Integration of a Circle in Polar Coordinates

In summary, integration of a circle in polar coordinates involves using the formula ∫ r^2 dθ to find the area enclosed by a circle. This formula takes into account the radius of the circle and integrates over the angle θ to find the total area. By converting the Cartesian coordinates (x,y) of a circle into polar coordinates (r,θ), the integral becomes simpler and can be easily solved. This method is useful in various fields of mathematics, such as physics and engineering, where circular shapes and polar coordinates are frequently encountered.
  • #1
dluu
15
0

Homework Statement


Hi, I'm trying to find the area of a circle in polar coordinates.I'm doing it this way because I have to put this into an excel sheet to have a matrix of areas of multiple circles.
Here is an example of the problem.
4rtWQ.png

a= radius of small circle
(gamma, r0) = polar coordinate system where phi is the angle and r0 is the distance to the center of the circle
gamma = phi (sorry for the confusion) just another constant
They are all constants except for theta.

Using the equation provided by wikimedia
d52097e7e82669349c8da7c64617c8f5.png

and the equation
eq0001MP.gif


Homework Equations


Integrating this would give a constant C, but this value does not matter, does it?
Also, are my trig identities correct?

The Attempt at a Solution


KcHwN.png

from interval a to b

*found a mistake in my work: should be sin^2 not sin
 
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  • #2
I asked a math professor at my university about this and we tried to do this converting a circle from cartesian to polar coordinates. The equation of the circle would result in a +- equation in front of the square root through quadratic formula.
He used the equation:
eq0008M.gif

and told me to use the positive answer for r0 and the negative for ri. I thought this applies to two different curves but in this case the circle should be one curve.

Am I doing this right?

If I am wrong, what equation do I use if the limits of integration only has one curve (side of the circle)
 
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  • #3
dluu said:

Homework Statement


Hi, I'm trying to find the area of a circle in polar coordinates.I'm doing it this way because I have to put this into an excel sheet to have an array of areas with multiple circles.
Here is an example of the problem.
hWJHV.png


Using the equation provided by wikimedia
d52097e7e82669349c8da7c64617c8f5.png

and the equation
eq0001MP.gif

I'm sorry, but we aren't mind readers here. What are ##a,r_0, \theta, \phi##? And is the ##r## in the first equation supposed to be the same ##r## in the second equation?
 
  • #4
LCKurtz said:
I'm sorry, but we aren't mind readers here. What are ##a,r_0, \theta, \phi##? And is the ##r## in the first equation supposed to be the same ##r## in the second equation?

Sorry I'll upload a better picture with a description

Yes the r's are the same. I changed phi to gamma. Not sure why I did.
 
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  • #5
dluu said:

Homework Statement


Hi, I'm trying to find the area of a circle in polar coordinates.I'm doing it this way because I have to put this into an excel sheet to have an array of areas with multiple circles.
Here is an example of the problem.
[ IMG]http://i.imgur.com/4rtWQ.png[/PLAIN]

Using the equation provided by wikimedia
d52097e7e82669349c8da7c64617c8f5.png

and the equation
eq0001MP.gif


Homework Equations


Integrating this would give a constant C, but this value does not matter, does it?
Also, are my trig identities correct?

The Attempt at a Solution

Last question first:

If you're asking,
"Do I need to include the Integration constant from the anti-derivative when evaluating a definite integral?"
The answer is, No.


Now that you've updated your figure:

For a circle of radius, a, centered at (r0, φ ) in polar coordinates, the equation in polar coordinates, (r, θ), is like the one you found on wikipedia:
[itex]\displaystyle
r=r_0\cos(\theta-\varphi)\pm\sqrt{a^2-r_0^2\sin^2(\theta-\varphi)\,}[/itex]​

Added in Edit:

If a ≤ r0, then the limits of integration for θ should be [itex]\displaystyle \varphi\pm\sin^{-1}\left(\frac{a}{r_0}\right)\ .[/itex]
 
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  • #6
SammyS said:
For a circle of radius, a, centered at (r0, φ ) in polar coordinates, the equation in polar coordinates, (r, θ), is like the one you found on wikipedia:
[itex]\displaystyle
r=r_0\cos(\theta-\varphi)\pm\sqrt{a^2-r_0^2\sin^2(\theta-\varphi)\,}[/itex]​

Yes they're the same.
 
  • #7
dluu said:
Yes they're the same.

No they aren't. Your original post left off the ##\pm## sign. You can use your professor's suggestion for ##r_{outer}^2-r_{inner}^2## using the plus or minus sign accordingly (I think).
 
  • #8
LCKurtz said:
No they aren't. Your original post left off the ##\pm## sign. You can use your professor's suggestion for ##r_{outer}^2-r_{inner}^2## using the plus or minus sign accordingly (I think).

