91Ó°ÊÓ

The Divergence Theorem relates the flow (or divergence) of a vector field \( \mathbf{F} \) through a closed surface to the divergence of \( \mathbf{F} \) over the volume \( E \) it encloses. Mathematically, it states \( \iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iiint_{E} abla \cdot \mathbf{F} \, dV \), where \( S \) is the boundary surface of the volume \( E \).

The divergence of \( \mathbf{F} = 3x \mathbf{i} + xy \mathbf{j} + 2xz \mathbf{k} \) is found by computing \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(3x) + \frac{\partial}{\partial y}(xy) + \frac{\partial}{\partial z}(2xz) \). Calculating, we have \( 3 + x + 2x = 3 + 3x \).

To verify the theorem, calculate the volume integral \( \iiint_{E} (3 + 3x) \, dV \) where \( E \) is the cube defined by \( 0 \leq x, y, z \leq 1 \). The integral becomes \[ \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (3 + 3x) \, dz \, dy \, dx = \int_{0}^{1} \int_{0}^{1} [3 + 3x]_{z=0}^{z=1} \, dy \, dx = \int_{0}^{1} \int_{0}^{1} 3 + 3x \, dy \, dx. \]

Next, solve the integral \( \int_{0}^{1} \int_{0}^{1} 3 + 3x \, dy \, dx = \int_{0}^{1} [3y + 3xy]_{y=0}^{y=1} \, dx = \int_{0}^{1} (3 + 3x) \, dx \). This evaluates to \( [3x + \frac{3}{2}x^2]_{0}^{1} = 3 + \frac{3}{2} = \frac{9}{2} \).

Now, compute \( \iint_{S} \mathbf{F} \cdot d\mathbf{S} \) across each face of the cube. Break it down for each face:1. **Face \( x = 0 \):** \( \mathbf{F} \cdot \mathbf{i} = 0 \).2. **Face \( x = 1 \):** \( \mathbf{F} \cdot \mathbf{i} = 3 \cdot 1 = 3 \), surface area = 1.3. **Face \( y = 0 \):** \( \mathbf{F} \cdot \mathbf{j} = 0 \).4. **Face \( y = 1 \):** \( \mathbf{F} \cdot \mathbf{j} = x \), surface area = 1.5. **Face \( z = 0 \):** \( \mathbf{F} \cdot \mathbf{k} = 0 \).6. **Face \( z = 1 \):** \( \mathbf{F} \cdot \mathbf{k} = 2x \), surface area = 1.Calculate these contributions: Integration over each face:- For \( x = 1 \) and \( z = 1 \), the net flux contributes \( 3 \) and \( \int_{0}^{1}\int_{0}^{1} 2x \, dy \, dx = \int_{0}^{1} 2x \, dx = 1 \);- \( y = 1 \) contributes \[ \int_{0}^{1}\int_{0}^{1} x \, dz \, dx = \int_{0}^{1} x \, dx = \frac{1}{2} \].Total: \( 3 + 1 + \frac{1}{2} = \frac{9}{2} \).

Both the surface integral and the volume integral agree with the Divergence Theorem: \( \frac{9}{2} = \frac{9}{2} \), confirming that the theorem is satisfied for this vector field and volume.