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Change of variables in multiple integrals. What am I doing wrong?

  1. Jan 22, 2010 #1
    the problem:

    evaluate the following integral by making appropriate change of variables.

    double integral, over region R, of xy dA

    R is bounded by lines:
    2x - y = 1
    2x - y = -3
    3x + y = 1
    3x + y = -2

    my attempt:

    let 2x - y = u, and let 3x + y = v

    then the new region in (u,v) coordinates is bounded by the following lines:
    u = 1
    u = -3
    v = 1
    v = -3

    I calculated the Jacobian of u and v, with respect to x and y, and got 5. The area in (u,v) coordinates also looks like it could be 5 times the area in (x,y) coordinates.

    Solving for x and y, I get x = (u + v)/5, y = (2v - 3u)/5

    Then I integrated [(-uv - 3u2 + 2v2)/125] du dv, where u is from -3 to 1, and v is from -2 t o 1.

    I got -102/125. The correct answer is -66/125.

    What am I doing wrong?
     
  2. jcsd
  3. Jan 23, 2010 #2

    CompuChip

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    Science Advisor
    Homework Helper

    I agree with everything you said up to the last line, which means the error must be in your carrying out the integration.

    The integral of -3u2 gives -3 (13 - (-3)3) = -84, the v2-integral gives 24 and the u v integral contributes -6. This adds up to -66 for me.
     
  4. Jan 26, 2010 #3
    thank you so much :)
     
  5. Jan 27, 2010 #4
    Uh ... no!

    [tex]\int^{1}_{-3} -3u^2 du = -1\times\int^{1}_{-3} 3u^2 du = -\left( u^3 \right) \Big|^{1}_{-3} = -\big(1^3 - (-3)^3\big) = -\big(1 - (-27)\big) = -(1 + 27) = -28[/tex]
     
  6. Jan 27, 2010 #5
    Here's what I came up with:

    [tex]\int^{1}_{-2} \int^{1}_{-3} \frac{-uv - 3u^2 + 2v^2}{125}\,du\,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{1}{125}\int^{1}_{-3} -uv-3u^2+2v^2\,du\,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{1}{125}\left( -\frac{1}{2}u^2v - u^3 + 2uv^2 \right) \bigg|^{1}_{-3} \,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{1}{125}\Bigg( \bigg(-\frac{1}{2}v(1)^2 - (1)^3 + 2v^2(1)\bigg) - \bigg(-\frac{1}{2}v(-3)^2 - (-3)^3 + 2v^2(-3)\bigg)\Bigg)\,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{1}{125}\Bigg( \bigg(-\frac{1}{2}v - 1 + 2v^2\bigg) - \bigg(-\frac{1}{2}v(9) - (-27) - 6v^2\bigg)\Bigg)\,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{1}{125}\Bigg( \bigg(-\frac{1}{2}v - 1 + 2v^2\bigg) - \bigg(-\frac{9}{2}v + 27 - 6v^2\bigg)\Bigg)\,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{1}{125}\bigg( -\frac{1}{2}v - 1 + 2v^2 + \frac{9}{2}v - 27 + 6v^2\bigg)\,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{1}{125}\Big( 8v^2 + 4v - 28 \Big)\,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{1}{125}\Big( 4(2v^2 + v - 7) \Big)\,dv[/tex]

    [tex]= \int^{1}_{-2} \frac{4}{125}\Big( 2v^2 + v - 7 \Big)\,dv[/tex]

    [tex]= \frac{4}{125} \int^{1}_{-2}\Big( 2v^2 + v - 7 \Big)\,dv[/tex]

    [tex]= \frac{4}{125} \Bigg( \frac{2}{3}v^3 + \frac{1}{2}v^2 - 7v \Bigg) \Bigg|^{1}_{-2}[/tex]

    [tex]= \frac{4}{125} \Bigg( \bigg( \frac{2}{3}(1)^3 + \frac{1}{2}(1)^2 - 7(1)\bigg) - \bigg( \frac{2}{3}(-2)^3 + \frac{1}{2}(-2)^2 - 7(-2) \bigg) \Bigg)[/tex]

    [tex]= \frac{4}{125} \Bigg( \bigg( \frac{2}{3} + \frac{1}{2} - 7\bigg) - \bigg( \frac{2}{3}(-8) + \frac{1}{2}(4) + 14 \bigg) \Bigg)[/tex]

    [tex]= \frac{4}{125} \Bigg( \bigg( \frac{4}{6} + \frac{3}{6} - \frac{42}{6}\bigg) - \bigg( -\frac{16}{3} + \frac{4}{2} + 14 \bigg) \Bigg)[/tex]

    [tex]= \frac{4}{125} \Bigg( \bigg( \frac{4}{6} + \frac{3}{6} - \frac{42}{6}\bigg) - \bigg( -\frac{32}{6} + \frac{12}{6} + \frac{84}{6} \bigg) \Bigg)[/tex]

    [tex]= \frac{4}{125} \Bigg( \frac{4 + 3 - 42 + 32 - 12 - 84}{6}\Bigg)[/tex]

    [tex]= \frac{4}{125} \Bigg( -\frac{99}{6}\Bigg)[/tex]

    [tex]= \frac{4}{125} \Bigg( -\frac{33}{2}\Bigg)[/tex]

    [tex]= -\frac{66}{125}[/tex]
     
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