Hence the volume is 2 2 2 2 0 2 2 0 0

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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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