# Arc Length.

1. Feb 5, 2008

### azatkgz

It's easy question,but I don't know whether I solved it correctly.
1. The problem statement, all variables and given/known data
Calculate the length of the curve given by
$$r=a\sin^3 \frac{\theta}{3}$$
in polar coordinates. Here, a > 0 is some number.

2. Relevant equations

$$l=\int \sqrt{r^2(\theta)+(\frac{dr}{d\theta})^2}d\theta$$

3. The attempt at a solution

$$l=\int \sqrt{a^2 \sin^6\frac{\theta}{3}+a^2\sin^4\frac{\theta}{3}\cos^2\frac{\theta}{3}}\theta$$

$$l=a\int \sin^2\frac{\theta}{3}d\theta$$
for $$0<\frac{2\theta}{3}<2\pi$$

$$l=\frac{a}{2}\int_{0}^{3\pi}(1-\cos\frac{2\theta}{3})d\theta$$

$$l=\frac{3\pi}{2}$$

2. Feb 5, 2008

### foxjwill

It's all correct except for the very last line. You forgot to include $$a$$. Your answer should be:

$$s = \frac{3a\pi}{2}$$

p.s. Use $$s$$ for arclength--it's more widely used and recognized. Also, you can include limits of integration like this: \int^b_a Always put the ^ first, though. Otherwise it doesn't work right.

Last edited: Feb 5, 2008
3. Feb 5, 2008

### HallsofIvy

Staff Emeritus
It doesn't? What the difference between
$$\int_0^1 f(x)dx$$
and
$$\int^1_0 f(x)dx$$

4. Feb 5, 2008

### foxjwill

hmm. That's odd. I guess it just didn't work right when I tried it. Ah, well. Not a very scientific conclusion, eh?

5. Feb 6, 2008

### azatkgz

Thanks a lot!!!

6. Feb 6, 2008

### HallsofIvy

Staff Emeritus
That's alright. There are millions of thing that work for everyone except me!