Tutorial06-solutions

# Further let cos cos sin sin sin this is a one to one

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Unformatted text preview: the integral 2+ 2− 2− 2 2 + 2 + 2 + 2 ≤1 Hint: use two copies of polar coordinates for S Let Jacobian is = cos θ , = sin θ , = cos υ, and for = sin υ with cos θ − sin θ 0 0 sin θ cos θ 0 0 0 0 cos υ − sin υ 0 0 sin υ cos υ 11 respectively. = ≥ 0 and ≥ 0. The Hence the integral is √ 2π 2π 0 υ θ = 4π 2 + 2 ≤1 0 1 2π 0 = 2π 2 2− 2 = 0 1 =0 √ = 1− 2 2 0 2− 2 2 1− 1 2− 2 2 2 2 2 −1 − = 0 1 π E 2 2 1 − 2 2 −1− = π2 2 + 2 0 1 2 = π 2 (cosh 1 − 1) :) Modify spherical substitution to ﬁnd the volume of the solid 2 2 2 + 2 4 + 4 ≤1 S Notice that the solid is symmetric under reﬂection from each of the coordinate planes O , O , and O and hence it’s enough to ﬁnd the volume for ≥ 0 and multiply with 8. Further, let √ ρ cos = ρ cos θ sin = ρ sin θ sin = This is a one-to-one map for > 0 since it’s a composition of spherical substitution, stretching, and square root. Thus we can use it for the given volume integral. Further, the Jacobian is cos...
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## This document was uploaded on 02/10/2014.

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