- #1
Yami
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Homework Statement
Evaluate [tex]\iiint_\textrm{V} |xyz|dxdydz[/tex]
where [tex]V = \{(x,y,z) \in ℝ^3:\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} ≤ 1\}[/tex]
Homework Equations
Change of Variables Theorem:
[tex]\int_\textrm{ψ(u)} f(x)dx = \int_\textrm{K} f(\Psi(u))|detD\Psi(u)|du[/tex]
Examples:
1)
For a ball of radius a,
[tex]B(a) = \{(x,y,z) \in ℝ^3:x^2 + y^2 + z^2 ≤ a^2\}[/tex]
[tex]vol B(a) = \int_\textrm{B(a)}1 dxdydz[/tex]
[tex]= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{a} \rho^2 sin \phi d \rho d \phi d \theta[/tex](change of variables to spherical coodinates)
2)
For a continuous function f: D → ℝ where
[tex]D = \{(x,y) :\frac{x^2}{a^2} + \frac{y^2}{b^2} ≤ 1\}[/tex]
define ψ: ℝ^2 → ℝ^2 by ψ(au, bv) for all u,v in ℝ^2. ψ is a smooth change of variables.
Then
[tex]\int_\textrm{D}f(x,y)dxdy = ab\int_\textrm{u^2 + v^2 ≤ 1}f(au, bv) du dv[/tex]
[tex] = ab\int_{0}^{2\pi} \int_{0}^{1}f(ar cos \theta, br sin \theta) r dr d \theta . [/tex]
(change of variables to polar coordinates)
The Attempt at a Solution
Based on those examples above in the book, I set this up:
Let ψ(u,v,w) = (au, bv, cw)
Then
[tex]\iiint_\textrm{V} |xyz|dxdydz = abc\int_\textrm{u^2 + v^2 + w^2≤ 1}|abcuvw| du dvdw[/tex]
change of variables to spherical coordinates:
[tex] = abc\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} |abc\rho^3 cos \phi sin^2 \phi cos \theta sin \theta|\rho^2 sin \phi d \rho d \phi d \theta . [/tex]
[tex] = (abc)^2\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \rho^5 cos \phi sin^3 \phi cos \theta sin \theta d \rho d \phi d \theta . [/tex]
p^3 is positive on [0, 1] so I ignored the absolute value lines.
This eventually led to an answer of 0 since one of the antiderivates is sin^4(phi)on [0,pi] which is zero. This is wrong. So I'm guessing the integral I set up or the way I evaluated it is wrong. But I got half credit for it, so I assume some part of it is right.