1. Lesson 6-2a Volumes Known Cross-sectional Areas

2. Ice Breaker Find 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 - 2x 2 - x 4 ) dx = π ( 3x - (2/3)x³ - (1/5)x 5 | = π (3 - (2/3) - 1/5) – (0) = 32 π /15 = 6.702 x = 0 x = 1 x = 0 x = 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 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?

6. Example 1 ∆Volume = Area Thickness Area = s² squares! = (1-¼x 2 ) 2 Thickness = ∆x X ranges from 0 out to 2 Volume = ∫ ((1- ¼x 2 ) 2 ) dx x = 0 x = 2 = ∫ (1 – ½x² + (1/16)x 4 ) dx = (x – (1/6)x 3 + (1/80)x 5 ) | = (2 – (1/6)(8) + (1/80)(32)) – (0) = 16/15 = 1.067 x = 0 x = 2

7. 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 Sin n x is a special integral – that works out very easy on your calculator! sin x 60° h = ½  3 sin x

8. 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

9. Example 4 Volume = ∫ (π/8) ((1-x²)-(x 4 -1))² dx x = -1 x = 1 = (π/8) ∫ x 8 + 2x 6 – 3x 4 – 4x 2 + 4) dx = (π/8) (x 9 /9 +2x 7 /7 – 3x 5 /5 – 4x 3 /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)-(x 4 -1))² = (π/8) ((1-x²)-(x 4 -1))² Thickness = ∆x X ranges from -1 out to 1 Use calculator to expand the square and to evaluate integral y = 1 – x 2 and y = x 4 – 1. d = f(x) – g(x).

10. 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