Lesson 6-2a

VolumesKnown Cross-sectional Areas

Ice BreakerFind the volume of the region bounded by y = 1, y = x² and the y-axis revolved about the line y = -1.

dx

1 (1,1)

Volume = ∫ π(R² - r²) dx = π ∫ [(1+1)² – (1+x²)² ] dx x = 0

x = 1

= π ∫ (3 - 2x2 - x4) dx

= π (3x - (2/3)x³ - (1/5)x5 |

= π (3 - (2/3) - 1/5) – (0)

= 32π/15 = 6.702

x = 0

x = 1

x = 0

x = 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 of a Known Cross-Sectional Area

Volume = ∑ Area • thickness (∆variable)

V = ∫ Area • dx or V = ∫ Area • dy

Integration endpoints are based on the ranges x or y can have based on the picture of the area. Where does ∆x or ∆y go from and to?

Example 1∆Volume = Area • Thickness

Area = s² squares! = (1-¼x2)2

Thickness = ∆x

X ranges from 0 out to 2

Volume = ∫ ((1- ¼x2)2 ) dx x = 0

x = 2

= ∫ (1 – ½x² + (1/16)x4) dx

= (x – (1/6)x3 + (1/80)x5) |

= (2 – (1/6)(8) + (1/80)(32)) – (0)

= 16/15 = 1.067

x = 0

x = 2

Example 2

Volume = ∫ (√3/4) sin² x dx x = 0

x = π

∆Volume = Area • Thickness

Area = equilateral triangles! = ½ bh = ½ (sin x)(½√3sin x) = (√3/4) sin² x

Thickness = ∆x

X ranges from 0 out to π

= (√3/4) ∫ (sin x)² dx

= (√3/4) (½)(x – sin(x) cos(x)) |

= (√3/8) (π – (0)(-1)) – (0 – (0)(1))

= (√3/8) π = 0.6802

x = 0

x = π

Sinn x is a special integral – that works out very easy on

your calculator!

sin x60°

h = ½ 3 sin x

Example 3

= 2 ∫ (4 - x²) dx

= 2 (4x – ⅓x³) |

= 2 (8– (⅓)(8)) – (0)

= 2(8 – (8/3)) = 32/3 = 10.667

x = 0

x = 2

∆Volume = Area • Thickness

Area = two right isosceles triangles! = 2(½ bh) = 2 (½) (√4 - x²) (√4 - x²) = (4 - x²)

Thickness = ∆x

X ranges from -2 out to 2

Volume = 2 ∫ (4 - x²) dx x = 0

x = 2

The 2 comes from changing the limits of integration

(from symmetry)

2-2

dx x² + y² = 4

Right isosceles: b=h

1 2

Example 4

Volume = ∫ (π/8) ((1-x²)-(x4-1))² dx x = -1

x = 1

= (π/8) ∫ x8 + 2x6 – 3x4 – 4x2 + 4) dx

= (π/8) (x9/9 +2x7/7 – 3x5/5 – 4x3/3 +4x) |

= (π/8) (1/9+2/7-3/5-4/3+4) – (-1/9-2/7+3/5+4/3-4)

= (π/8) (1552/315) = 194π/315 = 1.93482

x = -1

x = 1

∆Volume = Area • Thickness

Area = semi-circles! (½ circles) = ½ πr² = (π/2)(½(1-x²)-(x4-1))² = (π/8) ((1-x²)-(x4-1))²

Thickness = ∆x

X ranges from -1 out to 1

Use calculator to expand the square

and to evaluate integral

y = 1 – x2 and y = x4 – 1.

d = f(x) – g(x).

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, 1, 2, 5, 9, 10