Download presentation

Presentation is loading. Please wait.

Published byVance Ingrum Modified over 2 years ago

1
Lesson 6-2c Volumes Using Washers

2
Ice Breaker Volume = ∫ π(15 - 8x² + x 4 ) dx x = 0 x = √3 = π ∫ (15 - 8x² + 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 ∆Volume = Area Thickness Area = Outer circle – inner circle = washers! = π(R² - r²) = π[(4 - x²)² - (1)²] = π(15 - 8x² + x 4 ) 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 – x 2, the line y = 1, y-axis and the x-axis around the x-axis. 2 dx y = 4 – x 2 x = √4-y 4 1

3
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 method Shell method

4
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).

5
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 r R R f(x) g(y)

6
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 – x 4 ) dx x = 0 x = 2 = π ∫ (8x – x 4 ) dx = π (4x² - (1/5)x 5 ) | = 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 = x 2 and y 2 = 8x about the x-axis.

7
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.

8
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 y 2 = x, the x-axis and x = 4. Find the volume when the region is revolved about the x-axis.

9
Example 6b Volume = ∫ π (16– y 4 ) dy y = 0 y = 2 = π ∫ (16 - y 4 ) dy = π (16y – (1/5) y 5 ) | = 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 – y 4 ) Thickness = ∆y y ranges from 0 up to 2 2 dy x = y² √x = y 4 Consider the first quadrant region bounded by y 2 = x, the x-axis and x = 4. Find the volume when the region is revolved about the y-axis.

10
Example 6c Volume = ∫ π (32 – 12y² + y 4 ) dy y = 0 y = 2 = π ∫ (32 – 12y² + y4) dy = π (32y – 4y³ + (1/5) y 5 ) | = 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 + y 4 ) Thickness = ∆y y ranges from 0 up to 2 2 dy x = y² √x = y 4 6 6-y² Consider the first quadrant region bounded by y 2 = x, the x-axis and x = 4. Find the volume when the region is revolved about the line x = 6.

11
Example 6d Volume = ∫ π ( 4√x - x ) dx x = 0 x = 4 = π ∫ ( 4√x - x ) dx = π (8/3)x 3/2 - 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 y 2 = x, the x-axis and x = 4. Find the volume when the region is revolved about the line y = 2.

12
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)

13
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

Similar presentations

OK

Volume: The Shell Method Lesson 7.3. Find the volume generated when this shape is revolved about the y axis. We can’t solve for x, so we can’t use a horizontal.

Volume: The Shell Method Lesson 7.3. Find the volume generated when this shape is revolved about the y axis. We can’t solve for x, so we can’t use a horizontal.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on mass energy equation Ppt on railway track and structures Ppt on causes of second world war Ppt on diode transistor logic Ribosome display ppt online Ppt on famous indian entrepreneurship Ppt on world book day 2016 Ppt on hydraulic power steering system Ppt on bluetooth technology free download Ppt on second law of thermodynamics example