A3W08-138Sol

# A3W08-138Sol - Math 138 Assignment 3 Solutions Fall 2008 1...

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Unformatted text preview: Math 138 Assignment 3 Solutions Fall 2008 1. Determine the volume obtained by rotating the region bounded by y = sec x, y = cos x, ≤ x ≤ π 3 about the line y =- 1. Solution 1 y x y=cos x y=sinx 3 On the interval 0 ≤ x ≤ π 3 , sec x ≥ cos x thus the volume, V , is given by (using circular disks): V = Z π 3 [ π (sec x- (- 1)) 2- π (cos x- (- 1)) 2 ] dx = Z π 3 π [sec 2 x + 2 sec x + 1- cos 2 x- 2 cos x- 1] dx = Z π 3 π [sec 2 x + 2 sec x- 1 2 (1 + cos(2 x ))- 2 cos x ] dx = π tan x + 2 ln | sec x + tan x |- x 2- 1 4 sin(2 x )- 2 sin x π 3 = π " √ 3 + 2 ln | 2 + √ 3 |- π 6- √ 3 8- √ 3- (0 + 2 ln | 1 + 0 |--- 0) # = 2 π ln(2 + √ 3)- π 2 6- √ 3 8 π The volume of the solid is 2 π ln(2 + √ 3)- π 2 6- √ 3 8 π . 2. Evaluate the following a) Z sec 4 x 2 dx Solution Math 138 Assignment 3 Solutions Fall 2008 Let u = x 2 so du = 1 2 dx , and substitution gives: Z sec 4 x 2 dx = Z 2 sec 4 u du = 2 Z (1 + tan 2 u ) sec 2 u du = 2 Z sec 2 u + tan 2 u sec 2 u du = 2 tan u + 1 3 tan 3 u + C = 2 tan x 2 + 1 3 tan 3 x 2 + C = 2 tan x 2 + 2 3 tan 3 x 2 + C b) Z π 2 e 2 x cos xdx Solution Let I = Z e 2 x cos xdx , and use integration by parts with u = e 2 x , dv = cos xdx , so du = 2 e 2 x dx and v = sin x . This gives: I = e 2 x sin x- Z 2 e 2 x sin xdx Use integration by parts again with u = 2 e 2 x and dv = sin xdx , so du = 4 e 2 x and v...
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A3W08-138Sol - Math 138 Assignment 3 Solutions Fall 2008 1...

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