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Unformatted text preview: nce cos 2 ≤ 0, which implies
π
2 √ π
2 8 ρ sin 0 π
4 π
4 ≤ − cos 2 0 4π
3 π
2 3
π
4 (1 − 2 cos Further, substituting cos 0 (c) First, notice that ( 2 +
following four cases:
≥ 0; 4 0
π
2 ≥0 ) 3/ 2 + π
2 3
π
4 4π
=
3 sin 4π 3
=√
32 1 + 2 cos 2 +
4
2 cos 2 (− cos 2 )3/2 sin
1
√
2 3 (1 − 2 = 2 3/ 2 ) 0 , we obtain π
2 4π 3
√
32
4π 3
√
32 sin
√
2 = 2 2 ≤ π . Thus the answer is
2 4π
θ=
3 ρ 2 ≤ π . The volume
2 ≥ 0 implies that 0 ≤ θ Further, in spherical coordinates,
we are considering is given by 1+cos 4
2 ) ≤3 2 1 + cos 2
2 0 = = 4π 3 3π
π2 3
√·
=√
3 2 16
42 implies ≤ 0; ≥ 0 and hence there are the ≤0 22 ≤0 π
2 ≤ 0; ≥0 ≤0 ≤0 ≥0 Also notice that changing signs of two of the three coordinates
doesn’t change
the inequality. Hence we have for equal parts of the region, so we can ﬁnd the
volume for
≥ 0 and multiply with 4. Further, in spherical coordinates we 8 have ρ ≤ 3 cos cos θ sin θ and hence the answer is sin2 π
2 π
2 sin2 3 c...
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This document was uploaded on 02/10/2014.
 Spring '13

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