Answer
The flux of the vector field \(\mathbf{F}\) across the portion of the cone \(\sigma\) is \(\frac{14\pi}{3}\)
Work Step by Step
Given vector field \(\mathbf{F} = xi + yj + 2zk\). The flux across the portion of the cone \(\sigma\) between the planes \(z = 1\) and \(z = 2\) is given by: \[ \Phi = \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS, \] where \(\mathbf{n}\) is the unit normal vector. Using the formula (5), we have: \[ \Phi = \iint_{\mathbf{R}} \mathbf{F} \cdot \left(\frac{\partial\mathbf{r}}{\partial r} \times \frac{\partial\mathbf{r}}{\partial \theta}\right) \, dA, \] where \(\mathbf{r} = r\cos\theta i + r\sin\theta j + rk\) represents the position vector. The partial derivatives are: \[ \frac{\partial\mathbf{r}}{\partial r} = \cos\theta i + \sin\theta j + k, \] \[ \frac{\partial\mathbf{r}}{\partial \theta} = -r\sin\theta i + r\cos\theta j, \] The cross product is: \[ \frac{\partial\mathbf{r}}{\partial r} \times \frac{\partial\mathbf{r}}{\partial \theta} = -r\cos\theta i - r\sin\theta j + rk, \] So, the dot product becomes: \[ \begin{aligned} \mathbf{F} \cdot \left(\frac{\partial\mathbf{r}}{\partial r} \times \frac{\partial\mathbf{r}}{\partial \theta}\right) &= \left(xi + yj + 2zk\right) \cdot \left(-r\cos\theta i - r\sin\theta j + rk\right) \\ &= -r^2 \left(\cos^2\theta + \sin^2\theta\right) + 2r^2 \\ &= -r^2 + 2r^2 \\ &= r^2. \end{aligned} \] Finally, the flux of the vector field \(\mathbf{F}\) across \(\sigma\) is: \[ \begin{aligned} \Phi &= \int_{0}^{2\pi} \int_{1}^{2} r^2 \, dr \, d\theta \\ &= 2\pi \left[\frac{1}{3}r^3\right]_1^2 \\ &= \frac{2\pi}{3} \left(2^3 - 1^3\right) \\ &= \frac{14\pi}{3}. \end{aligned} \] So, the flux of the vector field \(\mathbf{F}\) across the portion of the cone \(\sigma\) is \(\frac{14\pi}{3}\).
