91Ó°ÊÓ

The divergence theorem states that \( \int_{S} \mathbf{F} \cdot \mathbf{dS} = \int_{V} abla \cdot \mathbf{F} \, dV \) where \( V \) is the volume enclosed by \( S \). This theorem allows us to convert a surface integral into a volume integral over the region inside the surface.

Calculate \( abla \cdot \mathbf{F} \). For \( \mathbf{F}(x, y, z) = (2x + y \cos z)\mathbf{i} + (x^2 - y) \mathbf{j} + (y^2 z) \mathbf{k} \), the divergence is:\[ abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(2x + y \cos z) + \frac{\partial}{\partial y}(x^2 - y) + \frac{\partial}{\partial z}(y^2 z) \].Calculating each term, we have:- \( \frac{\partial}{\partial x}(2x + y \cos z) = 2 \)- \( \frac{\partial}{\partial y}(x^2 - y) = -1 \)- \( \frac{\partial}{\partial z}(y^2 z) = y^2 \).Thus, \( abla \cdot \mathbf{F} = 2 - 1 + y^2 = 1 + y^2 \).

Consider \( V \), the interior of the sphere \( x^2 + y^2 + z^2 = 4 \). Substitute \( abla \cdot \mathbf{F} = 1 + y^2 \) into the volume integral:\[ \int_{V} (1 + y^2) \, dV \].Since the region of integration is a sphere of radius 2, it's convenient to use spherical coordinates \( (\rho, \theta, \phi) \):\( x = \rho \sin(\phi) \cos(\theta) \),\( y = \rho \sin(\phi) \sin(\theta) \),\( z = \rho \cos(\phi) \), where \( 0 \leq \rho \leq 2 \), \( 0 \leq \theta < 2\pi \), and \( 0 \leq \phi \leq \pi \).The differential volume element is \( dV = \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \).

Substituting into the integral, we have:\[ \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{2} (1 + (\rho^2 \sin^2(\phi) \sin^2(\theta))) \cdot \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \].Separate the integrals:\[\int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta + \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{2} \rho^4 \sin^3(\phi) \sin^2(\theta) \, d\rho \, d\phi \, d\theta\].

1. Evaluate \( \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta \): - Integrating over \( \rho \): \[ \int_{0}^{2} \rho^2 \, d\rho = \left. \frac{\rho^3}{3} \right|_{0}^{2} = \frac{8}{3} \]. - Integrating over \( \phi \): \[ \int_{0}^{\pi} \sin(\phi) \, d\phi = 2 \]. - Integrating over \( \theta \): \[ \int_{0}^{2\pi} \, d\theta = 2\pi \]. - Combine: \[ 2\pi \times 2 \times \frac{8}{3} = \frac{32\pi}{3} \].2. Evaluate \( \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{2} \rho^4 \sin^3(\phi) \sin^2(\theta) \, d\rho \, d\phi \, d\theta \): - Integrating over \( \rho \): \[ \int_{0}^{2} \rho^4 \, d\rho = \left. \frac{\rho^5}{5} \right|_{0}^{2} = \frac{32}{5} \]. - Integrating over \( \phi \): \[ \int_{0}^{\pi} \sin^3(\phi) \, d\phi = \frac{4}{3} \]. - Integrating over \( \theta \): \[ \int_{0}^{2\pi} \sin^2(\theta) \, d\theta = \pi \]. - Combine: \[ \frac{32}{5} \times \frac{4}{3} \times \pi = \frac{128\pi}{15} \].

Combine both results from Step 5:- \( \frac{32\pi}{3} + \frac{128\pi}{15} \). Convert to a common denominator and add:- Common denominator is 15: \( \frac{32\pi}{3} = \frac{160\pi}{15} \).- Thus, \[ \frac{160\pi}{15} + \frac{128\pi}{15} = \frac{288\pi}{15} = \frac{96\pi}{5} \].