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Evaluate double integral using transformations

  1. Jan 7, 2012 #1

    sharks

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    The problem statement, all variables and given/known data
    http://s2.ipicture.ru/uploads/20120107/vVVkUT7f.jpg

    The attempt at a solution
    I plotted the graph x-y:
    http://s2.ipicture.ru/uploads/20120107/ja3V9aSV.jpg

    [itex]y=\frac{1}{2}(u+v)[/itex] and [itex]x=\frac{1}{2}(u-v)[/itex]

    So, after finding the Jacobian, the double integral becomes: [itex]\int\int \frac{1}{1+u^2}\,.\frac{1}{2}dvdu[/itex]

    Now, to find the transformed limits, i plot another graph u-v:
    http://s2.ipicture.ru/uploads/20120107/RrA8UKQ1.jpg

    From the graph above, the shaded area is what i need to find. Since the area is symmetrical about the v-axis, i decided to evaluate the lower half of the right-hand side and then multiply by 4.

    So, the limits for the lower right-half (meaning the section found under the line v=a) is:
    For u fixed, v varies from v=u to v=a
    u varies from u=0 to u=a

    The double integral becomes:
    [tex]4\int^a_0 \int^a_u \frac{1}{1+u^2}\,.dvdu[/tex]
    The answer to that integral above is ultimately incorrect, hence why i have no idea where i messed up. The answer i got is:
    [tex]2a\arctan a - ln(1+a^2)[/tex]
     
    Last edited: Jan 7, 2012
  2. jcsd
  3. Jan 7, 2012 #2

    Dick

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    How can your variable transformation turn a triangle into a quadrilateral? Which point in the triangle is going to map to (2a,0)?
     
  4. Jan 7, 2012 #3

    sharks

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    OK, i have fixed the graph of u-v:
    http://s2.ipicture.ru/uploads/20120108/cvc1aJm9.jpg

    Since the area of region is symmetrical about the u-axis, taking limits of upper triangular section and then multiply by 2.

    The limits:
    For u fixed, v varies from v=0 to v=u
    u varies from u=0 to u=a

    [tex]2\int^a_0 \int^u_0 \frac{1}{1+u^2}\,.\frac{1}{2}.dvdu[/tex]

    The answer is [itex]\frac{1}{2}ln (1+a^2)[/itex]
     
  5. Jan 8, 2012 #4

    ehild

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    What did you got for the first integral with respect to v? Take care, the integrand does not depend on v!

    ehild
     
  6. Jan 9, 2012 #5

    sharks

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    For the first integral w.r.t.v. i got (after putting in the limits for v and then u):
    [tex]\frac{u}{1+u^2}\,.\frac{1}{2}du=\frac{1}{2}ln (1+a^2)[/tex]
     
    Last edited: Jan 9, 2012
  7. Jan 9, 2012 #6

    Dick

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    Your answer look ok to me.
     
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