Answer
$W = \frac{92}{9} + \ln(3)$
Work Step by Step
Step 1: Given the vector field: \[ \mathbf{F}(x, y) = (x^2 + xy)\mathbf{i} + (y - x^2y)\mathbf{j} \] and the parameterization of the curve \(C\): \[ x = t, \quad y = \frac{1}{t} \quad (1 \leq t \leq 3) \] 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: \[ \begin{aligned} W &= \int_C \mathbf{F} \cdot d\mathbf{r} \\ &= \int_{1}^{3} \left[(x^2 + xy)\mathbf{i} + (y - x^2y)\mathbf{j}\right] \cdot \left(dt\mathbf{i} - \frac{t^2}{1}\,dt\mathbf{j}\right) \\ &= \int_{1}^{3} \left(t^2 + 1 - \frac{1}{t^3} + \frac{1}{t}\right)\,dt \\ &= \left[\frac{1}{3}t^3 + t + 2\left(\frac{1}{t^2}\right) + \ln|t|\right]_{1}^{3} \\ &= \left(\frac{1}{3}(3^3) + 3 + 2\left(\frac{1}{3^2}\right) + \ln(3)\right) - \left(\frac{1}{3}(1^3) + 1 + 2\left(\frac{1}{1^2}\right) + \ln(1)\right) \\ &= \frac{92}{9} + \ln(3) \end{aligned} \] Result: \[ W = \frac{92}{9} + \ln(3) \]
