Tutorial06-solutions

# Thus the answer is 2 4 3 2 the volume 2 0 implies

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: 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...
View Full Document

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

Ask a homework question - tutors are online