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Unformatted text preview: θ sin
ρ cos θ cos
−ρ sin θ sin
sin θ sin
ρ sin θ cos
ρ cos θ sin
cos
sin
ρ
1
−2
0
2
ρ
cos
Expanding along the last row, we obtain
ρ2 cos
ρ 2 Thus the answer is
π
2 π
2 1 4
0 0 0 sin cos + √
ρρ
sin
√
cos ρ ρ sin
2 ρ
cos θ=4 · 12 2 sin √
ρρ
=√
sin
2 cos π
√
π
2
8π
2
· [−2 cos ]0 · =
2
5
5 :) 5
E Extra examples Find the volume of the wedge cut from the cylinder
= − and = 0 for ≥ 0. E S + 2 = 1 by planes Applying polar coordinates, ﬁnd the area given by
2 Here 2 + 22 ≤2 2 2 − 2 2 + 2 ≥ 2 > 0.
In polar coordinates, the region can be rewritten as
4 ≤2 2 ( 2 cos2 θ − 2 sin2 θ ); cos 2θ ≥ 2 ≥ 2 which is equivalent to
2 ≤2 2 Thus, we must have
2 2 cos 2θ ≥ 2 Or equivalently,
cos 2θ ≥ 1
2 which implies that
− π
π
+ 2 π ≤ 2θ ≤ + 2 π
3
3 for any ∈ Z. We can take θ ∈ [−π π ] so that 2θ ∈ [−2π 2π ]). In this case, the above
inequalities imply that the only possible ’s we can take here are = 0 ±1 and we have
− π
π
≤ 2θ ≤
3
3 or...
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This document was uploaded on 02/10/2014.
 Spring '13

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