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

[Eclair_de_XII]

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 A_circle = ∫√(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?

 

[blue_leaf77]

$f(x)=∫\sqrt{a^2-x^2}dx$

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.

 

[Eclair_de_XII]

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?

 

[blue_leaf77]

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.

 

[Eclair_de_XII]

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

 

[blue_leaf77]

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.

 

[Eclair_de_XII]

What do you usually call for aa and bb in a definite integral ∫baf(x)dx\int_a^b f(x) dx?

https://www.physicsforums.com/help/latexhelp/

So... $\int_b^a \sqrt{a^2-x^2}$? Where $b=-a$?

 

[Eclair_de_XII]

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?

 

[blue_leaf77]

https://www.physicsforums.com/help/latexhelp/

So... $\int_b^a \sqrt{a^2-x^2}$? Where $b=-a$?

Yes, but again, that will give you the area of a semicircle. The full area is then just the twice of it.

How do you figure? Why?

Try to plot it using whatever function plotter you want for a given value of $a$ and see if the curve can reach $\pi a^2$.

 

[Eclair_de_XII]

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

 

[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?

 

[blue_leaf77]

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.

 

[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.

 

[Eclair_de_XII]

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: $θ=arcsin(\frac{x}{a})$?

 

[Eclair_de_XII]

Nobody post any hints; I think I got it.

In the next post...

 

[blue_leaf77]

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.

 

[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.

 

[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.

 

[Samy_A]

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

 

[Eclair_de_XII]

Thanks!

 

[blue_leaf77]

I can't figure out how to post negatives in the integral,

\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.

 

[Eclair_de_XII]

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.