Tutorial06-solutions

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 02/10/2014.

Ask a homework question - tutors are online