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Double integral problem

  • #1
8yv6i.png








My idea was that the limits are
BrQKm.png


and that the anti-derivative of dy was

xlog(1+y^2)

but that seems wrong...

maybe use these limits instead


YKvJd.png


and start with dx?

gives us
rDFwG.png


then we take dy

rgKWx.png


guess, i figured it out eventually with the help of wolfram with the last integration
 
Last edited:

Answers and Replies

  • #2
SammyS
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The reason your 1st attempt didn't work is that [tex]\int\frac{1}{1+y^2}\,dy=\tan^{-1}(y)+C\,.[/tex]
 
  • #3
HallsofIvy
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8yv6i.png








My idea was that the limits are
BrQKm.png


and that the anti-derivative of dy was

xlog(1+y^2)

but that seems wrong...
Yes, it is wrong. It is standard "Calculus I" mistake to treat a function of the variable as if it were just the variable but you should be past that by the time you are doing multiple integrals. "[itex]1/(1+ y^2)[/itex]" is NOT the same as [itex]1/y[/itex] and its anti-derivative is not a logarithm. The anti-derivative of [itex]1/(1+ y^2)[/itex] is [itex]arctan(y)+ C[/itex]. That's a standard anti-derivative that you should have memorized.

maybe use these limits instead


YKvJd.png


and start with dx?

gives us
rDFwG.png


then we take dy

rgKWx.png


guess, i figured it out eventually with the help of wolfram with the last integration
Yes, reversing the order of integration is the best way to handle this one. Integerating [itex]arctan(1)- arctan(x^2)= \pi/4- arctan(x^2)[/itex] is likely to be very difficult!
 
  • #4
SammyS
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Yes, reversing the order of integration is the best way to handle this one. Integerating [itex]arctan(1)- arctan(x^2)= \pi/4- arctan(x^2)[/itex] is likely to be very difficult!
Yes. I agree !
 

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