volumes by slicing. disk find the volume of revolution using the disk method washer find the volume...

Volumes by Slicing

Find the Volume of revolution using the diskdisk methodFind the volume of revolution using the washerwasher methodFind the volume of revolution using the shellshell methodFind the volume of a solid with known cross sectionsknown cross sections

Volume of REVOLUTIONVolume of REVOLUTION

These are a few of the many industrial uses for volumes of revolutions.

Volume of the washerR

r

ExampExample:le:

Find the volume of the solid formed by revolving the region bounded by y = x and y = x² over the interval [0, 1] about the x – axis.

The region between the curve , and the

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

1xy

1 4y

y x

1 12

3

4

1 .7072

1 .5773

12

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 dyy

volume of disk

4

1

1 dyy

4

1ln y ln 4 ln1

02ln 2 2 ln 2

Example of rotating the region about y-axis

Find the volume of the solid of revolution formed by rotating the finite region bounded by the graphs of about the x-axis.  

Example: rotate it Example: rotate it around around xx = axis = axis

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

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 dx 1

22

0

dx xx

Find the volume of the solid generated by revolving the regionsabout the x-axis.2y 2sin2x, 0 x bounded by

/ 22

0

xr d

/ 2

2

0

dx2sin2x

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 dy 2

2

0

d2y y

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

2 22

2

3 x 1 dx

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

2y x

2y x

y x

2y x

2y x

2

24

0 2yV y dy

4 2

0

14

V y y dy

4 2

0

1 4

V y y dy 4

2 3

0

1 12 12y y

1683

83

Example of a washer

2y x

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

2y x

The outer radius is:

22yR

R

The inner radius is:

2r y

r

2y x

2y x

2y x

y x

4 2 2

0V R r dy

2

24

02 2

2y y dy

24

04 2 4 4

4yy y y dy

24

04 2 4 4

4yy y y dy

14 2 20

13 4 4

y y y dy 43

2 3 2

0

3 1 82 12 3y y y

16 64243 3

83

The volume of the solid generated by revolving the first quadrantregion bounded by the curve and the lines x = ln 3 andy = 1 about the x-axis is closest to

x / 2y e

a) 2.79 b) 2.82 c) 2.85 d) 2.88 e) 2.91

ln3

2 2x / 2

0

e 1 dx

CALCULATOR REQUIRED

2

322

0

222

0

24 2

0

2 4

The volume of the solid generated by rotating about the x-axis

the region enclosed between the curve y 3x and the line y 6x is given by

A. 6x 3x dx

B. 6x 3x dx

C. 9x 36x dx

D. 36x 9x d

2

0

22

0

x

E. 6x 3x dx

CALCULATOR REQUIRED

Let R be the region in the first quadrant above by the graph of f x 2Arc tanx and below by the graph of y = x. What is the volume

of the solid generated when R is rotated about the x-axis?A. 1.21

B. 2.28 C. 2.69 D. 6.66 E. 7.15

CALCULATOR REQUIRED

Let R be the region in the first quadrant that is enclosed by thegraph of f x ln x 1 , the x-axis and the line x = e. What is

the volume of the solid generated when R is rotated about theline y = -1?A.

5.037 B. 6.545 C. 10.073 D. 20.146 E. 28.686

CALCULATOR REQUIRED

e

2 2

0

ln x 1 1 1 dx 20.14627352 D

Volumes by Cylindrical Shells

Shell Method

• Based on finding volume of cylindrical shells– Add these volumes to get the total volume

• Dimensions of the shell– Radius of the shell– Thickness of the shell– Height

The Shell• Consider the shell as one of many of a solid

of revolution

• The volume of the solid made of the sum of the shells

f(x)

g(x)xf(x) – g(x)

dx

2 ( ) ( )b

a

V x f x g x dx

Hints for Shell Method

• Sketch the graph over the limits of integration• Draw a typical shell parallel to the axis of

revolution• Determine radius, height, thickness of shell• Volume of typical shell

• Use integration formula 2 radius height thickness

2b

a

Volume radius height thickness

• Consider the region bounded by x = 0, y = 0, and

28y x

2 22

0

2 8V x x dx

Rotation About x-Axis• Rotate the region bounded by y = 4x and

y = x2 about the x-axis

• What are the dimensions needed?– radius– height– thickness

radius = y

height = 4yy

thickness = dy

16

0

24yV y y dy

Rotation About Non-coordinate Axis

• Possible to rotate a region around any line

• Rely on the basic concept behind the shell method

x = a

f(x) g(x)

2sV radius height thickness

Rotation About Non-coordinate Axis

• What is the radius?

• What is the height?

• What are the limits?

• The integral:

x = a

f(x) g(x)a – x

f(x) – g(x)x = c

r

c < x < a

( ) ( ) ( )a

c

V a x f x g x dx

• Rotate the region bounded by 4 – x2 , x = 0 and, y = 0 about the line x = 2

• Determine radius, height, limits

4 – x2 r = 2 - x

2

0

22

0

2 (2 ) (4 )V x x dx

u = x – 1 x = 1 u=0

x = u + 1 x = 2 u=1 du = dx

Blobs in Space

Volume of a blob:

Cross sectional area at height h: A(h)

Volume =

Volumes with cross-sections:

• We will be given a “boundary” for the base of the shape which will be used to find a length.

