# Lesson 6-2b Volumes Using Discs. Ice Breaker Homework Check (Section 6-1) AP Problem 1: A particle moves in a straight line with velocity v(t) = t². How.

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Lesson 6-2b Volumes Using Discs

Ice Breaker Homework Check (Section 6-1) AP Problem 1: A particle moves in a straight line with velocity v(t) = t². How far does the particle move between times t = 1 and t = 2? AP Problem 2: ∫ sin(2x+3) dx

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).

Lesson 6-2 - Discs Finding Volume of Rotated Areas using Discs Volume = ∑ Region of Revolution thickness (∆) V = ∫ πr(x)² dx or V = ∫ πr(y)² dy Region is the area of a circle, πr², where r is a function of our variable of integration. Integration endpoints are the same as before. r = f(x) dx dy r = f(y) Area of a circle ┴ to axis of revolution

Example 1 ∆Volume = Area Thickness Area = circles! = πr² = π(√x)² = π(x) Thickness = ∆x X ranges from 0 out to 4 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 Find the volume of the solid of revolution obtained by revolving the plane region R bounded by, the x-axis, and the line x = 4 about the x-axis.

Example 3a Volume = ∫ π(16 - 8x² + x 4 ) dx x = 0 x = 2 = π ∫ (16 - 8x² + x 4 ) dx = π (16x – (8/3)x 3 + (1/5)x 5 ) | = π ((32 – (64/3) + (32/5)) – (0)) = (256/15) π = 53.617 x = 0 x = 2 ∆Volume = Area Thickness Area = circles! = πr² = π(4 - x²)² = π(16 - 8x² + x 4 ) Thickness = ∆x x ranges from 0 out to 2 2 dx y = 4 – x 2 x = √4-y 4 Find the volume of the solid generated by revolving the first quadrant region bounded by y = 4 – x 2, the x-axis and the y-axis around the x-axis.

Example 2 ∆Volume = Area Thickness Area = circles! = πr² = π(y 1/3 )² = π(y 2/3 ) Thickness = ∆y y ranges from 0 out to 3 Volume = ∫ π(y 2/3 ) dy y = 0 y = 3 = π ∫ (y 2/3 ) dy = π (3/5)y 5/3 | = π ((3/5)(6.2403) – (0) ) = 3.7442π = 11.763 y = 0 y = 3 Find the volume of the solid generated by revolving the region bounded by y = x 3, the y-axis, and the line y = 3 about the y-axis.

Example 3b Volume = ∫ π (4 - y) dy y = 0 y = 4 = π ∫ (4 - y) dy = π (4y - ½y²) | = π ((16 - 8) – (0)) = 8π = 25.133 y = 0 y = 4 2 dy y = 4 – x 2 x = √4-y 4 ∆Volume = Area Thickness Area = circles! = πr² = π(√4-y)² = π (4 - y) Thickness = ∆y y ranges from 0 up to 4 Find the volume of the solid generated by revolving the first quadrant region bounded by y = 4 – x 2, the x-axis and the y-axis around the y-axis.

Example 4 Volume = ∫ π(9 - 6x² + x 4 ) dx x = 0 x =√3 = π ∫ (9 - 6x² + x 4 ) dx = π (9x – (6/3)x 3 + (1/5)x 5 ) | = π ((9√3 – (6√3) + (9√3/5)) – (0)) = (24√3/5) π = 26.119 x = 0 x = √3 ∆Volume = Area Thickness Area = circles! = πr² = π(4 - x² - 1)² = π(3 - x²)² = π(9 - 6x² + 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 y-axis and the line y = 1 around the line y = 1. 2 dx y = 4 – x 2 x = √4-y 4 1

Example 5 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

In-Class Quiz Friday Covering 6-1 and 6-2 –Area under and between curves –Volumes Disc method Washer method Know cross-sectional areas (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, 3, 7, 13, 14, 17

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