Surface area of a revolution

I'm having some trouble with a question, I've got a lot of the difficult stuff down but I can't get the limits right and my answer is twice what it should be.

Quote:

Curve C: $\displaystyle x=cos^3 \theta \: , \: y= sin^3 \theta \: , \: 0 \leq \theta \leq \frac{\pi}{2} $

The curve C is rotated through 360 degrees about the x-axis. Show that the curved surface area of the solid of revolution generated by:

$\displaystyle

6 \pi \int^{ \frac{\pi}{2} }{0} }_0 sin^4 \theta cos \theta d\theta$

Hence find this curved surface area.

So I know that $\displaystyle A = \int 2 \pi y ds $ and that $\displaystyle ds = \sqrt{ (\frac{dx}{d \theta})^2 + (\frac{dy}{d \theta})^2 } = \frac{3}{2}sin 2 \theta $.

So:

$\displaystyle A = 2\pi \int 2 sin^3 \theta \frac{3}{2} \sin 2 \theta d \theta $

$\displaystyle sin 2 \theta \equiv 2 \sin \theta \cos \theta $

$\displaystyle A = 6\pi \int sin^4 \theta cos \theta d \theta $

So I have the right integral but I can't show that the limits are 0 to $\displaystyle \frac{\pi}{2} $.

I don't think it wants me to use symmetry as that would make it $\displaystyle 12 \pi$, I tried $\displaystyle \theta = arcos( x^{\frac{1}{3} } )$ but I can't get the right answers.

Cheers