Lebesgue with substitution

maxandri

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Dec 13, 2015
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I can't solve this integral

\(\displaystyle \displaystyle \int_{D} \, \dfrac{1}{|x^2\,+\,y^4|^{\,a}}\, dx\,dy \)

with D = { (x, y) | x > 1, and x/(x+1) < y < x }

I can do y^2 = v, but with polar coordinates: It comes x^2/(x^2+1) < v <x^2, r^2 cosTh^2 / (1 + r^2cosTh^2) and so on.

Could you help me? :sad:

It is ok if you send me a picture/photo/scan of solution with all steps.

Thank you
 
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I can't solve this integral

\(\displaystyle \displaystyle \int_{D} \, \dfrac{1}{|x^2\,+\,y^4|^{\,a}}\, dx\,dy \)

It is ok if you send me a picture/photo/scan of solution with all steps.
Um... no.

This forum's policy remains the same as you'd seen in the "Read Before Posting" thread (that you read before posting, right?). Whoever told you that this had been cancelled and that this is now one of those disreputable "cheetz" sites was very much mistaken. :shock:
 
lebesgue integral with D = { (x, y) | x > 1, and x/(x+1) < y < x }

I can't solve this integral

[FONT=MathJax_Size2]∫[/FONT][FONT=MathJax_Math-italic]D[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main]4[/FONT][FONT=MathJax_Main]|[/FONT][FONT=MathJax_Math-italic]a[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Math-italic]d[/FONT][FONT=MathJax_Math-italic]y[/FONT]

with D = { (x, y) | x > 1, and x/(x+1) < y < x }

I can do y^2 = v, but with polar coordinates: It comes x^2/(x^2+1) < v <x^2, r^2 cosTh^2 / (1 + r^2cosTh^2) and so on.

Could you help me? :sad:


Thank you
 
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