Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

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Section 7.2 Solids of Revolution

Transcript of Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Page 1: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Section 7.2

Solids of Revolution

Page 2: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

1st Day

Solids with Known Cross Sections

Page 3: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Method of Slicing

1

Find a formula for A(x) = the area of a cross section.(When you multiply A(x) by dx, you will have the third length necessary to find the volume)

Sketch the solid and a typical cross section.

2

3 Find the limits of integration.

4 Integrate A(x) to find volume.

Page 4: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Examples

Page 5: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

1. Find the volume of the solid whose base is bounded by the circle x2 + y2 = 9 and each cross section perpendicular to the x-axis is a square.

x

y

Page 6: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

x

y

side of each square = 2y

Area = (2y)2 = 4y2

29y x

224 9A x x

23 2

02 4 9V x dx

3 2

08 9V x dx

33

0

8 93

xV x

3144 uV

Page 7: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

2. Find the volume of the solid whose base is bounded by the graphs f(x) = x + 1 and g(x) = x2 – 1 whose cross sections perpendicular to the x-axis are equilateral triangles. Area of an equilateral triangle is given by 2 3

4

sA

Page 8: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Side of equilateral triangle =

f(x) – g(x)

Page 9: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

2side 1 1x x 22 x x

2232

4A x x x

4 3 232 3 4 4

4x x x x

2 4 3 2

1

32 3 4 4

4V x x x x dx

381 3 u

40

Page 10: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

3. The region bounded by the graph of y = 2x – x2 and the x-axis is the base of a solid. For this solid each cross section perpendicular to the x-axis is a semicircle.

Page 11: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Diameter of semicircle = 2x – x2

21radius of semicircle 2

2x x

2

21 12

2 2A x x x

Page 12: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

2

21 12

2 2A x x x

2212

8A x x x

2 2 3 4

0

14 4

8V x x x dx

32 u

15

Page 13: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

2nd Day

Disk Method

Page 14: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

y x Suppose I start with this curve.

How did we first find the approximation for area under this curve?

Drawing rectangles and finding the area of each rectangle and adding them together.

Page 15: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

If I take the graph and rotate it about the x-axis, I will get a 3-dimensional solid that is cone shaped.

y x

Page 16: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

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.

Page 17: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Where do these flat cylinders (disks) come from?

They are created by rotating an infinitely many rectangles around the axis of revolution.

y x

Page 18: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Volume of a Cylinder = r2hWhat is the radius of each cylinder(disk) in the cone?What is the height of each cylinder(disk) in the cone?

The radius of each cylinder(disk) is y coordinate of its rectangle.The height of each cylinder(disk) is Δx.Volume of each cylinder(disk) = (y2)Δ x

y x

Page 19: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

How do we find the volume of the solid?

Add the volume of all the cylinders(disks).

y x

24

0πV x dx 4

0π x dx

42

0

π

2V x 8π cubic units

Page 20: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Examples

Page 21: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

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

Page 22: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

24

1

1π V dy

y

volume of disk

4

1

1π dy

y

4

1π ln y

π ln 4 ln1

2π ln 232π ln 2 u

Page 23: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Find the volume of the solid formed by rotating the region bounded by the x-axis (0 ≤ x ≤ ) and the graph of

about the x-axis.

sinf x x

Page 24: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.
Page 25: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

sin x

2

sin sinx x

radius of a disk

Area of a disk

Page 26: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Volume of the solid 0

sin x dx

0cos x

cos cos0 32 u

Page 27: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Find the volume of the solid by revolving by

f(x) = 2 – x2 and g(x) = 1 about the line y = 1.

radius of a disk 2 22 1 1x x

Area of a disk 221 x

Page 28: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Volume of the solid 1 22

11 x dx

1 2 4

11 2x x dx

153

1

2

3 5

xx x

316

u15

Page 29: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

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.

Solid of Revolution Formula with a vertical axis:

2π b

aV y dx

2π b

aV x dy

Solid of Revolution Formula with a horizontal axis:

End of 2nd Day

Page 30: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

3rd Day

Washer Method

Page 31: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

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

Page 32: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

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

Page 33: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

4 2

0

1

4 V y y dy

4 2

0

1

4V y y dy

4

2 3

0

1 1

2 12

y y

168 0

3

38 u

3

2

24

0 2

yV y dy

Page 34: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

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

Page 35: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

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

2y x

The outer radius is:

22

y

RR

The inner radius is:

2 r y

r

2y x

2

yx

2y x

y x

Page 36: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

4 2 2

0 V R r dy

2

24

02 2

2

yy dy

2

4

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

4 y y y dy

Page 37: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

432 3 2

0

3 1 8

2 12 3

y y y

16 6424 0

3 3

38 u

3

14 2 2

0

13 4

4 V y y y dy

Page 38: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Examples

Page 39: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Find the volume of the solid by revolving the region bounded by the graphs of

about the x-axis.

2 and y x y x

Page 40: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.
Page 41: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

outer radius

inner radius

x

2x

Page 42: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Volume of the solid 21 22

0x x dx

1 4

0x x dx

12 5

02 5

x x

33 u

10

Page 43: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Find the volume of the solid formed by the equations:

rotated about the line x = 6.

, 1, a1 nd 4yx xy

1y x

1x y

21x y

Page 44: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.
Page 45: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

outer radius

inner radius

26 1y

2

21x y

Page 46: Section 7.2 Solids of Revolution. 1 st Day Solids with Known Cross Sections.

Volume of the solid 23 2 2

16 1 2y dy

3 4 3 2

14 6 20 21y y y y dy

54 3 22 10 21

5

yy y y y

1

3

3192 u

5