Integrating Polar Curves over Period

[samtouchdown]

Hello. I am having trouble conceptualizing and/or decisively arriving to a conclusion to this question. When finding the area enclosed by a closed polar curve, can't you just integrate over the period over the function, for example: 3 cos (3θ), you would integrate from 0 to 2pi/3? It intuitively seems so but graphically I am integrating where the curve is not there . Thanks in advance for the help.

 

[LCKurtz]

The answer is no. You need to examine the graph. Part of the problem is that r can be negative so the graph isn't where you would expect for a given ##\theta##. If you draw the graph of your example, you will find that it is a 3 leaved rose which is completed as ##\theta## goes from ##0## to ##\pi##.

 

[Chirag B]

No. It will not always be from 0 to what makes the inside of the trigonometric function [itex]2\pi[/itex]. Here, 0 to [itex]\frac{2\pi}{3}[/itex] will only give one loop. Since [itex]cos\left(n\theta\right)[/itex] gives n loops when n is odd, there are three loops, or petals, here. Going from 0 to [itex]2\pi[/itex] will only give you one loop. Thus, you would have to go from 0 to [itex]6\pi[/itex] to get the whole function. But because the inside of the function is [itex]3\theta[/itex], plugging in [itex]2\pi[/itex] will give you the whole thing.