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Why do you get the area of a whole circle when you integrate

  1. Mar 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Okay, I don't actually have a homework problem that explicitly tells me to find the reason why; it's just something I was wondering while I was trying to integrate circles using trigonometric substitution. But just for reference's sake: "Why does Acircle = ∫√(a2-x2)dx?" Additionally, when I try to take the integral of the negative formula for a semicircle, it ends up with a negative area; it's as if the circle is below the x-axis.

    2. Relevant equations
    ##x^2+y^2=a^2##
    ##a=radius##
    ##y=±\sqrt{a^2-x^2}##

    3. The attempt at a solution
    ##f(x)=∫\sqrt{a^2-x^2}dx##
    ##x=asinθ##
    ##dx=acosθdθ##
    ##θ=arcsin(\frac{x}{a})##
    ##f(θ)=∫\sqrt{a^2-a^2sin^2θ}⋅acosθdθ=∫\sqrt{a^2cos^2θ}⋅acosθdθ=a^2∫cosθ⋅cosθdθ##
    ##f(θ)=\frac{1}{2}a^2∫(1+cos2θ)dθ=\frac{1}{2}a^2(θ+\frac{1}{2}sin2θ)##
    ##f(θ)=\frac{1}{2}a^2(θ+sinθ⋅cosθ)##
    ##f(x)=\frac{1}{2}a^2(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))##

    Now how do I equate...

    ##πa^2=\frac{1}{2}a^2(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))##

    If I do some manipulation, I apparently get the the circumference of a circle.

    ##2πa=a(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))##

    I tried doing this with real numbers earlier, which is why I'm asking this question.

    ##f(x)=∫\sqrt{16-x^2}dx##
    ##x=4sinθ##
    ##dx=4cosθdθ##
    ##θ=arcsin(\frac{x}{4})##
    ##f(θ)=16∫cos^2θdθ=8∫(1+cos2θ)dθ=8θ+8sinθ⋅cosθ##
    ##f(x)=(8arcsin(\frac{x}{4})+\frac{1}{2}x\sqrt{16-x^2})## on the interval (-4,4)
    ##f(4)=4π##
    ##f(-4)=12π##
    ##f(4)+f(-4)=16π##

    Which is clearly the area of a circle with a radius of 4. So I integrated a semicircle and got the area of a circle. What gives?
     
  2. jcsd
  3. Mar 23, 2016 #2

    blue_leaf77

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    You should specify the integral limits if you want to calculate the area of your circle.
    No, it cannot be satisfied unless ##x## is allowed to be complex.
     
  4. Mar 23, 2016 #3
    I don't know how to add the limits onto the integrals on here. Anyway, though, the limits are (-a,a).

    I don't understand. Where did you get that x was larger than a?
     
  5. Mar 23, 2016 #4

    blue_leaf77

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    No, that's not true either. If you calculate the right hand side, you will get a half of the full area at most. If you want to calculate the full area, you should sum the areas under both ##+\sqrt{a^2-x^2}## and ##-\sqrt{a^2-x^2}##. Alterntively, you can work in the polar coordinates where only one integral is involved.
     
  6. Mar 23, 2016 #5
    Did you read the latter half of my post; the one with actual numbers?
     
  7. Mar 23, 2016 #6

    blue_leaf77

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    What do you usually call for ##a## and ##b## in a definite integral ##\int_a^b f(x) dx##?
    It means, that equation cannot be satisfied by a real number ##x##. In regard with the present discussion, simply put, it cannot satisfied by any ##x## within the range where the function
    $$\frac{1}{2}a^2(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))$$
    is defined.
     
  8. Mar 23, 2016 #7
  9. Mar 23, 2016 #8
    How do you figure? Why?
     
  10. Mar 23, 2016 #9

    blue_leaf77

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    Yes, but again, that will give you the area of a semicircle. The full area is then just the twice of it.
    Try to plot it using whatever function plotter you want for a given value of ##a## and see if the curve can reach ##\pi a^2##.
     
