6. (15 pts) Find the work done by a force field $\vec{F}(x, y)=\sqrt{x y} \vec{\imath}+e^{y} \vec{\jmath}+x z \vec{k}$ to move an object along a curve $C$ with vector equation $\vec{r}(t)=t^{4} \vec{\imath}+t^{2} \vec{\jmath}+t^{3} \vec{k}, \quad 0 \leq t \leq 1$.

Step 1 ：The work done by a force field \(\vec{F}\) to move an object along a curve \(C\) is given by the line integral of \(\vec{F}\) along \(C\). In this case, we need to compute the line integral of \(\vec{F}(x, y, z) = \sqrt{x y} \vec{\imath}+e^{y} \vec{\jmath}+x z \vec{k}\) along the curve \(C\) with vector equation \(\vec{r}(t)=t^{4} \vec{\imath}+t^{2} \vec{\jmath}+t^{3} \vec{k}, \quad 0 \leq t \leq 1\).

Step 2 ：The line integral of a vector field \(\vec{F}\) along a curve \(C\) parametrized by \(\vec{r}(t)\), \(a \leq t \leq b\) is given by: \(\int_{C} \vec{F} \cdot d\vec{r} = \int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt\) where \(\vec{r}'(t)\) is the derivative of \(\vec{r}(t)\) with respect to \(t\).

Step 3 ：So, we first need to compute \(\vec{r}'(t)\), then substitute \(\vec{r}(t)\) and \(\vec{r}'(t)\) into \(\vec{F}\), and finally compute the integral.

Step 4 ：Let's compute \(\vec{r}'(t)\): \(\vec{r}'(t) = \begin{bmatrix} 4t^{3} \\ 2t \\ 3t^{2} \end{bmatrix}\)

Step 5 ：Substitute \(\vec{r}(t)\) and \(\vec{r}'(t)\) into \(\vec{F}\): \(\vec{F}(\vec{r}(t)) = \begin{bmatrix} \sqrt{t^{6}} \\ e^{t^{2}} \\ t^{7} \end{bmatrix}\)

Step 6 ：Compute the dot product of \(\vec{F}(\vec{r}(t))\) and \(\vec{r}'(t)\): \(\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) = 3t^{9} + 4t^{3}\sqrt{t^{6}} + 2t e^{t^{2}}\)

Step 7 ：Finally, compute the integral from 0 to 1: \(\int_{0}^{1} \vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt = -\frac{9}{70} + e\)

Step 8 ：Final Answer: The work done by the force field to move the object along the curve is \(\boxed{-\frac{9}{70} + e}\)