Rotate about the $x$ axis
$x = 2t-2/3t^3$
$y = 2t^2$
$0 \leq t \leq 1$
I did the integral of $\sqrt{(2-2t^2)^2+(4t)^2}$ and got $(2x(x^2+3))/3$ and then I did the integral of $2\pi 2t^2 ((2x(x^2+3))/3)$ and I get $22\pi/9$ but apparently that's not the correct answer
The surface area integral for a figure revolved about the $ \ x-$ axis is
$$ S \ = \ 2 \pi \ \int \ y \ \ ds \ , $$
which, for a parametric curve, will be
$$ S \ = \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dx}{dt} \right)^2} \ \ dt \ \ = \ \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( 2 - 2t^2 \right)^2 + \left( 4t \right)^2} \ \ dt , $$
$$ = \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( 2 + 2t^2 \right)^2} \ \ dt \ \ = \ 2 \pi \ \int \ y(t) \ \cdot \ \underbrace{(2 + 2t^2) \ \ dt}_{ds} \ \ , $$
as you have set up. However, while your arclength has been determined correctly, you have to multiply the factors of the integrand first before you set about performing the integration itself. So you want
$$ S \ = \ 2 \pi \ \int_0^1 \ ( \ 2t^2 \ ) \ \cdot \ (2 + 2t^2) \ \ dt \ \ = \ \ \ 2 \pi \ \int_0^1 \ 4t^2 \ + \ 4t^4 \ \ dt $$
$$ = \ \ 2 \pi \ \left .\left( \ \frac{4}{3}t^3 \ + \ \frac{4}{5}t^5 \right) \ \right \vert_0^1\ \ \ = \ \frac{64 \pi}{15} . $$
[corrected from what I did late last night]