Volume by disk v π ˆb a r 2 dxdy v π ˆ 23 x 2 dxv

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Volume by disk
V = π ˆ b a r 2 dx/dy V = π ˆ 2 0 [3 x ] 2 dx V = π ˆ 2 0 9 x 2 dx V = 3 π ( x 3 ) | 2 0 = 24 π y y = 2 x 4 , y = 2 x 0 y = 2 x 4 1 2 y = x 4 y 2 1 4 = x V = π ˆ 2 0 y 2 1 4 2 dy = π ˆ 2 0 y 2 1 2 dy u = 1 2 y du = 1 2 dy 2 du = dy a = 0 b = 1 2 π ˆ 1 0 u du = 2 π 2 3 u 3 2 | 1 0 = 4 π 3 (1) = 4 π 3
x y = x, y = x 3 x = x 3 x - x 3 = 0 x 1 2 - x 3 = 0 x 1 2 1 - x 5 2 = 0 x = 0 x = 1 V = π ˆ b a r 2 o - r 2 i dx/dy r o y = x r i y = x 3 Volume by Washer
V = π ˆ 1 0 ( x ) 2 - ( x 3 ) 2 dx = π ˆ 1 0 x - x 6 dx = π 1 2 x 2 - 1 7 x 7 | 1 0 = π 1 2 - 1 7 = 5 π 14 x y = 6 - x 2 ,y = 2 V = π ˆ 2 0 r 2 o - r 2 i dx V = π ˆ 2 0 (6 - x ) 2 - (2) 2 dx V = π ˆ 2 0 x 4 - 12 x + 32 dx V = π 1 5 x 5 - 4 x 3 + 32 x | 2 0 V = π 32 5 - 32 + 64 V = π 32 + 90 5 V = 112 π 5
D efinitions Volume: Let be a solid that lies between and . If the cross­sectional area of S x = a x = b in the plane , through and perpendicular to the x­axis, is , where A is a S P x x ( x ) A continuous function, then the volume of is S ( x ) Δ x ( x ) dx V = lim n →∞ n i =1 A i * = b a A Volume of a cylindrical shell: rh Δ r V = Can be remembered as [ circumference ][ height ][ thickness ] V = Shell Method: the volume of the solid, obtained by rotating about the y­axis the region under the curve from to , is where f ( x ) y = a b π xf ( x ) dx V = b a 2 b 0 ≤ a < Examples Find the volume of the solid obtained by rotating about the y­axis the region bounded by and 2 x x y = 2 3 0 y = Understand that a typical shell has radius , circumference , and height x π x 2 ( x ) 2 x x f = 2 3 Apply the information to the Shell Method to get the volume (2π x ) (2 x x ) dx π (2 x ) dx V = 2 0 2 3 = 2 0 2 x 4 Simplify the volume 2π [ x x ] 2π (8 ) V = 2 1 4 5 1 5 2 0 = 5 32 This yields a final volume of π V = 5 16 Find the volume of the solid obtained by rotating about the y­axis the region between and x y = x y = 2 Understand that a typical shell has radius , circumference , and height x π x 2 x x 2 Apply the information to the Shell Method to get the volume (2π x ) ( x ) dx ( x ) dx V = 1 0 x 2 = 1 0 2 x 3 Simplify the volume Volumes by Cylindrical Shells
(2π y ) (1 ) dy ( y ) dy V = 1 0 y 2 = 1 0 y 3 Simplify the volume 2π [ ] V = 2 y 2 4 y 4 1 0 This yields a final volume of V = 2 π V = 2π [ x x 3 x 4 4 ] 0 1 This yields a final volume of V = π 6 Use the cylindrical shells to find the volume of the solid obtained by rotating about the x­axis the region under the curve y = √ x from [0, 1] To use shells, relabel the curve y = √ x as x = y 2 Understand that a typical shell has radius y , circumference y , and height 1 − y 2 Apply the information to the Shell Method to get the volume
Work & Energy Example 1: Carrying sack of concrete up a ladder while it spills concrete: Weight (Force) 80 pounds at bottom 50 pounds at top 20 feet elevation charge -Write a formula F=80-1.5x x = distance above ground ∑ 𝐹𝐹 . ∆𝑥𝑥 → 𝑤𝑤 = 𝐹𝐹 . 𝑑𝑑𝑥𝑥 20 0 𝑤𝑤 = 80 3 2 𝑥𝑥� 𝑑𝑑𝑥𝑥 20 0 𝑤𝑤 = 80 𝑥𝑥 − ( 3 2 )( 𝑥𝑥 2 2 ) 20 0 𝑤𝑤 = 1600 3 4 (400) 0 Work=(Force)(Distance) Foot-Pounds Pounds Feet Newton Meters (Nm) Newtons Meters 𝑘𝑘𝑘𝑘 . 𝑚𝑚 2 𝑠𝑠 2 ( 𝑘𝑘𝑘𝑘 )( 𝑚𝑚 ) 𝑠𝑠 2 English Units Metric Units w = 1,300 Foot-Pounds
Example 2: Chain with bucket lowered down a well. Lift it.

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