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Lesson 6-2c Volumes Using Washers

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Lesson 6-2c. Volumes Using Washers. Ice Breaker. x = √3. Volume = ∫ π (15 - 8x ² + x 4 ) dx. = π ∫ (15 - 8x ² + x 4 ) dx = π (15x – (8/3)x 3 + (1/5)x 5 ) | = π ((15√3 – (8√3) + (9√3/5)) – (0)) = (44 √3 /5) π = 47.884. x = 0. x = √3. x = 0. y = 4 – x 2 x = √4-y. - PowerPoint PPT Presentation

### Transcript of Lesson 6-2c Lesson 6-2c

VolumesUsing Washers Ice Breaker

Volume = ∫ π(15 - 8x² + x4) dx x = 0

x = √3

= π ∫ (15 - 8x² + x4) dx

= π (15x – (8/3)x3 + (1/5)x5) |

= π ((15√3 – (8√3) + (9√3/5)) – (0))

= (44√3/5) π = 47.884

x = 0

x = √3

∆Volume = Area • Thickness

Area = Outer circle – inner circle = washers! = π(R² - r²) = π[(4 - x²)² - (1)²] = π(15 - 8x² + x4)

Thickness = ∆x

x ranges from 0 out to √3 (x = √4-1)

Find the volume of the solid generated by revolving the first quadrant region bounded by y = 4 – x2, the line y = 1, y-axis and the x-axis around the x-axis.

2dx

y = 4 – x2 x = √4-y

4

1 Objectives

• Find volumes of non-rotated solids with known cross-sectional areas

• Find volumes of areas rotated around the x or y axis using

Disc/Washer methodShell method Vocabulary

• Cylinder – a solid formed by two parallel bases and a height in between

• Base – the bottom part or top part of a cylinder

• Cross-section – a slice of a volume – an area – obtained by cutting the solid with a plane

• Solids of revolution – volume obtained by revolving a region of area around a line (in general the x or y axis). Volume using Washers

Finding Volume of Rotated Areas using Washers

Volume = ∑ Outer – Inner Region • thickness (∆)

V = ∫ π(R² - r²) • dx or V = ∫ π(R² - r²) • dy

Where outer radius R and the inner radius r are functions of the variable of integration. Integration endpoints are the same as before.

dx

dy

Area of a outer circle – inner circle

r rR

R

f(x)

g(y) Example 4

∆Volume = Area • Thickness

Area = washers (outer - inner)! = π[(√8x)2 – (x²)²]

Thickness = ∆x

X ranges from 0 out to 2

Volume = ∫ (π(8x – x4) dx x = 0

x = 2

= π ∫ (8x – x4) dx

= π (4x² - (1/5)x5) |

= π [(16) – (32/5)]

= 48π/5 = 30.159

x = 0

x = 2

Find the volume of the solid generated by revolving the region bounded by the parabolas y = x2 and y2 = 8x about the x-axis. Example 5

= π ∫ (4 + 2√4-y² + y²) dy

∆Volume = Area • Thickness

Area = washers (outer - inner)! = π((1+√4-y²)2 – (1)²) = π (4 + 2√4-y + y²)

Thickness = ∆y

y ranges from 0 up to 2

Volume = ∫ (π) (4 + 2√4-y² + y²) dy y = 0

y = 2

y = 0

y = 2

= π (4y + 4sin-1(y/2) + y√4-y² + ⅓y³)|

= π ((32/3 + 2π) – (0))

= π (32/3 + 2π) = 53.2495

The semicircular region bounded by the y-axis and x = √4-y² is revolved about the line x = -1. Setup the integral for its volume. Example 6a

Volume = ∫ π (x) dx x = 0

x = 4

= π ∫ (x) dx

= π (½x²) |

= π ((½ 16) – (0))

= π (8 – (0) = 8π

= 25.133

x = 0

x = 4

∆Volume = Area • Thickness

Area = discs (not washers)! = πr² = π(√x)² = π (x)

Thickness = ∆x

x ranges from 0 out to 4

2

dx

x = y²√x = y

4

Consider the first quadrant region bounded by y2 = x, the x-axis and x = 4. Find the volume when the region is revolved about the x-axis. Example 6b

Volume = ∫ π (16– y4) dy y = 0

y = 2

= π ∫ (16 - y4) dy

= π (16y – (1/5) y5) |

= π (32 – (32/5) – (0))

= 128π/5 = 80.425

y = 0

y = 2

∆Volume = Area • Thickness

Area = washers (outer - inner)! = π((4)² – (y²)²) = π (16 – y4)

Thickness = ∆y

y ranges from 0 up to 2

2

dy

x = y²√x = y

4

Consider the first quadrant region bounded by y2 = x, the x-axis and x = 4. Find the volume when the region is revolved about the y-axis. Example 6c

Volume = ∫ π (32 – 12y² + y4) dy y = 0

y = 2

= π ∫ (32 – 12y² + y4) dy

= π (32y – 4y³ + (1/5) y5) |

= π (64 – 32 + 32/5) – (0)

= 192π/5 = 120.64

y = 0

y = 2

∆Volume = Area • Thickness

Area = washers (outer - inner)! = π ((6-y²)² – (2²)) = π (32 – 12y² + y4)

Thickness = ∆y

y ranges from 0 up to 2

2

dy

x = y²√x = y

46

6-y²

Consider the first quadrant region bounded by y2 = x, the x-axis and x = 4. Find the volume when the region is revolved about the line x = 6. Example 6d

Volume = ∫ π (4√x - x) dx x = 0

x = 4

= π ∫ (4√x - x) dx

= π (8/3)x3/2 - ½x² |

= π ((8/3)(8) – (½)(16)) – (0))

= π ((64/3) - 8) = 40π/3

= 41.888

x = 0

x = 4

∆Volume = Area • Thickness

Area = washers (outer - inner)! = π ((2²) - (2-√x)²) = π (4√x - x)

Thickness = ∆x

x ranges from 0 out to 4

2

dx

x = y²√x = y

4

Consider the first quadrant region bounded by y2 = x, the x-axis and x = 4. Find the volume when the region is revolved about the line y = 2. In-Class Quiz

• Friday

• Covering 6-1 and 6-2– Area under and between curves– Volumes

• Know cross-sectional areas• Disc method• Washer method (extra-credit) Summary & Homework

• Summary:– Area between curves is still a height times a width– Width is always dx (vertical) or dy (horizontal)– Height is the difference between the curves– Volume is an Area times a thickness (dy or dx)

• Homework: – pg 452-455, 4, 19, 23, 35, 66