How do I calculate the arc length of a polar curve?

[azatkgz]

It's easy question,but I don't know whether I solved it correctly.

Homework Statement

Calculate the length of the curve given by
r=a\sin^3 \frac{\theta}{3}
in polar coordinates. Here, a > 0 is some number.

Homework Equations

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

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}

 

[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.

 

[HallsofIvy]

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.

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

 

[foxjwill]

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

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

 

[azatkgz]

Thanks a lot!

 

[HallsofIvy]

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

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