrate this one!

Volumes of Solids with Known Cross Sections

1. For cross sections of area A(x) taken perpendicular to the x-axis, Volume =

2. For cross sections of area A(y) taken perpendicular to the y-axis,

Volume =

b

adxxA )(

d

cdyyA )(

Volume of Revolution - XLet’s look at the cross

section or slice. What is it?

It’s a circle.

What is the area?

What is the radius?

f(x).

A = r2 A = (f(x))2

Volume of Revolution - X

How wide (thick) is the disc?

dx

How do we add up all the disks from x = 1 to x = 4?

The volume of the disk is

V = (f(x))2dx

Find the volume of the solid whose base is the region in the xy-plane bounded by the given curves and whose cross-sections perpendicular to the x-axis are (a) squares; (b) semicircles; and (c) equilateral triangles.

Example 1

2,0,0,2 xyxxy

Example 1(a) Square Cross Sections

My Advice???

1) Draw the 2-D picture and imagine the cross-sectional shape coming out in the third dimension!

2) Try and think of how to write the area of these cross sections using the given function.

3) Set up the integral and integrate.

Think: SKATE BOARD RAMP!!!

!2 curvextheisHere

Example 1(b) Semicircular Cross Sections

Think: CORNICOPIA!!!

!2 curvextheisHere

Example 1(c) Equilateral Triangular

Cross SectionsThink: PYRAMID, kinda??

!2 curvextheisHere

Find the volume of the solid whose base is the region in the xy-plane bounded by the given curves and whose cross-sections perpendicular to the x-axis are (a) squares; (b) semicircles; and (c) equilateral triangles.

Example 2

0,2

1,2

1 xx

yx

y

Example 2 (a) Square Cross Sections

!2

1 curvex

theisHere

Think: PYRAMID!!!

!2

1 curvex

theisHere

Example 2 (b) Semicircular Cross Sections

!2

1 curvex

theisHere

Think: CORNICOPIA AGAIN!!!

!2

1 curvex

theisHere

Example 2 (c) Equilateral Triangular

Cross Sections

!2

1 curvex

theisHere Think: PYRAMID!!!

!2

1 curvex

theisHere

Theorem: The volume of a solid with cross-section of area A(x) that is perpendicular to the x-axis is given by

b

a

V A x dx

Finding the volume of a solid with known cross-section is a 3-step process:

a b

Theorem: The volume of a solid with cross-section of area A(x) that is perpendicular to the x-axis is given by

b

a

V A x dx

Finding the volume of a solid with known cross-section is a 3-step process:

Step #1: Find the length (S) of the rectangle used to create the base of the figure.

Step #2: Find the area A(x) of each cross-section (in terms of this rectangle).

Step #3: Integrate the area function from the lower to the upper bound.

***This circle is the base of a 3-D figure coming out of the screen.

***This rectangle is a side of a geometric figure (a cross-section of the whole).

a b

Volume of a solid with cross-sections of area A(y) and perpendicular to the y-axis:

d

c

V A y dyc

d

S

S

Find the volume of the solid whose base is bounded by the circle with cross-sections perpendicular to the x-axis. These cross-sections are

a. squaresStep #1: Find S.

Step #2: Find A(x).

Step #3: Integrate the area function.

2 2 4x y


a. squaresStep #1: Find S.

Step #2: Find A(x).

Step #3: Integrate the area function.

2 2 4x y

S

2 24 4S x x

2 24y x 24y x

22 4 x 2 2 4x y

-2 2

2squareA S

222 4A x x

24 4 x 216 4x S

2

2

2

16 4V x dx

2

2

0

2 16 4x dx 128

3 42.667


Step #1: Find S.

Step #2: Find A(x).

Step #3: Integrate.

b. equilateral triangles

2 2 4x y


Step #1: Find S.

Step #2: Find A(x).

Step #3: Integrate.

b. equilateral triangles

2 2 4x y

S

22 4S x -2 2

2

equilateral triangle

3

4

SA

223

2 44

A x x

23 4 x

S

23

4S

2

2

2

3 4V x dx

2

2

0

2 3 4 x dx 32 3

3 18.475

Volumes by SlicingSolids with known cross sectionsFind the volume of a solid where the base is an ellipse with semi-major axis of 4 units and semi-minor axis of 3 units if all cross sections perpendicular to the major axis are semicircles.

x4

3y

1916

22

yx

y

y2y

x)16(

16

9, 22 xyNow

xy

V .2

2

4

4

2 .1632

9dxxV

4

0

2 .1616

9dxx

4

0

3

316

16

9

xx

324 units

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

Volume: The Disc Method Section 6.2. If a region in the plane is revolved about a line, the...

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