142p23s11solution[1]

# 4 3 12 n n n 1 then v 1 w2 h 1 1 h h 32 h 2 h

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: of the pyramid as zero, wh using similar triangles, one would get = ⇒ w= 1h. 4 3 12 n n n 1 Then, V ≈ ∑ 1 w2 ∆h =∑ 1 ( 1 h ) ∆h = ∑ 32 h 2 ∆h . 2 24 i =1 2 i =1 n i =1 1 1 So, V = lim ∑ 32 h 2 ∆h = 32 ∫ h 2 dh = 18 . n →∞ ∆h → 0 i =1 12 0 The parabola and the line intersect at the points ( −1,1) and (1,1) . 2.3 y = x2 y=1 w ∆y yi yi 1 –1 Each cross section perpendicular to the y-axis is a square. The strategy is to sum up the volumes of each cross section with thickness ∆y . The length of each side of the square is w = 2 ⋅ y V ≈ ∑ Volume of cross section at height yi V ≈ ∑ Area yi ∆y = ∑ w2 ∆y = ∑ 2 y ( ) 2 ∆y = ∑ 4 y∆y 1 So, V = 4 ∫ y dy = 2 . 0 2.4 2 2 y=1 y = sin x 1 r π 2 π Notice that r = 1 − sin x sin x –1 π 2 Notice that r = 1 − sin x . V ≈ ∑ π r 2 ∆x = ∑ π (1 − sin x ) ∆x . 2 V =π∫ π 2 0 π (1 − 2sin x + sin 2 x ) dx = π ⋅ x + 2 cos x + 12 ( − sin x cos x + x ) = 2 0 π ( 3π − 8 ) 4 2.5 2 y = 1 − x9 r V ≈ ∑ π r 2 ∆x = ∑ π V =π∫ 2.6 3 −3 ( 2 1 − x9 2 ) ∆x = ∑...
View Full Document

## This note was uploaded on 02/04/2014 for the course MAT 132 taught by Professor Poole during the Fall '08 term at SUNY Stony Brook.

Ask a homework question - tutors are online