91Ó°ÊÓ

We can first find the normal to the surface by taking the gradient of the implicit function $$\frac{x^2}{4} + \frac{y^2}{8} + \frac{z^2}{12} = 1$$. The normal vector is then given by: $$\mathbf{n} = \langle\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\rangle\left(\frac{x^2}{4} + \frac{y^2}{8} + \frac{z^2}{12}\right)$$ $$\mathbf{n} = \left\langle\frac{1}{2} x, \frac{1}{4} y, \frac{1}{6} z\right\rangle$$

Now, let's evaluate $$\iint_{\partial D} \mathbf{F} \cdot \mathbf{n} dS$$ where $$\mathbf{n}$$ is the normal to the surface at each point. To set up the integral, we convert to spherical coordinates (\(\rho\), \(\theta\), \(\phi\)), noting that $$x = 2\rho \sin\phi\cos\theta$$, $$y = 2\sqrt{2}\rho\sin\phi\sin\theta$$, and $$z = 2\sqrt{3}\rho\cos\phi$$. We also have that $$dS = \rho^{2}\sin\phi d\phi d\theta$$. Then we have: $$\iint_{\partial D} \mathbf{F} \cdot \mathbf{n} dS = \int_{0}^{2\pi}\int_{0}^{\pi}\left(\langle\frac{2\sqrt{3}}{3}(1-\rho\cos\phi), \rho\sin\phi\cos\theta, -2\rho\sin\phi\cos\theta\rangle\right) \cdot \left\langle\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi\right\rangle \rho^{2}\sin\phi d\phi d\theta$$ Now, integrating with respect to \(\phi\) and \(\theta\), we get: $$\iint_{\partial D} \mathbf{F} \cdot \mathbf{n} dS = -\frac{256\pi}{3}$$

Now, we need to calculate the divergence of vector field F: $$\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(z-y) + \frac{\partial}{\partial y}(x) + \frac{\partial}{\partial z}(-x)$$ $$\nabla \cdot \mathbf{F} = 0 - 1 - 1$$ $$\nabla \cdot \mathbf{F} = -2$$

Now we can compute the volume integral of the divergence of the vector field over the region D: $$\iiint_{D} (\nabla\cdot \mathbf{F}) dV = \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} (-2) \cdot (32\rho^{2}\sin\phi) d\rho d\phi d\theta$$ Integrating with respect to \(\rho\), \(\phi\), and \(\theta\), we get: $$\iiint_{D} (\nabla\cdot \mathbf{F}) dV = -\frac{256\pi}{3}$$

As we can see, both the flux through the surface $$\iint_{\partial D} \mathbf{F} \cdot \mathbf{n} dS = -\frac{256\pi}{3}$$ and the volume integral of the divergence $$\iiint_{D} (\nabla\cdot \mathbf{F}) dV = -\frac{256\pi}{3}$$ agree. Therefore, the Divergence Theorem holds for this vector field and region.