surface area of a parametric curve or revolution

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I have the equations $x=9t-3t^3 $ and $ y=9t^2$ , $0\le t\le 2$ rotated about the x-axis

what I did to try and solve this was:

$ S= \int^2_0 2\pi(9t^2) \sqrt{(9-9t^2)^2+(18t)^2}dt $

$S= \int^2_0 18\pi t^2 \sqrt{81t^4 +162t^2+81}dt $

$ S=\int^2_0 18\pi t^2 (9-9t^2)dt $

$S= \int^2_0 162\pi t^2 (t^2+1) dt $

$ S= 162\pi [\frac{t^5}{5} +\frac{t^3}{3}] $

$ S= \frac{7344\pi}{5}$

I'm not sure where I went wrong , any help pointing that out is greatly appreciated.