• We will use that length to find the area of a figure generated from the slice .

• The dy or dx will be used to represent the thickness.

• The volumes from the slices will be added together to get the total volume of the figure.

Procedure: volume by slicing

o sketch the solid and a typical cross sketch the solid and a typical cross sectionsection

o find a formula for the area, find a formula for the area, A(x),A(x), of of the cross sectionthe cross section

o find limits of integrationfind limits of integration

o integrate integrate A(x)A(x) to get volume to get volume

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.

x2 y2 1

Bounds?

Top Function?

Bottom Function?

[-1,1]

y 1 x2

y 1 x2

Length? 1 x2 1 x2 2 1 x2

x2 y2 1

We use this length to find the area of the square.

Length? 2 1 x2

Area? 2 1 x2 24 1 x2

4 1 x2 dx 1

1

Volume?

Using the half circle [0,1] as the base slices perpendicular to the x-axis are isosceles right triangles.

x2 y2 1

Length? 2 1 x2

Area?12

2 1 x2 2 1 x2 Volume? 2 1 x2 dx

0

1

Bounds? [0,1]

Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all squares whose sides lie on the base of the circle.

First, find the length of a side of the squarethe distance from the curve to the x-axis is half the length of the side of the square … solve for y

2 2

2 2

2

4 4 4

x yy x

y x

22 4 x

length of a side is :

2

2 2

2

2 4 4 4 16 4Area x x

x

22

216 4 Volume x dx

Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all squares whose sides lie on the base of the circle.

The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections

perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve.

y 2 sin x

Bounds:Top Function: Bottom Function:

[0,π]y 2 sin x y 0

Length: 2 sin x

Area of an equilateral triangle:

34

(2 sin x )2

Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all equilateral triangles whose sides lie on the base of the circle.

Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all semicircles whose sides lie on the base of the circle.

Find the volume of a solid whose base is the circle x2 + y2 = 4 and where cross sections perpendicular to the x-axis are all Isosceles right triangles whose sides lie on the base of the circle.

Let R be the region marked in the first quadrant enclosed bythe y-axis and the graphs of as shown in the figure below

2y 4 x and y 1 2sinx

R

a) Setup but do not evaluate the integral representing the volume of the solid generated when R is revolved around the x-axis.

b) Setup, but do not evaluate the integral representing the volume of the solid whose base is R and whose cross sections perpendicular to the x-axis are squares.

1.102

2 22

0

4 x 1 2sinx dx

1.102 22

0

4 x 1 2sinx dx

CALCULATOR REQUIRED

2 2

The region S is represented by the area between the graphs of

f x 0.5x 2x 4 and g x 2 4 4x x . Write, but do

not evaluate, a definite integral which represents:a. the volume of a solid with base S if eac

h cross section ofthe solid perpendicular to the x-axis is a semi-circle.

b. the volume generated by rotating region S around the line y = 5.

24

0

g x f xdx

2 2

4

2 2

0

5 f x 5 g x dx

NO CALCULATORThe base of a solid is the region in the first quadrant bounded by

the curve y sinx for 0 x . If each cross section of the solid perpendicular to the x-axis is a square, the volume of thesolid, in cu

bic units, is:A. 0 B. 1 C. 2 D. 3 E. 4

2

00 0

sinx dx sinxdx cos x | 1 1 2 C

NO CALCULATORThe base of a solid is a right triangle whose perpendicular sideshave lengths 6 and 4. Each plane section of the solid perpendicularto the side of length 6 is a semicircle whose diameter lies in theplane of the triangle. The volume, in cubic units, of the solid is:A. 2 B. 4 C. 8 D. 16 E. 24

2

6 62

0 0

3 60

2 x1 13 dx x dx2 2 18

x |544 B

Let R be the region in the first quadrant under the graph of

3

8y for 1, 8x

a) Find the area of R.

b) The line x = k divides the region R into two regions. If the part of region R to the left of the line is 5/12 of the area of the whole region R, what is the value of k?

c) Find the volume of the solid whose base is the region R and whose cross sections cut by planes perpendicular to the x-axis are squares.

CALCULATOR REQUIRED

Let R be the region in the first quadrant under the graph of

3

8y for 1, 8x

a) Find the area of R.

8

31

8 dx 36x

Let R be the region in the first quadrant under the graph of

3

8y for 1, 8x

b) The line x = k divides the region R into two regions. If the part of region R to the left of the line is 5/12 of the area of the whole region R, what is the value of k?

k

31

8 dxx

A

5 3612

2/ 3 k112x | 15

2/32k 11 12 5

k 3.375

Let R be the region in the first quadrant under the graph of

3

8y for 1, 8x

c) Find the volume of the solid whose base is the region R and whose cross sections cut by planes perpendicular to the x-axis are squares.

28

21

8 dx 192x

Let R be the region in the first quadrant bounded above by thegraph of f(x) = 3 cos x and below by the graph of 2xg x e

a) Setup, but do not evaluate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the x-axis.

b) Let the base of a solid be the region R. If all cross sections perpendicular to the x-axis are equilateral triangles, setup, but do not evaluate, an integral expression of a single variable for the volume of the solid.

20.836 22 x

0

3cosx e dx

20.836 2

x

0

3 3cosx e dx4