# Limits of integration with polar coordinates.

• Nov 4th 2009, 07:01 PM
RockHard
Limits of integration with polar coordinates.
Ok, I have the following problem find the area inside the cardioid

$r = 1 -sin\theta$

and outside the circle

$r = \frac{1}{2}$

I know you check for the limits of integration by finding where the two graphs intersect which is

$\frac{\pi}{6} and \frac{5\pi}{6}$

However they show how the solution is done im a trying to re work it and they say you can just take the area of half the region and multiply it by two which gives the these limits of integration

$\frac{\pi}{6} and- \frac{\pi}{2}$

How the hell that came up? I cant see where the right half the graph starts at -pi/2
• Nov 4th 2009, 08:16 PM
tonio
Quote:

Originally Posted by RockHard
Ok, I have the following problem find the area inside the cardioid

$r = 1 -sin\theta$

and outside the circle

$r = \frac{1}{2}$

I know you check for the limits of integration by finding where the two graphs intersect which is

$\frac{\pi}{6} and \frac{5\pi}{6}$

However they show how the solution is done im a trying to re work it and they say you can just take the area of half the region and multiply it by two which gives the these limits of integration

$\frac{\pi}{6} and- \frac{\pi}{2}$

How the hell that came up? I cant see where the right half the graph starts at -pi/2

Do you even know how this cardioid look like? Take a peek here:

File:CardioidsLabeled.PNG - Wikipedia, the free encyclopedia

Your cardioid is the green one, and then you can see why the area wanted can be divided when the angle is taken from Pi/6 to -pi/2 and multipled by two (it is symmetric with respect to the y-axis)

Tonio
• Nov 4th 2009, 10:34 PM
RockHard
Yes I have seen the graph
• Nov 5th 2009, 03:04 AM
tonio
Quote:

Originally Posted by RockHard
Yes I have seen the graph

Good. Then you see now why they told you to do what they did and why it works.

Tonio
• Nov 5th 2009, 06:27 AM
RockHard
Would doing it from the limits of integrations found by the intersection poiints not work? 5pi/6 and pi/6
• Nov 5th 2009, 09:19 AM
tonio
Quote:

Originally Posted by RockHard
Would doing it from the limits of integrations found by the intersection poiints not work? 5pi/6 and pi/6

Yes, but you have to go from below! Looking at the picture one sees that you better integrate from $\frac{5\pi}{6}\,\,\to\,\,\frac{13\pi}{6}$, to force the path to "run" below the x-axis (most of the time),
If you do as you say the path will go above the x-axis and you'll be evaluating something wrong.

Tonio