Answer
$\Phi = 18\pi$
Work Step by Step
Given the vector field \(\mathbf{F}(x, y, z) = xi + yj + zk\) and the vector-value function \(\mathbf{r}(u, v) = u\cos(v)\mathbf{i} + u\sin(v)\mathbf{j} + (1 - u^2)\mathbf{k}\) with \(\sigma\) defined as: \[ \sigma: \{(u, v) \mid 1 \leq u \leq 2, \, 0 \leq v \leq 2\pi\} \] The flux across \(\sigma\) in the direction of the positive orientation is given by: \[ \Phi = \int_{\sigma} \mathbf{F} \cdot \mathbf{n} \,d\sigma = \iint_{R} \mathbf{F} \cdot \left(\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\right) \,dA \] where: \[ \frac{\partial \mathbf{r}}{\partial u} = \cos(v)\mathbf{i} + \sin(v)\mathbf{j} - 2u\mathbf{k} \] \[ \frac{\partial \mathbf{r}}{\partial v} = -u\sin(v)\mathbf{i} + u\cos(v)\mathbf{j} \] And their cross product results in: \[ \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = 2u^2\cos(v)\mathbf{i} + 2u^2\sin(v)\mathbf{j} + u\mathbf{k} \] Thus, \[ \mathbf{F} \cdot \left(\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\right) = (xi + yj + k) \cdot (2u^2\cos(v)\mathbf{i} + 2u^2\sin(v)\mathbf{j} + u\mathbf{k}) = 2u^3\cos^2(v) + 2u^3\sin^2(v) + u = 2u^3 + u \] Using this, the flux across \(\sigma\) is calculated as: \[ \Phi = \int_{0}^{2\pi} \int_{1}^{2} (2u^3 + u) \,du\,dv = 2\pi \left[ \frac{u^4}{2} + \frac{u^2}{2} \right]_1^2 = 18\pi \] Finally, \[ \Phi = 18\pi \]
