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How does this integral come out to be zero?

  1. Apr 21, 2012 #1
    The question is:
    ∫ (y(4-x^2-y^2)^(3/2) dy with y=(-sqrt(4-x^2)) to sqrt(4-x^2)) dy

    I evaluated it online and it came out to be zero which was quite astonishing since both limits of y are not the same. The limits are the additive inverses of each other. Does this mean that regardless of the integrand if the limits are additive inverses of each other the integral will evaluate to be zero?
     
  2. jcsd
  3. Apr 21, 2012 #2

    I really don't understand what you mean by "[itex]y=-\sqrt{4-x^2}\,\, to\,\, \sqrt{4-x^2}\, dy[/itex]" , but

    [itex]\int y\left(4-x^2-y^2\right)^{3/2}\,dy=-\frac{1}{2}\int -2y\left(4-x^2-y^2\right)^{3/2}\, dy=\frac{2}{5}\left(4-x^2-y^2\right)^{5/2}+C\,,\,\,C=[/itex] constant in y .

    DonAntonio
     
  4. Apr 21, 2012 #3
    My bad. The limits of y were -sqrt(4-x^2) to sqrt(4-x^2)).

    (And by the way you forgot to multiply the -1/2 factor in your last step).
     
  5. Apr 21, 2012 #4


    Indeed I did, but then we're done as your integral equals

    [itex]-\frac{1}{5}\left[\left(4-x^2-(\sqrt{4-x^2})^2\right)-\left(4-x^2-(-\sqrt{4-x^2})^2\right)\right]=0[/itex] , and there's anything surprising in this as the integrand

    function is an even one (in y, of course) and thus it vanishes in any symmetric interval.

    DonAntonio
     
  6. Apr 21, 2012 #5
    Oh. Oh I see. Thanks DonAntonio.
     
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