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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.
 Spring '13

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