# Thread: Find length of polar curve

1. ## Find length of polar curve

How do I find the length of the polar curve
r = 4sin(3theta)
I know what the formula for the length of a polar curve is, but the answer I get seems to be too small.

2. You should have the basic formula:

$\displaystyle s=\int_{\theta_1}^{\theta_2}\sqrt{r^2+\left(\frac{ dr}{d\theta}\right)^2}d\theta$

The curve has six lobes, so your answer should be greater than 6 times (4 out and 4 in) = 48. And it should be less than 6 times the perimeter of an equilateral triangle with height 4, or 6 times 3 times 8/sqrt(3) = 48sqrt(3).

If you show me your attempt at the problem I can probably help more. Post in this thread and I'll try to answer as soon as possible.

- Hollywood

3. Sorry, I'm new here, so I'm not sure how to put my reply in equations, but what I did was:
Integral from 0 to pi of [16sin^2(3theta) + 144cos^2(3theta)]^(1/2)

I'm not even sure if the end points for the integral are right. The more I try it the more lost I get

4. I made a mistake in my last post - the graph has 3 lobes, not 6. So the length should be between 24 and 24sqrt(3). Sorry about that.

I think your setup is correct. I thought the integral would be easy, but I can't seem to get it. Does anyone know how to integrate this function?

$\displaystyle \int_{0}^{\pi}\sqrt{16sin^23x+144cos^23x}\ dx$

5. For what it's worth, I was able to get a numerical value: 26.73.

- Hollywood

6. I can pull some constants out of the integral:

$\displaystyle \frac{4}{3}\int_{0}^{3\pi}\sqrt{sin^2u+9cos^2u}\ du$

I've tried changing the integrand to $\displaystyle \sqrt{1+8\cos^2u}$ and $\displaystyle \sqrt{5+4cos{2u}}$, but that doesn't seem to make it any easier.

- Hollywood

7. In response to your PM I got it from my calc text book: Stewart essential Calculus. I too got the numerical answer, but I'm sure it must be able to be done by hand. But it was for an assignment, and I've already handed it in, so it's not a huge deal anymore. Thanks for your help

8. Ok. Let us know if you get the solution.

- Hollywood