On wikipedia it says: "the solution with a minus sign in front of the square root giving the same curve"
http://en.wikipedia.org/wiki/Polar_coordinate_system

If they're the same curve then I wouldn't need to use ##r_{outer}^2-r_{inner}^2## would I?

What if my limits of integration was between these points? Which equations would I use?
vRW08.png
 
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  • #9
Yes. It's just like when you have a circle ##x^2+y^2=4## and you solve for the upper and lower curves ##y = \pm\sqrt{4-x^2}## for the separate equations of the upper and lower curve if you were going to find its area with a dydx integral. It's all the same circle, but different parts. Your r is apparently double valued function of ##\theta## just like a circle in xy is a double valued function of x. I say "apparently" because I have not myself derived that equation.
 
  • #10
LCKurtz said:
Yes. It's just like when you have a circle ##x^2+y^2=4## and you solve for the upper and lower curves ##y = \pm\sqrt{4-x^2}## for the separate equations of the upper and lower curve if you were going to find its area with a dydx integral. It's all the same circle, but different parts. Your r is apparently double valued function of ##\theta## just like a circle in xy is a double valued function of x. I say "apparently" because I have not myself derived that equation.

Okay I'll try that way. So the picture of my previous post, would I use the -sqrt since its the lower curve?
 
  • #11
dluu said:
Okay I'll try that way. So the picture of my previous post, would I use the -sqrt since its the lower curve?
As that math prof. said,

Using the plus sign of ±√, gives part (at least a little more than half) of the circle, which he called ro, for rOUTER . The minus sign of ±√, gives part of the circle he called ri for rINNER .

Find the area from the origin to rOUTER. Subtract from that the area from the origin to rINNER

Below is rOUTER .

attachment.php?attachmentid=51455&stc=1&d=1349217383.gif


Below is rINNER .

attachment.php?attachmentid=51456&stc=1&d=1349217383.gif


(From WolframAlpha)
 

Attachments

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    WA_r_inner.gif
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  • WA_r_outer.gif
    WA_r_outer.gif
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  • #12
Okay I understand now. This is the general equation I should be using throughout excel.

After working this out, I came to this equation. How do I integrate this?

i85gJ.png
 
  • #13
SwPVm.png

Is this correct? Not sure where to go after this.

*ignore 2nd to last line*
 
  • #14
dluu said:
SwPVm.png

Is this correct? Not sure where to go after this.

*ignore 2nd to last line*
It seems crazy, but the next thing to do is a trig substitution.

Let u = sin(t) ...
 
  • #15
Thanks, I see how this works.

One more question. Are my trig identities correct where:
u = sin(theta-phi)
du = cos(theta-phi)
 
  • #16
dluu said:
Thanks, I see how this works.

One more question. Are my trig identities correct where:
u = sin(theta-phi)
du = cos(theta-phi)
They're not really identities, and they're almost correct.

du = cos(theta-phi) d(theta)
 
  • #17
SammyS said:
It seems crazy, but the next thing to do is a trig substitution.

Let u = sin(t) ...

I tried to do this and checked my math with Wolfram Alpha. I'm confused on one of the steps

BzeDd.png

*for some reason I input x-1 but it kept putting 1-x*

My question is how does this work?
sin(2p) ----> s√(1-s^2 )
 
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  • #18
So I checked my work on Wolfram and I think this is correct. But when I try to plug in values, they don't work.

I'm not sure what I'm doing wrong.

Here is my work

7BSL6.png

CLU0R.png
 
  • #19
dluu said:
I tried to do this and checked my math with Wolfram Alpha. I'm confused on one of the steps
...

My question is how does this work?
sin(2p) ----> s√(1-s^2 )
If you're asking in regards to sin(2sin-1(s)), then the double angle formula for sine.

cos(sin-1(s))=√(1-s2) .
 

FAQ: Integration of a Circle in Polar Coordinates

What is the formula for finding the area of a circle in polar coordinates?

The formula for finding the area of a circle in polar coordinates is A = ½r²θ, where r is the radius and θ is the angle in radians.

How do you convert Cartesian coordinates to polar coordinates?

To convert Cartesian coordinates (x,y) to polar coordinates (r,θ), use the formulas r = √(x²+y²) and θ = tan⁻¹(y/x).

What is the difference between polar coordinates and Cartesian coordinates?

Polar coordinates use a distance and angle to locate a point, while Cartesian coordinates use x and y coordinates on a grid.

What is the relationship between polar and rectangular equations?

Polar equations can be converted to rectangular equations using the formulas x = rcosθ and y = rsinθ. Rectangular equations can also be converted to polar equations using the formulas r = √(x²+y²) and θ = tan⁻¹(y/x).

How is integration of a circle in polar coordinates used in real life?

Integration of a circle in polar coordinates is often used in physics and engineering to calculate the areas of circular objects, such as wheels and gears. It is also used in calculating the volumes of cylindrical objects, such as pipes and tanks.

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