Set6 - Math 21B-B Homework Set 6 Section 6.5 1 y = tan x ≤ x ≤ π 4 x-axis a Area = Z π 4 2 π tan x s 1 d dx tan x 2 dx = Z π 4 2 π tan x p

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Unformatted text preview: Math 21B-B - Homework Set 6 Section 6.5: 1. y = tan x , ≤ x ≤ π 4 ; x-axis a. Area = Z π/ 4 2 π tan x s 1 + d dx tan x 2 dx = Z π/ 4 2 π tan x p 1 + sec 4 xdx 4. x = sin y , hskip .1in 0 ≤ y ≤ π ; y-axis a. Area = Z π 2 π sin y s 1 + d dy sin y 2 dy = Z π 2 π sin y p 1 + cos 2 y dy 9. y = x 2 , ≤ x ≤ 4; x-axis Integral Formula: Area = Z 4 2 π · x 2 s 1 + 1 2 2 dx = Z 4 π √ 5 2 xdx = π √ 5 4 x 2 4 = π √ 5 4 · 16 = 4 √ 5 π Geometry Formula: Base Circumference = 2 π · 4 2 = 4 π Slant Height = √ 4 2 + 2 2 = √ 20 = 2 √ 5 Area = 1 2 · 4 π · 2 √ 5 = 4 √ 5 π 1 13. y = x 3 9 , ≤ x ≤ 2; x-axis Area = Z 2 2 π · x 3 9 s 1 + x 2 3 2 dx = 2 π 9 Z 2 x 3 r 1 + x 4 9 dx = 2 π 9 Z 25 / 9 1 9 4 √ udu u = 1 + x 4 9 , du = 4 9 x 3 dx = π 2 Z 25 / 9 1 √ udu = π 2 · 2 3 u 3 / 2 25 / 9 1 = π 3 " 5 3 3- 1 # = π 3 125 27- 1 = 98 π 81 14. y = √ x, 3 4 ≤ x ≤ 15 4 ; x-axis Area = Z 15 / 4 3 / 4 2 π √ x · s 1 + 1 2 √ x 2 dx = 2 π Z 15 / 4 3 / 4 √ x · r 1 + 1 4 x dx = 2 π Z 15 / 4 3 / 4 r x + 1 4 dx = 2 π Z 4 1 √ udu u = x + 1 4 , du = dx = 2 π · 2 3 u 3 / 2 4 1 = 4 π 3 (2 3- 1 3 ) = 28 π 3 2 21. x = e y + e- y 2 , ≤ y ≤ ln2; y-axis Area = Z ln 2 2 π · e y + e- y 2 · s 1 + e y- e- y 2 2 dy = π Z ln 2 ( e y + e- y ) · r 1 + e 2 y- 2 + e- 2 y 4 dy = π Z ln 2 ( e y + e- y ) · r e 2 y + 2 + e- 2 y 4 dy = π Z ln 2 ( e y + e- y ) · s e y + e- y 2 2 dy = π Z ln 2 ( e y + e- y ) 2 2 dy = π 2 Z ln 2 e 2 y + 2 + e- 2 y dy = π 2 1 2 e 2 y + 2 y- 1 2 e- 2 y ln 2 = π 2 2 + 2ln2- 1 8- 1 2- 1 2 = π · 15 16 + ln2 24. y = cos x,- π 2 ≤ x ≤ π 2 ; x-axis Area = Z π/ 2- π/ 2 2 π cos x p 1 + (- sin x ) 2 dx = 2 π Z π/ 2- π/ 2 cos x p 1 + sin 2 xdx 28. We want to consider the area of a surface gotten by revolving the arc AB about the x-axis (the area corresponds to the amount of crust for that slice of bread). Therefore we will consider the following area problem: y = √ r 2- x 2 , a ≤ x ≤ a + h ; x-axis where a and h are numbers such that 0 < h < 2 r and- r ≤ a ≤ r- h . 3 Area = Z a + h a 2 π p r 2- x 2 · s 1 + d dx p r 2- x 2 2 dx = 2 π Z a + h a p r 2- x 2 · s 1 +- x √ r 2- x 2 2 dx = 2 π Z a + h a p r 2- x 2 · r 1 + x 2 r 2- x 2 dx = 2 π Z a + h a p ( r 2- x 2 ) + x 2 dx = 2 π Z a + h a r dx = 2 πrx | a + h a = 2 π [( ar + hr...
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This note was uploaded on 11/17/2010 for the course MAT 21B MAT 21B taught by Professor Mat21b during the Spring '07 term at UC Davis.

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Set6 - Math 21B-B Homework Set 6 Section 6.5 1 y = tan x ≤ x ≤ π 4 x-axis a Area = Z π 4 2 π tan x s 1 d dx tan x 2 dx = Z π 4 2 π tan x p

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