91Ó°ÊÓ

The Divergence Theorem relates the flux of a vector field through a closed surface to a volume integral over the region bounded by the surface. It states: \[ \iint_{\partial D} \mathbf{F} \cdot \mathbf{n}\, dS = \iiint_{D} (abla \cdot \mathbf{F})\, dV \] where \(\partial D\) is the boundary of \(D\), \(\mathbf{F}\) is the vector field, and \(abla \cdot \mathbf{F}\) is the divergence of \(\mathbf{F}\).

For the vector field \(\mathbf{F} = x^{2} \mathbf{i} + xz \mathbf{j} + 3z \mathbf{k}\), the divergence is given by: \(abla \cdot \mathbf{F} = \frac{\partial }{\partial x}(x^2) + \frac{\partial }{\partial y}(xz) + \frac{\partial }{\partial z}(3z)\). Calculate each partial derivative: \(\frac{\partial }{\partial x}(x^2) = 2x\), \(\frac{\partial }{\partial y}(xz) = 0\), and \(\frac{\partial }{\partial z}(3z) = 3\). Thus, \(abla \cdot \mathbf{F} = 2x + 0 + 3 = 2x + 3\).

The region \(D\) is the solid sphere given by \(x^2 + y^2 + z^2 \leq 4\). In spherical coordinates, this is described by \(0 \leq \rho \leq 2\), \(0 \leq \theta \leq 2\pi\), and \(0 \leq \phi \leq \pi\). The divergence \(2x + 3\) in spherical coordinates becomes \(2\rho \cos(\theta)\sin(\phi) + 3\). The volume element in spherical coordinates is \(\rho^2 \sin(\phi)\, d\rho \, d\theta \, d\phi\). The integral setup is: \[ \iiint_{D} (2x + 3)\, dV = \int_{0}^{\pi} \int_{0}^{2\pi} \int_{0}^{2} \left(2\rho \cos(\theta)\sin(\phi) + 3\right)\rho^2 \sin(\phi)\, d\rho \, d\theta \, d\phi \]

First integrate with respect to \(\rho\): \[ \int_{0}^{2} \left(2\rho \cos(\theta)\sin(\phi) + 3\right)\rho^2 \sin(\phi)\, d\rho = \int_{0}^{2} (2\rho^3 \cos(\theta) \sin^2(\phi) + 3\rho^2 \sin(\phi))\, d\rho \] Calculate it: \[= \left. \frac{1}{2} \rho^4 \cos(\theta) \sin^2(\phi) + \rho^3 \sin(\phi) \right|_{0}^{2} = 8\cos(\theta) \sin^2(\phi) + 24\sin(\phi)\] Integrate with respect to \(\theta\): \[ \int_{0}^{2\pi} (8\cos(\theta) \sin^2(\phi) + 24\sin(\phi)) \, d\theta = \int_{0}^{2\pi} 8\cos(\theta) \sin^2(\phi)\, d\theta + \int_{0}^{2\pi} 24\sin(\phi) \, d\theta \] \(\int_{0}^{2\pi} 8\cos(\theta) \sin^2(\phi)\, d\theta\) evaluates to zero, and \(\int_{0}^{2\pi} 24\sin(\phi)\, d\theta = 48\pi\sin(\phi)\). Now integrate with respect to \(\phi\): \[ \int_{0}^{\pi} 48\pi\sin(\phi)\, \sin(\phi)\, d\phi = 48\pi \int_{0}^{\pi} \sin^2(\phi)\, d\phi \] Use the identity \(\sin^2(\phi) = \frac{1 - \cos(2\phi)}{2}\): \[= 48\pi \int_{0}^{\pi} \frac{1 - \cos(2\phi)}{2}\, d\phi = 48\pi\left[\frac{\phi}{2} - \frac{\sin(2\phi)}{4}\right]_{0}^{\pi} = 48\pi\left(\frac{\pi}{2}\right) = 24\pi^2\]

The outward flux of the vector field \(\mathbf{F}\) through the boundary of the region \(D\) is \(24\pi^2\).