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

  • #1
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Homework Statement


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.

Homework Equations


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

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?
 

Answers and Replies

  • #2
blue_leaf77
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##
f(x)=∫\sqrt{a^2-x^2}dx##
You should specify the integral limits if you want to calculate the area of your circle.
##
πa^2=\frac{1}{2}a^2(arcsin(\frac{x}{a})+(\frac{x}{a})(\frac{\sqrt{a^2-x^2}}{a}))##
No, it cannot be satisfied unless ##x## is allowed to be complex.
 
  • #3
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You should specify the integral limits if you want to calculate the area of your circle.
I don't know how to add the limits onto the integrals on here. Anyway, though, the limits are (-a,a).

No, it cannot be satisfied unless ##x## is allowed to be complex.
I don't understand. Where did you get that x was larger than a?
 
  • #4
blue_leaf77
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Why does Acircle = ∫√(a2-x2)dx?
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.
 
  • #5
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Did you read the latter half of my post; the one with actual numbers?
 
  • #6
blue_leaf77
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I don't know how to add the limits onto the integrals on here.
What do you usually call for ##a## and ##b## in a definite integral ##\int_a^b f(x) dx##?
I don't understand. Where did you get that x was larger than a?
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
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In regard with the present discussion, simply put, it cannot satisfied by any xx 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.
How do you figure? Why?
 
  • #10
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tiYdGoM.png


Huh. You're right. I wonder what I did wrong.
 
  • #11
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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?
 
  • #12
blue_leaf77
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According to that site, its range is ##(-\frac{π}{4},\frac{π}{4})##, and its domain is ##(-1,1)##.
The range depends on the chosen value for ##a##.
So does this mean that my bottom-most calculation is also wrong?
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.
 
  • #13
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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.
 
  • #14
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But the point of your mistake is that you treat the area under a curve as an indefinite integral.
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})##?
 
  • #15
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Nobody post any hints; I think I got it.

In the next post...
 
  • #16
blue_leaf77
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But if I put it within its proper range, would it not just be zero?
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.
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.
Yes, that's the point.
 
  • #17
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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.
 
  • #18
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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.
 
  • #19
Samy_A
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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##
##\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
blue_leaf77
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I can't figure out how to post negatives in the integral,
Code:
\int_{-a}^a
And then just add that to the integral of the other semicircle.
There is a tricky part here, as you have noticed in the first post, the integral of the negative semicircle yields a negative value. If you bluntly add the areas of the two semicircles together you will get zero.
A more neat way to calculate the area of a full circle is by using polar coordinate.
 
  • #22
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I haven't gotten to polar coordinates yet. Thanks for suggesting it, though.

I feel more satisfied that I was able to derive the area formula for a circle by integration, anyway.
 

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