# Homework Help: Integration of a Circle in Polar Coordinates

1. Oct 2, 2012

### dluu

1. The problem statement, all variables and given/known data
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.

(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

and the equation

2. Relevant equations
Integrating this would give a constant C, but this value does not matter, does it?
Also, are my trig identities correct?
3. The attempt at a solution

from interval a to b

*found a mistake in my work: should be sin^2 not sin

Last edited: Oct 2, 2012
2. Oct 2, 2012

### dluu

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:

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)

Last edited: Oct 2, 2012
3. Oct 2, 2012

### LCKurtz

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. Oct 2, 2012

### dluu

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.

Last edited: Oct 2, 2012
5. Oct 2, 2012

### SammyS

Staff Emeritus
Last question first:

"Do I need to include the Integration constant from the anti-derivative when evaluating a definite integral?"

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:
$\displaystyle r=r_0\cos(\theta-\varphi)\pm\sqrt{a^2-r_0^2\sin^2(\theta-\varphi)\,}$​

If a ≤ r0, then the limits of integration for θ should be $\displaystyle \varphi\pm\sin^{-1}\left(\frac{a}{r_0}\right)\ .$

Last edited: Oct 2, 2012
6. Oct 2, 2012

### dluu

Yes they're the same.

7. Oct 2, 2012

### LCKurtz

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. Oct 2, 2012

### dluu

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?

Last edited: Oct 2, 2012
9. Oct 2, 2012

### LCKurtz

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. Oct 2, 2012

### dluu

Okay I'll try that way. So the picture of my previous post, would I use the -sqrt since its the lower curve?

11. Oct 2, 2012

### SammyS

Staff Emeritus
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 .

Below is rINNER .

(From WolframAlpha)

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12. Oct 2, 2012

### dluu

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?

13. Oct 2, 2012

### dluu

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

*ignore 2nd to last line*

14. Oct 2, 2012

### SammyS

Staff Emeritus
It seems crazy, but the next thing to do is a trig substitution.

Let u = sin(t) ...

15. Oct 3, 2012

### dluu

Thanks, I see how this works.

One more question. Are my trig identities correct where:
u = sin(theta-phi)
du = cos(theta-phi)

16. Oct 3, 2012

### SammyS

Staff Emeritus
They're not really identities, and they're almost correct.

du = cos(theta-phi) d(theta)

17. Oct 4, 2012

### dluu

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

*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 )

Last edited: Oct 4, 2012
18. Oct 4, 2012

### dluu

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

19. Oct 4, 2012

### SammyS

Staff Emeritus
If you're asking in regards to sin(2sin-1(s)), then the double angle formula for sine.

cos(sin-1(s))=√(1-s2) .