Unformatted text preview: meaningful only when ≤ . In this case, we have
is
2− 2 ≤
2 − 2 . Note that we also have
≤+
−
√ ≥ and it is easy to
√
2− 2≤
2 − 2 , now squaring
(this is equivalent to (0 ≤) − ≤
see that −
both sides and simplifying the expressions, it’s easy to see that this inequality is
equivalent to ≤ ).
Hence the volume is √
+ 2π √ 2− 2 θ = 2π
0 2 − 2 − 0 0 = 2π 2 + − 2 1
3 2 2 3/ 2 − − 3 3 = 2π 3 +
2 0 2 3/ 2 1
3 − 3 =π 3 3 On the other hand, we can do the same in spherical coordinates. The region is
given by ρ ≤ 2 cos (and hence 0 ≤
≤ π ) and sin2 ≤ cos2 (and hence
2
π
0 ≤ ≤ 4 ). Thus the volume is
2π π /4 2 cos ρ2 sin
0 0 ρ θ= 3 16 cos3
3 0
3 16
=
3 π − cos4
4 π /4 π sin 0 π /4 =π 3 0 (b) First of all, notice that region is symmetric under reﬂection from each of the planes
O , O , and O
and hence it’s enough to compute the volume for
≥0
and multiply the answer with 8.
7 ρ2 ≤ 2 − cos2 ) = − (sin2 and he...
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This document was uploaded on 02/10/2014.
 Spring '13

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