# Find area bounded by loop

• Apr 18th 2012, 04:13 PM
Bowlbase
Find area bounded by loop
I can do the integral for these problems all day long but cannot determine the boundaries to save my life. Polar coordinates are my arch nemesis at the moment.

The problem:
find the area bounded by =$\displaystyle r=3sin(3\theta)$

1. $\displaystyle 1/2\int_a^b(3sin(3\theta))^2d\theta$
2. find boundaries
3. do integral
4. win

I must have been catatonic when boundaries were explained because I simply do not get how to find them.

thanks for any help!
• Apr 18th 2012, 04:45 PM
HallsofIvy
Re: Find area bounded by loop
First, note that $\displaystyle sin(\theta)$ goes from 0, up to +1, then down to 0 again as $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$. For $\displaystyle \theta$ from $\displaystyle \pi$ to $\displaystyle 2\pi$, sine is negative so you just go over the same path again. That is, $\displaystyle sin(\theta)$ covers one loop as $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$. Here, $\displaystyle \theta$ is replaced by $\displaystyle 3\theta$ so this function covers one loop as $\displaystyle 3\theta$ goes from 0 to $\displaystyle \pi$. That is, as $\displaystyle \theta$ goes from 0/3 to $\displaystyle \pi/3$.
• Apr 18th 2012, 05:00 PM
Bowlbase
Re: Find area bounded by loop
I follow that. Does 3 at the beginning of the function not make a difference then?
• Apr 18th 2012, 05:24 PM
HallsofIvy
Re: Find area bounded by loop
No, that does not affect the $\displaystyle \theta$ values. In fact, it could be taken outside the integral.
• Apr 18th 2012, 05:32 PM
Bowlbase
Re: Find area bounded by loop
Oh, okay. So if I have a function like$\displaystyle r=ksin(3\theta)cos(3\theta)$ I only need to look at $\displaystyle sin(3\theta)cos(3\theta)$? But this is only for multiplications of functions correct? In other words I could not pull out "n" or "k" in this example:$\displaystyle r=n-ksin(3\theta)cos(3\theta)$. Do I have this right?

And, in general form I could solve $\displaystyle r=n-ksin(3\theta)cos(3\theta)$ by $\displaystyle n=ksin(3\theta)cos(3\theta)$ and the solution to $\displaystyle \theta$ would just be between 0 and $\displaystyle 2\pi$.

Am I thinking of this correctly?