Problem 402 - Calculus Volume 3

Question

For the following exercises, use a CAS and the divergence theorem to compute the net outward flux for the vector fields across the boundary of the given regions \(D\). Let \(\mathbf{F}(x, y, z)=2 x \mathbf{i}-3 x y \mathbf{j}+x z^{2} \mathbf{k}\). Use the divergence theorem to calculate \(\iint_{S} \mathbf{F} \cdot d \mathbf{S}, \quad\) where \(S\) is the surface of the cube with corners at (0,0,0),(1,0,0),(0,1,0) \((1,1,0),(0,0,1),(1,0,1),(0,1,1),\) and (1,1,1) oriented outward.

Step-by-Step Solution

Answer

The net outward flux is \( \frac{5}{6} \).

Step 1: Understand the Problem

We are required to calculate the net outward flux of the vector field \( \mathbf{F}(x, y, z)=2x \mathbf{i} - 3xy \mathbf{j} + xz^2 \mathbf{k} \) across the surface of a cube using the Divergence Theorem. The cube is defined by its corners (0,0,0), (1,0,0), (0,1,0), (1,1,0), (0,0,1), (1,0,1), (0,1,1), and (1,1,1).

Step 2: Recall the Divergence Theorem

The Divergence Theorem states that \( \iint_{S} \mathbf{F} \cdot d \mathbf{S} = \iiint_{D} abla \cdot \mathbf{F} \, dV \), where \( S \) is the boundary surface of the region \( D \), and \( abla \cdot \mathbf{F} \) is the divergence of the vector field \( \mathbf{F} \).

Step 3: Calculate the Divergence

The divergence of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \). For \( \mathbf{F}(x, y, z) = 2x \mathbf{i} - 3xy \mathbf{j} + xz^2 \mathbf{k} \), we calculate: \[ \frac{\partial}{\partial x}(2x) + \frac{\partial}{\partial y}(-3xy) + \frac{\partial}{\partial z}(xz^2) = 2 - 3x + 2xz. \]

Step 4: Set up the Triple Integral

The divergence \( abla \cdot \mathbf{F} = 2 - 3x + 2xz \) must be integrated over the volume of the cube \( D \). The integral is set up as \( \iiint_{D} (2 - 3x + 2xz) \, dx \, dy \, dz \) where \( x, y, z \) each range from 0 to 1.

Step 5: Evaluate the Triple Integral

Evaluate the triple integral \( \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (2 - 3x + 2xz) \, dx \, dy \, dz \). Perform the integration with respect to \( x \), then \( y \), and finally \( z \). Integration steps are as follows:- Integrate with respect to \( x \): \( \int_{0}^{1} (2 - 3x + 2xz) \, dx = \left[ 2x - \frac{3}{2}x^2 + xz^2 \right]_{0}^{1} = 2 - \frac{3}{2} + z^2. \)- Integrate with respect to \( y \): since the result doesn't depend on \( y \), multiply by \( \int_{0}^{1} \, dy = 1 .\)- Integrate with respect to \( z \): \( \int_{0}^{1} (2 - \frac{3}{2} + z^2) \, dz = \left[ 2z - \frac{3}{2}z + \frac{1}{3}z^3 \right]_{0}^{1} = 2 - \frac{3}{2} + \frac{1}{3} = \frac{5}{6}. \)

Step 6: Interpret the Result

The result from the evaluation is \( \frac{5}{6} \). This means the net outward flux of the vector field \( \mathbf{F} \) across the surface of the cube is \( \frac{5}{6} \).

Key Concepts

Net Outward Flux

In mathematics, the concept of net outward flux is integral in understanding how a vector field flows out through a closed surface. Imagine a vector field as a collection of arrows representing both magnitude and direction in space.
For any surface in this vector field, the net outward flux measures how much of this field exits the surface. Net outward flux can be visualized by thinking of the vector field as a continuous flow of fluid. The flux tells us how much fluid is leaving the closed boundary. The divergence theorem simplifies the calculation of net outward flux.
Instead of calculating through the surface directly, you can use a volume integral of the divergence within it. This makes finding the net outward flux computationally easier, especially for complex surfaces.

Vector Fields

A vector field is a fascinating mathematical construct where every point in a space is assigned a vector. These vectors carry both direction and magnitude, allowing us to represent various physical phenomena, such as wind speed, magnetic forces, and even fluid flow.
In the exercise address in this article, the vector field is given as \( \mathbf{F}(x, y, z)=2x \mathbf{i} - 3xy \mathbf{j} + xz^2 \mathbf{k} \). Each component of the vector field correlates to its respective axis:

• The \( \mathbf{i} \) component corresponds to the x-axis, with a magnitude dependent on \( 2x \).
• The \( \mathbf{j} \) component corresponds to the y-axis, with its magnitude \( -3xy \).
• The \( \mathbf{k} \) component corresponds to the z-axis, with a magnitude of \( xz^2 \).

This vector field varies in space and thus showcases how the flow can change in different regions.

Triple Integral

A triple integral is a mathematical tool used for integrating a function over a three-dimensional region. In calculus, those regions are often complex shapes like cubes, spheres, or irregular volumes.
In the context of vector calculus, triple integrals help compute total quantities like mass or flux across a volume.For this exercise, the function being integrated was the divergence of the vector field and had to be performed over the volume of the cube. The integral \( \iiint_{D} (2 - 3x + 2xz) \, dx \, dy \, dz \) was calculated over a cube. Triple integrals require addressing one variable at a time, sequentially integrating with respect to x, y, and z.
Proper evaluation results in a single value, which, in this case, quantifies the net outward flux from the vector field through the volume of the cube.

Cube Volume

A cube is one of the simplest yet most essential shapes in geometry, with all sides of equal length. Its volume, found by the formula \( V = s^3 \), where \( s \) is the side length, is the space the cube occupies in three dimensions. In the exercise, the cube defined by corners from (0,0,0) to (1,1,1) has a unit side length of 1, resulting in a volume of 1 cubic unit.Understanding this volume is paramount when applying the divergence theorem. The cube’s volume acts as the domain for a triple integral, allowing for the computation of flux over the entire region. Cube volume properties simplify calculation steps, given the uniform range of integration limits from 0 to 1 for each x, y, and z coordinate.
Such properties make cubes ideal for theoretical calculations and practical applications in physics and engineering.