Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

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

Chemist
Member
Registered: 2005-12-12
Posts: 35

Area of a circle

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)


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

Offline

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

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Area of a circle

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.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Offline

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

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Area of a circle

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.


Why did the vector cross the road?
It wanted to be normal.

Offline

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

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Area of a circle

Found a nice page about it here


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Offline

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

Chemist
Member
Registered: 2005-12-12
Posts: 35

Re: Area of a circle

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.


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

Offline

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

Chemist
Member
Registered: 2005-12-12
Posts: 35

Re: Area of a circle

Ah just what I need, thanks man.


"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

Offline

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

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

Re: Area of a circle

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


igloo myrtilles fourmis

Offline

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

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,711

Re: Area of a circle

Integrate 2π r dr to get π r² ... ?

It might just work ...

... nah, too easy wink


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

Offline

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

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Area of a circle

That works with spheres too.

∫ 4πr²dr = 4/3π

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


Why did the vector cross the road?
It wanted to be normal.

Offline

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

God
Member
Registered: 2005-08-25
Posts: 59

Re: Area of a circle

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?

Offline

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

irspow
Member
Registered: 2005-11-24
Posts: 1,055

Re: Area of a circle

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 π


I am at an age where I have forgotten more than I remember, but I still pretend to know it all.

Offline

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

Author 007
Member
Registered: 2006-09-23
Posts: 1

Re: Area of a circle

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 π

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.

Offline

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

Dima
Guest

Re: Area of a circle

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)

Board footer

Powered by FluxBB