91Ó°ÊÓ

The divergence of a vector field \(\mathbf{F} = \langle P,Q,R \rangle\) is given by \(\textit{div} \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\). In our case: $$ \textit{div} \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial (-2y)}{\partial y} + \frac{\partial (3z)}{\partial z} = 1 - 2 + 3 = 2 $$

According to the Divergence Theorem: $$ \iint_S \mathbf{F} \cdot \mathbf{n} dS = \iiint_V (\textit{div} \mathbf{F}) dV $$ We've already found \(\textit{div} \mathbf{F} = 2\). To evaluate the triple integral, we use spherical coordinates \((\rho, \theta, \phi)\), where \(x = \rho \sin \phi \cos \theta\), \(y = \rho \sin \phi \sin \theta\), and \(z = \rho \cos \phi\). The volume element in spherical coordinates is given by \(dV = \rho^2 \sin \phi d\rho d\theta d\phi\).

Calculate the net outward flux of the field \(\mathbf{F}\) across the surface \(S\): $$ \iiint_V (\textit{div} \mathbf{F}) dV = \int_{0}^{\sqrt{6}} \int_{0}^{2\pi} \int_{0}^{\pi} 2 \rho^2 \sin \phi d\rho d\theta d\phi $$

Break down the integral into the product of three separate integrals: $$ \iiint_V (\textit{div} \mathbf{F}) dV = 2 \int_{0}^{\sqrt{6}} \rho^2 d\rho \cdot \int_{0}^{2\pi} d\theta \cdot \int_{0}^{\pi} \sin \phi d\phi $$ Compute the integral for each term: $$ \int_{0}^{\sqrt{6}} \rho^2 d\rho = \frac{\rho^3}{3} \Big|_{0}^{\sqrt{6}} = \frac{6\sqrt{6}}{3} $$ $$ \int_{0}^{2\pi} d\theta = \theta \Big|_{0}^{2\pi} = 2\pi $$ $$ \int_{0}^{\pi} \sin \phi d\phi = -\cos \phi \Big|_{0}^{\pi} = 2 $$ Multiply all the calculated integrals: $$ 2\cdot \frac{6\sqrt{6}}{3} \cdot 2\pi \cdot 2 = 16\pi\sqrt{6} $$ The net outward flux of the vector field \(\mathbf{F} = \langle x,-2 y, 3 z\rangle\) across the sphere with radius \(\sqrt{6}\) is \(16\pi\sqrt{6}\).