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Transformations in double integrals. (Jacobian)

  1. Apr 12, 2005 #1

    [tex]\int\int_{R} \left(2x^2 - xy - y^2\right) dx\;dy[/tex]

    by applying the transformation

    [tex]u = x - y , v = 2x + y[/tex]

    for the region R in the first quadrant bounded by the lines

    [tex]y = -2x + 4, y = -2x + 7, y = x - 2, y = x + 1[/tex]

    I don't even know where to start! Please help.

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    Last edited: Apr 12, 2005
  2. jcsd
  3. Apr 12, 2005 #2
    I have this so far

    [tex]y = x - u[/tex]

    [tex]v = 2x + x - u[/tex]

    [tex]v = 3x - u[/tex]

    [tex]3x = v + u[/tex]

    [tex]x = \frac{1}{3} \left(u + v\right)[/tex]

    [tex]y = \frac{1}{3} \left(-2u + v\right)[/tex]

    then, using partial derivatives to get the Jacobian to be [tex]\frac{1}{3}[/tex], right?

    Then I use those equations and the line equations to solve for the limits of u and v:

    [tex]u = -1\;to\;2[/tex]

    [tex]v = 4\;to\;7[/tex]

    The resulting transformed integral (with Jacobian in front)would then be:

    [tex]\frac{1}{3}\int_{-1}^{2}\int_{4}^{7} v\;u\;dv\;du = \frac{33}{4}[/tex]

    Is this correct?
  4. Apr 12, 2005 #3


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    I think you need to transform the integrand in terms of u and v, that is:

    [tex]\iint_R F(x,y)dxdy=\pm\iint_S F(f(u,v),g(u,v))\frac{\partial (x,y)}{\partial (u,v)}dudv[/tex]

    Note the plus/minus sign.
    Last edited: Apr 12, 2005
  5. Apr 12, 2005 #4
    Is this part correct?

    [tex]\frac{\partial (x,y)}{\partial (u,v)} = \frac{1}{3}[/tex]

    But what is this equal to, how do I find it?

    [tex]F(f(u,v),g(u,v)) =\;?[/tex]
  6. Apr 12, 2005 #5


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    The Jacobian is correct. It's just a 2x2 determinant (1/6+1/6) right?

    To transform the integrand, just substitute the expressions for x and y in terms of u and v. You know, x=1/3(u+v) and y=1/3(v-2u). So [itex]x^2[/itex] is . . . xy is . . . and so on. Then integrate over the transformed area in the u-v plane which is a nice square.
  7. Apr 12, 2005 #6
    After your advice, i get the following

    [tex]\frac{1}{3}\;\int_{-1}^{2}\int_{4}^{7} \left(-\frac{1}{9}\;u^2 - \frac{1}{9}\;v^2 + \frac{7}{9}\;u\;v\right)\;dv\;du[/tex]

    Is this right?
  8. Apr 12, 2005 #7


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    Well, I didn't simplify it but for:





    Isn't that just:

  9. Apr 12, 2005 #8
    Simplified and hopefully I am understanding the concept now!



    simplifies to

    [tex]-\frac{4}{9}\;u^2 + \frac{2}{9}\;v^2 + \frac{7}{9}\;u\;v[/tex]

    (I think :confused: )

    Now that the Jacobian has been checked and we just simplified the re-substitution (correct term? :confused: ), are the limits of integration correct?

    [tex]\frac{1}{3}\;\int_{-1}^{2}\int_{4}^{7} \left(-\frac{4}{9}\;u^2 + \frac{2}{9}\;v^2 + \frac{7}{9}\;u\;v\right)\;dv\;du[/tex]

    Does everything seem to be in order?
  10. Apr 12, 2005 #9


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    Vinny, sorry I got to you late. Probably you've figured it out by now but I get just uv for the integrand:

    Last edited: Apr 12, 2005
  11. Apr 13, 2005 #10
    just wondering

    Just wondering... I'm learning the change of variables Jacobian things too right now and also having some problems as well.

    Would someone tell me how to find the limits of integration for 'u' and 'v'?
    Like you got:

    u = -1 to 2
    v = 4 to 7

    How do you find those?
  12. Apr 13, 2005 #11
    The area is bounded by y=-2x+4 and y=-2x+7, so that's y+2x=4 and y+2x=7, or v=4 to v=7, so v goes from 4 to 7, and the same for u. With this change of variables, you're actually fitting the area of integration to a rectangle.
  13. Apr 13, 2005 #12
    yea. that makes so much sense now, putting it into a rectangle like that. thanks.
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