Answer
Result: \(\Phi = 0\)
Work Step by Step
Step 1: In this problem, we have to find the flux of \(\mathbf{F}\) through \(\sigma\), where \(\sigma\) is a hemisphere of radius \(3\) with its center at the origin. Step 2: In order to find the flux, we will be using the following formula: \[ \begin{align*} \Phi &= \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \quad \text{(1)} \\ &= \iint_{\sigma} \mathbf{F} \cdot \left(\frac{\partial \mathbf{r}}{\partial \phi} \times \frac{\partial \mathbf{r}}{\partial \theta}\right) dS \quad \text{(2)} \\ &= \iint_{\sigma} <−y,z,3x> \cdot 3\sin(\phi) \, dA \\ &= \iint_{\sigma} 3\sin(\phi)(-zy + zy + 3xz) \, dA \\ &= 9\iint_{\sigma} xz\sin(\phi) \, dA \\ &= 9\iint_{\sigma} (3\sin(\phi)\cos(\theta))(3\cos(\phi))\sin(\phi) \, dA \\ &= 81\iint_{\sigma} \sin^2(\phi)\cos(\phi)\cos(\theta) \, dA \quad \text{(3)} \end{align*} \] Now, integrate with respect to \(\theta\) and \(\phi\): \[ \begin{align*} \Phi &= 81\int_0^{2\pi} \int_0^{\pi/2} \sin^2(\phi)\cos(\phi)\cos(\theta) \, d\phi \, d\theta \quad \text{(3)} \\ &= 81\int_0^{2\pi} \cos(\theta) \, d\theta \int_0^{\pi/2} \sin^2(\phi)\cos(\phi) \, d\phi \end{align*} \] However, the blue integral is \(0\) because the net area under a sinusoidal curve is \(0\) if the length of the interval is a complete period. Result: \(\Phi = 0\)
