Finding area in polar coordinates

In summary, the question asks for finding the area in one loop for the function r = cos[3(theta)]. The solution uses limits of integration from -pi/6 to pi/6. The frequency of the function is 3, so intervals of 2pi/3 do not work. There is no easy way to find the angle for the loop without drawing it, but going from -pi/6 to pi/6 ensures that the point goes around the loop exactly once. Going from 0 to 2pi/3 does not bound any area and will not give the same integral.
  • #1
MathewsMD
433
7
I've attached the solution to this post. The question is essentially just asking to find the area in one loop for r = cos[3(theta)].

This seems like a fairly simple question (and answer). I've solved and understand the general integration, but I am just a little uncertain on why exactly different limits for integration for the variable theta do not work. For example, the limits used in the solution are -pi/6 to pi/6 which makes perfect sense. Since this is a sinusoidal function with the frequency 3, I was wondering why any interval 2pi/3 rad does not work. For example, the limits of integration I originally used was 0 to 2pi/3. But when I do this, my answer is incorrect. I don't quite seem to understand why this is exactly, but if I'm to wager a guess: is it b/c the new function has a frequency of 6 (i.e. the original function was only + values for r so it did not have its reflection)? If not, any explanation would be greatly appreciated!

Also, is there any easy way to find the angle for the loop without drawing it? Is there a particular pattern for where the loops begin and end depending on the sinusoid and its frequency?
 

Attachments

  • Screen Shot 2015-01-21 at 5.29.10 PM.png
    Screen Shot 2015-01-21 at 5.29.10 PM.png
    11.6 KB · Views: 526
Physics news on Phys.org
  • #2
Yes, as [itex]\theta[/itex] goes [itex]-\pi/6[/itex] to [itex]\pi/6[/itex] point goes exactly once around that loop. If, instead, you go from 0 to [itex]2\pi/3[/itex] the point goes from the right tip of the first loop to the tip of the loop on the bottom. That does NOT bound any area and I don't see why you expect it to give the same integral
 

FAQ: Finding area in polar coordinates

1. What is the formula for finding the area in polar coordinates?

The formula for finding the area in polar coordinates is A = ∫ 1/2r²dθ, where r is the distance from the origin and θ is the angle measured from the positive x-axis.

2. How do you convert polar coordinates to rectangular coordinates?

To convert polar coordinates (r, θ) to rectangular coordinates (x, y), you can use the following equations: x = rcosθ and y = rsinθ. Simply substitute in the values for r and θ to find the corresponding x and y values.

3. Can you find the area of a sector in polar coordinates?

Yes, you can find the area of a sector in polar coordinates by using the formula A = 1/2r²θ, where r is the radius of the sector and θ is the central angle in radians.

4. What is the difference between polar coordinates and rectangular coordinates?

Polar coordinates use the distance from the origin and the angle measured from the positive x-axis to locate a point, while rectangular coordinates use the horizontal and vertical distance from the origin to locate a point.

5. How do you find the area of a region bounded by multiple polar curves?

To find the area of a region bounded by multiple polar curves, you can split the region into smaller sections and use the formula A = ∫ 1/2r²dθ for each section. Then, add up the individual areas to find the total area.

Similar threads

Replies
5
Views
3K
Replies
10
Views
4K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
4
Views
2K
Replies
3
Views
2K
Replies
12
Views
2K
Replies
15
Views
3K
Back
Top