# Polar Calculus Help Needed

• Mar 20th 2007, 02:06 PM
phack
Polar Calculus Help Needed
I have a graph of a polar function, r=1-2cos(theta). The graph produces a limacon. I had no problem graphing it, but I'm a bit confused as to how to find the area. The area I need to find is the area within the larger, outer loop and outside the smaller, inner loop. The inner loop isnt really a loop at all, more of an ellipse, so I cant treat it as a circle of known radius and subtract the two functions within the integral. I'm confused because of the loop inside of the loop. Usually, this would happen with two seperate functions and it would just be a matter of subtracting the two areas. Intuition would tell me to integrate from 0 to 2pi for the area, but for some reason I'm not sure if this will work. :confused:

Help
• Mar 20th 2007, 02:53 PM
ecMathGeek
Someone else correct me if I'm wrong, but if I remember correctly, I believe this is what is happening in your problem:

As theta goes from 0 to 2pi, the polar graph makes two separate loops. The graph actually completes each loop in pi radians, so integrating from 0 to pi gives you the area of one loop (the outer loop, I believe). Integrating from pi to 2pi gives you the area of the second loop, but notice what is happening between pi and 2pi: the graph seems to loop back on its self (instead of the point at pi radians being located on the negative x axis, it appears on the positive x axis (note that I have not graphed this function so I may be mistaken on its orientation)). This represents a negative radius, which means the integrating from pi to 2pi gives a negative area. If you integrate from 0 to 2pi, you actually are getting the area of the first loop minus the magnitude of the area of the second loop. In other words, to get the area you said you need, just integrate from 0 to 2pi.
• Mar 20th 2007, 03:11 PM
ThePerfectHacker
Look at picture below.

Now it seems dangerous to compute the integral from [0,2pi] because we will count the inside loop twice.

Instead draw the limacon for 0<=theta <= pi
(Shown below).

That area is equal to,
INT[0,pi](1/2)r^2 d(theta)=4.71

Now draw the limacon for 0<=theta<=pi/3
(Shown below).

That area is equal to,
INT[0,pi/3](1/2)r^2 d(theta)=.2718

Then if you subtract the area in the first integration from this one you get the area without the half-loop.
That is, 4.43

By Symettry we have total area, 8.86

Note, if you find,
INT[0,2pi](1/2)r^2 d(theta) = 9.42
And if you subtract .2718 twice you get,
8.86 (because we counted the half-loops twice).
• Mar 20th 2007, 05:35 PM
ecMathGeek
Quote:

Originally Posted by ThePerfectHacker
Look at picture below.

Now it seems dangerous to compute the integral from [0,2pi] because we will count the inside loop twice.

Instead draw the limacon for 0<=theta <= pi
(Shown below).

That area is equal to,
INT[0,pi](1/2)r^2 d(theta)=4.71

Now draw the limacon for 0<=theta<=pi/3
(Shown below).

That area is equal to,
INT[0,pi/3](1/2)r^2 d(theta)=.2718

Then if you subtract the area in the first integration from this one you get the area without the half-loop.
That is, 4.43

By Symettry we have total area, 8.86

Note, if you find,
INT[0,2pi](1/2)r^2 d(theta) = 9.42
And if you subtract .2718 twice you get,
8.86 (because we counted the half-loops twice).

 I just noticed that you integrated 1/2r^2d(theta)... why?

The function is r=1-2cos(theta). Should we not be integrating 1-2cos(theta)?

Suppose we integrate 1-2cos(theta) from 0 to 2pi:

INT[0,2pi](1-2cos(theta))d(theta) = (theta)-2sin(theta)[0,2pi] = (2pi-0)-(0) = 2pi = 6.2831

If, instead, we integrate from pi/3 to pi, subtract the absolute value of the integration from 0 to pi/3, and multiply the result by 2, we get:

2{ INT[pi/3,pi](1-2cos(theta))d(theta) - | INT[0,pi/3](1-2cos(theta))d(theta) | }
2{ (theta-2sin(theta))[pi/3,pi] - | (theta-2sin(theta))[0,pi/3] | }
2{ [(pi-0)-(pi/3-sqrt(3))] - | [(pi/3-sqrt(3))-(0-0)] | } = 2{ (3.826446) - |-.68485|} = 6.2831

The two integratations are equal because the integration between 0 and pi/3 is negative, thus it subtracts from the total integration

[Edit 2] Given what I did just above, I noticed that I made two mistakes in my first post: 1) the integrations from 0 to pi gives the area of the inner loop, and 2) the inner loop has the negative radius and thus gives a negative area.
• Mar 20th 2007, 05:51 PM
ThePerfectHacker
• Mar 20th 2007, 06:14 PM
ecMathGeek
Quote:

Originally Posted by ThePerfectHacker

Thank you :D