# Thread: Find area bounded by loop

1. ## 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 = $r=3sin(3\theta)$

1. $1/2\int_a^b(3sin(3\theta))^2d\theta$
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!

2. ## Re: Find area bounded by loop

First, note that $sin(\theta)$ goes from 0, up to +1, then down to 0 again as $\theta$ goes from 0 to $\pi$. For $\theta$ from $\pi$ to $2\pi$, sine is negative so you just go over the same path again. That is, $sin(\theta)$ covers one loop as $\theta$ goes from 0 to $\pi$. Here, $\theta$ is replaced by $3\theta$ so this function covers one loop as $3\theta$ goes from 0 to $\pi$. That is, as $\theta$ goes from 0/3 to $\pi/3$.

3. ## Re: Find area bounded by loop

I follow that. Does 3 at the beginning of the function not make a difference then?

4. ## Re: Find area bounded by loop

No, that does not affect the $\theta$ values. In fact, it could be taken outside the integral.

5. ## Re: Find area bounded by loop

Oh, okay. So if I have a function like $r=ksin(3\theta)cos(3\theta)$ I only need to look at $sin(3\theta)cos(3\theta)$? But this is only for multiplications of functions correct? In other words I could not pull out "n" or "k" in this example: $r=n-ksin(3\theta)cos(3\theta)$. Do I have this right?

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

Am I thinking of this correctly?