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Lesson 6-2a Volumes Known Cross-sectional Areas. Ice Breaker Find the following: (without calculator) sec(π/6) = sin(4π/3) = cos²(π/8) + sin²(π/8) = 1.

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Presentation on theme: "Lesson 6-2a Volumes Known Cross-sectional Areas. Ice Breaker Find the following: (without calculator) sec(π/6) = sin(4π/3) = cos²(π/8) + sin²(π/8) = 1."— Presentation transcript:

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

2 Ice Breaker Find the following: (without calculator) sec(π/6) = sin(4π/3) = cos²(π/8) + sin²(π/8) = 1 + tan²(π/8) =

3 Objectives Find volumes of non-rotated solids with known cross-sectional areas

4 Vocabulary Cross-section – a slice of a volume – an area obtained by cutting the solid with a plane Typically we see: –Squares: Area = s² –Rectangles: Area = l w –Semi-circles: Area = ½ π r² –Triangles: Area = ½ b h

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 Test Question ∆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 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).

8 Quonset Hut Volume = ∫ (100π/2) dy y = -20 y = 20 = (100π/2) ∫ dy = (100π/2) (y) | = (100π/2) (20) – (-20) = (100π/2) (40) = 2000π = 6283.19 cubic feet y = -20 y = 20 ∆Volume = Area Thickness Area = semi-circles! (½ circles) = ½ πr² = (π/2)(½20)² = (100π/2) Thickness = ∆x y ranges from -20 out to 20 feet Finding the volume in one of our greenhouse bldgs! x = -10 and x = 10. d = g(y).– f(y) y = 20 ft x =10 ft

9 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, ?? 54, 58 –Read 6.3 for Wednesday


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