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 onetoone 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.
 Spring '13

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