Finding Limits for Polar Coordinate Area Integration

  • Thread starter Thread starter daster
  • Start date Start date
  • Tags Tags
    Polar
AI Thread Summary
The discussion focuses on finding the correct limits for integrating the area in polar coordinates, specifically for the curve defined by r=2cos(2t) and the half-line t=π/6. To determine the limits, it's essential to find the values of theta where r equals zero, which helps establish the bounds for integration. A sketch of the curve is recommended to visualize the area being calculated, which can clarify the integration limits. The area is computed using the formula A = 1/2 ∫ (r^2) dθ, with specific limits derived from the intersection points. The conversation also touches on potential arithmetic errors in calculations and the importance of consistency in equations used.
daster
I need help with finding areas. I'm having trouble picking the correct limits for my integration.

Say for example we had r=2cos2t and a half-line t=pi/6, and I want to find the small area bounded between them.

\frac{1}{2}\int (2\cos (2\theta))^{2}\,d\theta

I can do the integration, but what do I choose as its limits?

I'm not particularly interested in the answer to this question; I'm looking for a good explanation or maybe a couple of pointers.
 
Mathematics news on Phys.org
At what value(s) of theta does r become zero ? Set the r(theta) equation to zero and solve for theta. Use the 1st quadrant value of theta obtained as the upper bound with \frac{\pi}{6} as the lower.
 
Why do we want the value of theta at r=0?
 
daster said:
Why do we want the value of theta at r=0?

Have you even sketched the graph yet ? It will become clear as day if you do.
 
Can you please check my working?

(a) Sketch the curve with polar equation

r=3\cos 2\theta, \, -\frac{\pi}{4}\leq\theta<\frac{\pi}{4}.

The curve looks like 1 rose petal.

(b) Find the area of the smaller finite region enclosed between the curve and the half-line \theta=\frac{\pi}{6}.

Area = \frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (3\cos 2\theta)^2\,d\theta = \frac{9}{4}\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (1+\cos 4\theta)\,d\theta = \frac{9}{4}\left[\theta+\frac{1}{4}\sin 4\theta\right]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = \frac{3\pi}{16}- \frac{9\sqrt{3}}{32}

A couple of arithmetic slips probably made it through, since I did all of this using my keyboard & notepad.exe. :-p So I'm just wondering if my method was correct.
 
Last edited by a moderator:
The coefficient is 3 ? Not 2 ? If that's the case then your working in the last post is correct. But why did you give a different equation in your first post ? :confused:
 
In my first post I made up an equation and a line, and when I went to do a question from my textbook, it happened to be almost like the one I made up. :-p

Anyway, thanks for your help. I appreciate it. :smile:
 
Sure. :smile:
 

Similar threads

I Area Differential in Cartesian and Polar Coordinates
Replies
13
Views
3K
B To calculate the polar unit vectors using rotated coordinates
Replies
1
Views
2K
I Transforming coordinates between Cartesian and spherical
Replies
1
Views
2K
A How do I express an equation in Polar coordinates as a Cartesian one.
Replies
1
Views
2K
Double Integral of function in region bounded by two circles
Replies
19
Views
2K
I Gaussian Integral Coordinate Change
Replies
4
Views
1K
I Cartesian and polar terminology
Replies
7
Views
3K
B Question about the limits of integration in a change of variables
Replies
2
Views
3K
B Integrating to find surface area/volume of hemisphere
Replies
3
Views
3K
MHB How do I get the book's answer for finding the area of a gable?
Replies
4
Views
1K

Hot Threads

Recent Insights

Back
Top