Timeline for Are there any alternatives for NIntegrate to calculate the area for which $f(x,y)<0$?
Current License: CC BY-SA 4.0
11 events
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| Jun 18, 2021 at 22:11 | history | edited | eyorble | CC BY-SA 4.0 |
Add exact solution notes to the end.
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| Jun 18, 2021 at 19:12 | comment | added | eyorble |
A note: although it takes a while to evaluate, Integrate can find a closed form for the final expression as well.
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| Jun 18, 2021 at 12:54 | comment | added | yarchik | Thank you, much better now! | |
| Jun 18, 2021 at 12:11 | comment | added | eyorble | @yarchik I have modified the answer to hopefully clarify what's being integrated here. | |
| Jun 18, 2021 at 12:11 | history | edited | eyorble | CC BY-SA 4.0 |
Labelling the integrands to improve clarity.
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| Jun 18, 2021 at 12:02 | comment | added | eyorble |
@yarchik The y in the integration should be read as y /. sol[[2]] (which is the upper curve from the contour plot) or y /. sol[[4]] (which is the lower curve from the contour plot). These are just explicit forms (in x) of the curves where f[x,y]==0, selected to match the curves we are interested in.
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| Jun 18, 2021 at 11:51 | comment | added | yarchik |
There are a lot of inconsistencies in your answer. What is y that you are integrating?
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| Jun 18, 2021 at 1:49 | history | edited | eyorble | CC BY-SA 4.0 |
Adding Michael Seifert's helpful comments to the answer, removed a note that no longer applies since a typo in the question was fixed.
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| Jun 17, 2021 at 20:12 | comment | added | Michael Seifert | If nothing else, it shows that the intersection of the lower curve with the $x$-axis is when $2 \cos x + 1 = 0$, or $x = 2 \pi/3$. | |
| Jun 17, 2021 at 20:10 | comment | added | Michael Seifert |
Using TrigFactor on f[x,y] shows that the two curves are $\cos((x-y)/2) + 2 \cos((x+y)/2) = 0$ (upper) and $2 \cos (x+y/2)+ \cos(y/2) = 0$ (lower). I don't know if that really helps, though.
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| Jun 17, 2021 at 19:13 | history | answered | eyorble | CC BY-SA 4.0 |