- #1
DryRun
Gold Member
- 838
- 4
Homework Statement
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]
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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