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

1. Mar 23, 2016

### Eclair_de_XII

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. Mar 23, 2016

### blue_leaf77

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.

3. Mar 23, 2016

### Eclair_de_XII

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?

4. Mar 23, 2016

### blue_leaf77

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. Mar 23, 2016

### Eclair_de_XII

Did you read the latter half of my post; the one with actual numbers?

6. Mar 23, 2016

### blue_leaf77

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.

7. Mar 23, 2016

### Eclair_de_XII

8. Mar 23, 2016

### Eclair_de_XII

How do you figure? Why?

9. Mar 23, 2016

### blue_leaf77

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 $\pi a^2$.

10. Mar 23, 2016

### Eclair_de_XII

Huh. You're right. I wonder what I did wrong.

11. Mar 23, 2016

### Eclair_de_XII

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. Mar 23, 2016

### blue_leaf77

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.

13. Mar 23, 2016

### Eclair_de_XII

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. Mar 23, 2016

### Eclair_de_XII

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. Mar 23, 2016

### Eclair_de_XII

Nobody post any hints; I think I got it.

In the next post...

16. Mar 23, 2016

### blue_leaf77

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.

17. Mar 23, 2016

### Eclair_de_XII

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. Mar 23, 2016

### Eclair_de_XII

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. Mar 23, 2016

### Samy_A

$\int_{-4}^4\sqrt{16-x^2}dθ$
Put the -4 between curly brackets. The LaTeX expression is: \int_{-4}^4\sqrt{16-x^2}dθ

20. Mar 23, 2016

Thanks!