# Evaluating line integral on the curve

I have this problem that I think I'm doing right, but not getting the correct answer. Could someone point out where I'm screwing up here?

Evaluate $$\int_C G(x,y,z) \, ds$$ on the curve indicated by $$x= \frac13 t^3, \quad y=t^2, \quad z=2t,$$ where $t$ is on the closed interval $[0,1]$ for $G(x, y, z) = 4xyz$. Round in the hundredths place.

Following the line integral formula in the text book, I'm getting

$$\int_0^1 [\frac{8t^6}{3} \sqrt{t^4 + 4t^2 + 4}] \, dt.$$ Exact value is 1.06, which is not the correct answer.

\begin{align} \int_0^1 \frac83 t^6 \sqrt{t^4 + 4t^2 + 4}] \, dt&=\frac83\int_0^1 t^6 \sqrt{(t^2 + 2)^2}] \, dt \\ &= \frac83 \int_0^1 (t^8 +2t^6) dt \\ &= \frac83 \left(\frac19 +\frac27 \right)\\ &=\frac{200}{189} \end{align}