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Lesson 6-2c

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.

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Lesson 6-2c

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  1. Lesson 6-2c VolumesUsing Washers

  2. Ice Breaker x = √3 Volume = ∫π(15 - 8x² + x4) dx = 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 x = 0 y = 4 – x2 x = √4-y 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. 4 1 ∆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) dx 2

  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 Area of a outer circle –inner circle 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 f(x) dy R g(y) r r R

  6. x = 2 =π ∫ (8x – x4) dx = π (4x² - (1/5)x5) | = π [(16) – (32/5)] = 48π/5 = 30.159 Volume = ∫ (π(8x – x4) dx x = 0 x = 2 x = 0 Example 4 Find the volume of the solid generated by revolving the region bounded by the parabolas y = x2 and y2 = 8x about the x-axis. ∆Volume = Area • Thickness Area = washers (outer - inner)! = π[(√8x)2 – (x²)²] Thickness = ∆x X ranges from 0 out to 2

  7. 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. ∆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 y = 2 Volume = ∫ (π) (4 + 2√4-y² + y²) dy = π ∫(4 + 2√4-y² + y²) dy 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 y = 0 Example 5

  8. x = y² √x = y 2 dx 4 x = 4 Volume = ∫π (x)dx = π ∫(x)dx = π(½x²) | = π ((½ 16) – (0)) = π (8 – (0) = 8π = 25.133 x = 0 x = 4 x = 0 Example 6a 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. ∆Volume = Area • Thickness Area = discs (not washers)! = πr² = π(√x)² = π (x) Thickness = ∆x x ranges from 0 out to 4

  9. x = y² √x = y 2 dy 4 y = 2 Volume = ∫π (16– y4) dy = π ∫ (16 - y4) dy = π (16y – (1/5) y5) | = π (32 – (32/5) – (0)) = 128π/5 = 80.425 y = 0 y = 2 y = 0 Example 6b 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. ∆Volume = Area • Thickness Area = washers (outer - inner)! = π((4)² – (y²)²) = π (16 – y4) Thickness = ∆y y ranges from 0 up to 2

  10. x = y² √x = y 2 dy 6 4 6-y² y = 2 Volume = ∫π (32 – 12y² + y4) dy = 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 y = 0 Example 6c 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. ∆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

  11. x = y² √x = y 2 dx 4 x = 4 Volume = ∫π (4√x - x)dx = 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 x = 0 Example 6d 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. ∆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

  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

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