(adsbygoogle = window.adsbygoogle || []).push({}); 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]

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# Evaluate double integral using transformations

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