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Thread: integration...should i use polar transformation?

  1. #1
    Senior Member Sambit's Avatar
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    Question integration...should i use polar transformation?

    Evaluate $\displaystyle \iint\limits_R \frac{1}{(1 + x^2 + y^2)^\frac{3}{2}} dx dy$
    where $\displaystyle R $ is the region bounded by the straight lines $\displaystyle y=0 , x=1 , y=x $

    if i use polar transformation, what will be the ranges of integration? or is there any way other than polar transformation?
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  2. #2
    A Plied Mathematician
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    I think I'd stay in cartesian coordinates for this one. What is your integral, including limits?
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  3. #3
    Senior Member Sambit's Avatar
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    writing in terms of limits, the integration becomes $\displaystyle \int_0^1\int_0^x \frac{1}{(1 + x^2 + y^2)^\frac{3}{2}} dy dx $

    how to evaluate this?
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  4. #4
    A Plied Mathematician
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    Right. So the $\displaystyle y$ indefinite integral is of the form

    $\displaystyle \displaystyle\int\frac{dy}{(a^{2}+y^{2})^{3/2}}.$

    How would you compute that antiderivative?
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  5. #5
    Senior Member Sambit's Avatar
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    we substitute $\displaystyle y = a \tan\theta $, which gives the final value of the above integral as $\displaystyle \frac{1}{a^2}\sin y $, ie, $\displaystyle \frac{sin x}{1+x^2} $. right?? then what?
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  6. #6
    A Plied Mathematician
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    I think you may have skipped too many steps there. I agree with the substitution, but I get a different result. How do the integrand and differential change with your trig substitution?
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  7. #7
    Senior Member Sambit's Avatar
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    we substitute $\displaystyle y = a \tan\theta $, which gives the final value of the above integral as $\displaystyle \frac{1}{a^2} \frac{y}{\sqrt{y^2+a^2}} $. ok? that is $\displaystyle \frac{x}{(1+x^2)(\sqrt{1+2x^2})} $. right??
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  8. #8
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    The final result is correct. So, now what? You've got

    $\displaystyle \displaystyle\int_{0}^{1}\frac{x}{(1+x^2)(\sqrt{1+ 2x^2})}\,dx.$

    Ideas?
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  9. #9
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    here we substitute $\displaystyle x^2 = z $ then $\displaystyle 1 + 2z = t^2 $ this gives us the final result as $\displaystyle \arctan\sqrt{2x^2 + 1} $. am i right?
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  10. #10
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    Not sure I'm following your steps here. The final result is incorrect, actually. It should be arctan.
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  11. #11
    Senior Member Sambit's Avatar
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    ya...sorry..it will be tan inverse....just a little typing mistake.....sorry for that

    i have made the correction in the previous post. so the answer comes to be $\displaystyle \arctan \sqrt{3} = \frac{\pi}{3} $
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  12. #12
    A Plied Mathematician
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    Don't forget the lower limit.
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  13. #13
    Senior Member Sambit's Avatar
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    oooops....so much silly mistakes...... yeah. the answer is then $\displaystyle \frac{\pi}{3}-\frac{\pi}{4} = \frac{\pi}{12}$.....

    many many thanks:d
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  14. #14
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    There you go. Nice work.

    One way to avoid silly mistakes is to pretend that someone is holding a gun to your temple, and if you make a mistake, he's going to pull the trigger. Or, you could work at a high-energy physics lab, with all those 10,000V wires running around. If you make a mistake, you're dead.

    So there are some nice, cheerful ways to avoid mistakes.
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