Tutorial06-solutions

# Moreover the jacobian is given by 1 1 2 11 14 thus we

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Unformatted text preview: − 2π ≤ 2θ ≤ π − 2π 3 or − π + 2π ≤ 2θ ≤ 2π 3 So that − π π ≤θ≤ 6 6 −π ≤ θ ≤ − 13 5π 6 5π ≤θ≤π 6 It follows that the area of the region is given by √ 2 π 6 2 θ+ −π 6 First, we have π 6 2 = 1 2 π 6 cos 2θ θ= −π 6 2 2 sin 2θ 2 π 6 −π 6 Similarly, we have 2 − 2 2 π 6 2 2 2 2 π − 56 −π 2 sin 2θ 2 π 6 1 = 2 −π 6 2 (2 cos 2θ − )θ 2 2 )θ π 5π 6 π − 6 2 1 2 5π 6 (2 2 cos 2θ − √ 3 3−π = 12 2 2 )θ ; By adding up these results, we obtain that the area is √ 3 3−π 2 3 E 2 π θ= 2 θ 1 −6 (2 2 cos 2θ − θ= 2 −π √ 2 π 3 3−π 2 − ; = 6 12 cos 2θ 5π 6 1 = 2 cos 2θ cos 2θ 5π cos 2θ sin 2θ 2 √ 2 π θ( 2 2 2 5π 6 √ 2 √ 2 π θ+ ) 2 √ 3 3−π = 6 −π 6 −π 1 2 cos 2θ √ π − 56 = 2 2 −π √ 2 √ π − 56 cos 2θ :) Applying some clever substitution, compute A cos − + where A is the trapezoidal region with vertices (1 0), (2 0), (0 2), and (0 1). S We consider the substitution = − , = + so that = ( + )/2 = ( − )/2 and in terms of , the region is given by 1/2 ≤ ≤ 1 − ≤ ≤ . Moreover, the Jacobian is given by 1 −1 =2 11 14 Thus, we have A − + cos 1 =2 = 1 2 E faces =2 sin 1 2 − 3 sin 1 2 :) Applying cylindrical coordinates, evaluate the volume bounded by the sur+ and ( 2 + 2 )2 = 2 , = 0 for > 0 and > 0. = S − 1 2 1 =4 sin 1 1 cos First, note that the curve ( 4 =2 2 2 22 + ) =2 can be rewritten as cos θ sin θ = 2 sin 2θ which is equivalent to 2 = sin 2θ Thus, in the cylindrical coordinates, the volume is given by √ π /2 sin 2θ (cos θ +sin θ ) √ π /2 θ= 0 = 1 3 π /2 0 sin 2θ 2 0 0 (cos θ +...
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## This document was uploaded on 02/10/2014.

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