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Integral calculus involving Change of Variables Theorem

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data
    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]


    2. Relevant 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)

    3. 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.
     
  2. jcsd
  3. May 3, 2012 #2

    hunt_mat

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    Homework Helper

    I think it is your region of integration, the original integral had modulus signs in and that will restrict which regions you have for [itex]\theta[/itex] and [itex]\phi[/itex]. So there are 8 regions you need to consider what the integrand does, you need to think about this, as a simple warm up problem, think about what you would do with the following integral:
    [tex]
    \int_{-a}^{a}|x|dx
    [/tex]
    Then perhaps move up to the double integral where you have only 4 regions to consider.
     
  4. May 3, 2012 #3
    You can't remove the absolute values from the angles. [itex] \sin(\phi) [/itex]is always positive in the integral domain, but all the other trigonometric functions change signs.
     
  5. May 3, 2012 #4
    Okay so I can't ignore the absolute value...
    And you're saying these are the wrong regions?
    [tex]\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1}[/tex]?
     
  6. May 3, 2012 #5

    hunt_mat

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    Did you see my suggestion? It turns the integrand into an even function (see the example I gave) and as with the example:
    [tex]
    \int_{-a}^{a}|x|dx=2\int_{0}^{a}xdx=a^{2}
    [/tex]
    How does this example fit in with your question?

    To answer your question, yes, it is your region that you're integrating over.
     
  7. May 3, 2012 #6
    Okay I'm a little confused.

    I removed the abc from the integral for the sake of simplicity and tried plugging it into Matlab's triplequad() function. The answer is 1/6 according to Matlab, but obviously I need to know how to evaluate this by hand.

    So I'm trying to see how it got to 1/6:

    [tex]\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} |\rho^3 cos \phi sin^2 \phi cos \theta sin \theta|\rho^2 sin \phi d \rho d \phi d \theta[/tex]
    [tex]= \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} |\rho^3|| cos \phi sin^2 \phi cos \theta sin \theta|\rho^2 sin \phi d \rho d \phi d \theta = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \rho^5 |cos \phi sin^2 \phi cos \theta sin \theta| sin \phi d \rho d \phi d \theta[/tex]
    [tex]=\int_{0}^{2\pi} \int_{0}^{\pi} \frac{1}{6}|cos \phi sin^2 \phi ||cos \theta sin \theta| sin \phi d \phi d \theta = \frac{1}{6} \int_{0}^{2\pi} |cos \theta sin \theta|[\int_{0}^{0.5\pi} (cos \phi sin^2 \phi ) sin \phi d \phi + \int_{0.5\pi}^{\pi} -(cos \phi sin^2 \phi ) sin \phi d \phi ] d \theta[/tex]
    [tex]=\frac{1}{6}\int_{0}^{2\pi}2|cos \theta sin \theta| d\theta[/tex]
    which I'm guessing I would spit into
    [tex]=\frac{1}{3}[\int_{0}^{0.5\pi}cos \theta sin \theta d \theta + \int_{0.5\pi}^{\pi}-cos \theta sin \theta d \theta + \int_{\pi}^{1.5\pi}cos \theta sin \theta d \theta + \int_{1.5\pi}^{2\pi}-cos \theta sin \theta d \theta][/tex]
    But I don't think that's 1/6
    Is any of this correct at all?
     
    Last edited: May 3, 2012
  8. May 3, 2012 #7

    hunt_mat

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    Homework Helper

    My idea was that you know that the modulus makes the integrand an even function, so you can reduce the limits of the integral. For the 2D example over the disc, we can reduce the integral:
    [tex]
    \int\int_{D}|xy|dxdy=4\int\int_{D_{1}}xydxdy
    [/tex]
    Where [itex]D=\{ (x,y)|x^{2}+y^{2}\leqslant 1\}[/itex] and [itex]D_{1}=\{ (x,y):x^{2}+y^{2}\leqslant 1,x\geqslant 0,y\geqslant 0\}[/itex]
     
  9. May 4, 2012 #8
    I'm still confused... so none of what I did above is correct?

    Okay so you're saying this would work?
    [tex]\int\int_{D}|xyz|dxdydz=8\iiint_{D_{1}}xyzdxdydz[/tex]
    [tex]D=\{ (x,y,z)|x^2+y^2+z^2 ≤ 1\}[/tex] and [tex]D_{1}=\{ (x,y,z)|x^2+y^2+z^2 ≤ 1, x ≥ 0, y≥0, z ≥ 0\}[/tex]
     
  10. May 4, 2012 #9

    hunt_mat

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    Yes, you are beginning to use the symmetry now. Did you test it to see if you got the right answer?

    I think what you did is a good idea but it can lead to errors if you're not careful.

    Mat
     
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