Evaluate $\int_{C} y d s$, where $C$ is the parabola $\vec{r}=\left\langle 3 t^{2}, 2 t\right\rangle$, for $-1 \leq t \leq 0$. Answer exactly or round to 2 decimal places.

Step 1 ：We are given the integral \(\int_{C} y d s\), where C is the parabola \(\vec{r}=\left\langle 3 t^{2}, 2 t\right\rangle\), for \(-1 \leq t \leq 0\).

Step 2 ：The function to be integrated is y, which corresponds to the second component of the vector function \(\vec{r}(t)\).

Step 3 ：We parameterize the curve C using the vector function \(\vec{r}(t)\), then substitute y with the second component of \(\vec{r}(t)\), and ds with the magnitude of the derivative of \(\vec{r}(t)\) dt.

Step 4 ：The derivative of \(\vec{r}(t)\) is \(\vec{r}'(t) = \left\langle 6t, 2 \right\rangle\). The magnitude of \(\vec{r}'(t)\) is \(\sqrt{(6t)^{2} + 2^{2}} = \sqrt{36t^{2} + 4}\).

Step 5 ：The integral becomes \(\int_{-1}^{0} (2t \sqrt{36t^{2} + 4}) dt\).

Step 6 ：Evaluating the integral, we get \(\frac{4}{27} - \frac{40\sqrt{10}}{27}\).

Step 7 ：Final Answer: The value of the integral is \(\boxed{\frac{4}{27} - \frac{40\sqrt{10}}{27}}\).