# Lesson 6-2c Volumes Using Washers. 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.

## Presentation on theme: "Lesson 6-2c Volumes Using Washers. 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."— Presentation transcript:

Lesson 6-2c Volumes Using Washers

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

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

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 r R 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 – 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.

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

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.

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.

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.

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

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