Area of a circle

Chemist · 2005-12-19 09:14:39

Can anyone please show me , using simple integration, how the area of a circle turns out to be (Pi)r^2 ? I've tried many ways and sometimes I end up with 2(Pi)r or completely irrelevant answers ... :S

Thanks

Last edited by Chemist (2005-12-19 09:21:07)

MathsIsFun · 2005-12-19 09:33:35

Start with x²+y² = r²

Then integrate just the top half, which is y = √( r²-x² )

I think it gets a little complicated after that.

mathsyperson · 2005-12-19 09:36:13

I can do it with polar co-ordinates.

For a circle, the magnitude doesn't depend on θ, so the equation would just be r=a, where a is a constant and is also the radius of the circle.

It's a given formula that the area of a region using polar co-ordinates is (0,2π)∫1/2r² dθ. [That's meant to mean integrating between 0 and 2π, but I'm bad at LaTeX]

So, using r=a, that is (0,2π)∫1/2a²dθ

As this doesn't involve θ, integrating merely multiplies the whole thing by θ.

Area = [1/2a²θ](0,2π) = [1/2a²*2π] - [1/2a²*0] = πa².

MathsIsFun · 2005-12-19 09:45:30

Found a nice page about it here

Chemist · 2005-12-19 09:49:40

Then integrate just the top half, which is y = √( r²-x² )
I think it gets a little complicated after that.

Indeed I tried that, and yes the poblem is with the integration.

mathsyperson, your method is good, but since I'm doing this from scratch, just why is the area (0,2π)∫1/2r² dθ ?

Thank your both for your time.

Chemist · 2005-12-19 09:51:40

Ah just what I need, thanks man.

John E. Franklin · 2005-12-20 12:17:55

I didn't go to those web sites, so here is maybe another way.
Allow the knowledge of a circumference being around 3 times the diameter (2π r)
If you don't approve of knowing this, then don't read on.
Imagine many concentric circles around the origin varying from really small to
a radius of one or r, if you want the general case. I'll use 1 for largest r.
Now the concentric circles are evenly spaced apart, and adding up their circumferences
times the distance between the circles should be about right.
Hence

should be your result.
Integrate to r if want the r^2 answer instead of just pi for the area.
So

MathsIsFun · 2005-12-20 17:28:05

Integrate 2π r dr to get π r² ... ?

It might just work ...

... nah, too easy

mathsyperson · 2005-12-20 22:31:02

That works with spheres too.

∫ 4πr²dr = 4/3πr³

I don't think it means anything though. After all, you wouldn't say that the circumference of a sphere was 8π.

God · 2005-12-28 16:42:30

That works, yes, but if r is the principle variable, then 2pi r is a line and pi r^2 is a parabola, so you aren't dealing with a circle, are you?

irspow · 2005-12-30 07:37:10

Integrating a line equation gives the area below the curve (or line). The equation for the first quadrant is:
y = √r²-x²

Integrating this equation from zero to r gives:

x/2 (√r²-x²) + r²/2 (arcsin x/r) = πr²/4

Since this is only the area of one quadrant, you must multiply this by 4 which gives πr²

Author 007 · 2006-09-23 04:32:18

irspow wrote:

Integrating a line equation gives the area below the curve (or line). The equation for the first quadrant is:
y = √r²-x²
Integrating this equation from zero to r gives:
x/2 (√r²-x²) + r²/2 (arcsin x/r) = πr²/4
Since this is only the area of one quadrant, you must multiply this by 4 which gives πr²

Hi guys

Sorry to resurrect this thread but I really need to be able to derive the formula for the area of a circle using integration and I can't find anything useful on the internet. Please could you explain how it works without missing stages out so it's easier to follow. How do you integrate the "r"? I can't remember!

Thank you so much.

Dima · 2006-09-30 20:45:25

If you want to do it this way (which is rather silly) oh well...

x^2+y^2=r^2

Find the area of the upper half:

y=sqrt(r^2-x^2) <--- equation for the upper half

integrate(sqrt(r^2-x^2)dx,x=-r..r) <---- inegrate from x=-r to x=r

use x=r*cos(theta) substitution. Notice if we want to integrate over theta, theta has to change from PI to 0, so that x=r*cos(theta) changes from r*cos(PI)=-r to r*cos(0)=r

then dx =-r*sin(theta)*d(theta)
sqrt(r^2-x^2) = sqrt(r^2-r^2*(cos(theta))^2)=r*sqrt(1-(cos(theta))^2)=r*sin(theta)
(we can do this because theta changes from 0 to PI and sine is always positive there)

So we have the following:
integrate(sqrt(r^2-x^2)dx,x=-r..r) = integrate(-r^2*[sin(theta)]^2(dtheta),theta=PI..0)=
=-r^2*[-PI/2] = PI*r^2/2. The total area is area of the upper half times two, i.e.

2 * [PI*r^2/2] = PI*r^2

How to evaluate int((sin(x))^2,x=0..PI): integrate by parts

int((sin(x))^2,x=0..PI) = -sin(x)*cos(x)|0,PI + int(cos(x)*cos(x),x=0..PI)

so int((sin(x))^2 dx,x=0..PI) = int((cos(x))^2 dx,x=0..PI)

thus

int((sin(x))^2,x=0..PI) = 1/2*int((sin(x))^2+(cos(x))^2 dx,x=0..PI) = 1/2*int(1 dx,x=0..PI)=
= 1/2 * PI

Integration by parts follows from the fact that

[f(x)*u(x)]' = f'(x)*u(x)+u'(x)*f(x)

f(x)=int(f'(x)), so

int(f'(x)*u(x)) = f(x)*u(x) - u'(x)*f(x). In the example above, I used f(x)=sin(x) and u(x)=sin(x), so that int(u(x)) = -cos(x) and u'(x) = cos(x)

Math Is Fun Forum

#1 2005-12-19 09:14:39

Area of a circle

#2 2005-12-19 09:33:35

Re: Area of a circle

#3 2005-12-19 09:36:13

Re: Area of a circle

#4 2005-12-19 09:45:30

Re: Area of a circle

#5 2005-12-19 09:49:40

Re: Area of a circle

#6 2005-12-19 09:51:40

Re: Area of a circle

#7 2005-12-20 12:17:55

Re: Area of a circle

#8 2005-12-20 17:28:05

Re: Area of a circle

#9 2005-12-20 22:31:02

Re: Area of a circle

#10 2005-12-28 16:42:30

Re: Area of a circle

#11 2005-12-30 07:37:10

Re: Area of a circle

#12 2006-09-23 04:32:18

Re: Area of a circle

#13 2006-09-30 20:45:25

Re: Area of a circle

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