Need help for text tomorrow

Find the surface area of the solid of revolution over the interval

1.$\displaystyle y=x^3, [0,2]$

Answer: $\displaystyle \pi(145\sqrt{145}-1)/27$

What I tried:

$\displaystyle \frac{dy}{dx}=3x^2$

Plug into formula for Surface Area of a solid of revolution about the x-axis:

$\displaystyle \int_a^b 2\pi y \sqrt{1+(\frac{dy}{dx})^2}dx$

$\displaystyle 2\pi\int_0^2 x^3\sqrt{1+(3x^2)^2}dx$

$\displaystyle 2\pi\int_0^2\sqrt{1+9x^4}x^3dx$

let $\displaystyle u=1+9x^4, \frac{du}{dx}=36x^3$

Convert limits: when$\displaystyle x=0, u=1$, when $\displaystyle x=2, u=289$

$\displaystyle \frac{2\pi}{36}\int_1^{289} \sqrt{u} du$

Integrate:

$\displaystyle \frac{\pi}{18}(\frac{2}{3}u^{\frac{3}{2}})\bigg|^{ 289}_1$

Plug in $\displaystyle u$, evaluate at limits, simplify.

I don't seem to be getting the answer that's in my book.