Answer
$W = \frac{27}{28} $
Work Step by Step
Step 1: Given the vector field: \[ \mathbf{F}(x, y, z) = xy\mathbf{i} + yz\mathbf{j} + xz\mathbf{k} \] And the parameterization of the curve \(C\): \[ x = t, \quad y = t^2, \quad z = t^3, \quad (0 \leq t \leq 1) \] Step 2: To find the work performed by the force field \(\mathbf{F}\) along the curve \(C\), you can set up the line integral as follows: \[ \begin{aligned} W &= \int_C \mathbf{F} \cdot d\mathbf{r} \\ &= \int_{0}^{1} \left[xy\mathbf{i} + yz\mathbf{j} + xz\mathbf{k}\right] \cdot \left(dti\ + 2tdtj + 3t^2dtk\right) \\ &= \int_{0}^{1} \left(t^3 + 5t^6\right)dt \\ &= \left[\frac{1}{4}t^4 + \frac{7}{5}t^7\right]_{0}^{1} \\ &= \frac{27}{28} \end{aligned} \] Result: \[ W = \frac{27}{28} \]