  11. Mar 23, 2016 #10
    tiYdGoM.png

    Huh. You're right. I wonder what I did wrong.
     
  12. Mar 23, 2016 #11
    According to that site, its range is ##(-\frac{π}{4},\frac{π}{4})##, and its domain is ##(-1,1)##.

    So does this mean that my bottom-most calculation is also wrong?
     
  13. Mar 23, 2016 #12

    blue_leaf77

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    The range depends on the chosen value for ##a##.
    There, I see you define ##\arcsin (-1)=3\pi/2##, and the site defines ##\arcsin (-1)=-\pi/2##, it's just a matter of definition. But the point of your mistake is that you treat the area under a curve as an indefinite integral.
     
  14. Mar 23, 2016 #13
    Oh, now I see it. I went beyond the range of the arcsin function, which is why I got the area of a full circle... somehow. But if I put it within its proper range, would it not just be zero?

    And when I tried to equate the area of a circle, and the integral of a semicircle, it was only natural that I failed, because they're not the same.
     
  15. Mar 23, 2016 #14
    I originally wanted it as ##(-a,a)##. I see now that I neglected to switch it perhaps, with the previous definition: ##θ=arcsin(\frac{x}{a})##?
     
  16. Mar 23, 2016 #15
    Nobody post any hints; I think I got it.

    In the next post...
     
  17. Mar 23, 2016 #16

    blue_leaf77

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    When you define an indefinite integral ##F(x) = \int f(x) dx##, then a definite integral ##I## calculated between the range ##[-a,a]## will be ##I=F(a)-F(-a)##. They will obviously not cancel to zero in the case of ##F(-a)=-F(a)## like the one you considered.
    Yes, that's the point.
     
  18. Mar 23, 2016 #17
    I can't figure out how to post negatives in the integral, so I'll just have capital "A" be the variable for "-4", and "B" for "-π/2".

    ##\int_A^4\sqrt{16-x^2}dθ##
    ##x=4sinθ##
    ##dx=4cosθdθ##
    ##θ=arcsin(\frac{1}{4}x)##
    ##⋅4→\frac{π}{2}##
    ##⋅-4→-\frac{π}{2}##

    ##\int_A^4\sqrt{16-x^2}dθ##
    ##\int_B^\frac{π}{2}\sqrt{16-16sinθ}4cosθdθ=16\int_B^\frac{π}{2}cos^2θdθ##
    ##8\int_B^\frac{π}{2}(1+cos2θ)dθ=(8θ+4sin2θ)|_B^\frac{π}{2}=(8θ+8sinθcosθ)|_B^\frac{π}{2}##
    ##(8θ+8sinθcosθ)|_B^\frac{π}{2}=(4π-(-4π))=8π##

    And then just add that to the integral of the other semicircle.
     
  19. Mar 23, 2016 #18
    Where ##b=-a##,

    ##\frac{1}{2}a^2(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))|_b^a=\frac{1}{2}a^2(arcsin(\frac{a}{a})+(\frac{a}{a})(\frac{\sqrt{a^2-a^2}}{a}))-\frac{1}{2}(-a)^2(arcsin(\frac{-a}{a})+(\frac{-a}{a})(\frac{\sqrt{a^2-(-a)^2}}{-a}))##
    ##\frac{1}{2}a^2(\frac{π}{2}-\frac{1}{2}a^2(\frac{-π}{2}))=\frac{1}{2}a^2(\frac{π}{2}+\frac{π}{2})##

    Now you have one side of the semicircle; just add it to the integral of the other side with the same limits, and one should end up with ##πa^2##. Thank you very much, blue, for your patience.
     
  20. Mar 23, 2016 #19

    Samy_A

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    ##\int_{-4}^4\sqrt{16-x^2}dθ##
    Put the -4 between curly brackets. The LaTeX expression is: \int_{-4}^4\sqrt{16-x^2}dθ
     
  21. Mar 23, 2016 #20
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