Problem 17 - Calculus Early Transcendentals

Question

Let \(\sigma\) be the surface of the cube bounded by the planes \(x=\pm 1, y=\pm 1, z=\pm 1,\) oriented by outward unit normals. In each part, find the flux of \(\mathbf{F}\) across \(\sigma .\) $$ \begin{array}{l}{\text { (a) } \mathbf{F}(x, y, z)=x \mathbf{i}} \\ {\text { (b) } \mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}} \\ {\text { (c) } \mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}}\end{array} $$

Step-by-Step Solution

Answer

(a) 0, (b) 0, (c) 24

Step 1: Understanding the Problem

The problem asks us to calculate the flux of vector fields \( \mathbf{F} \) across the surface \( \sigma \) of a cube oriented by outward unit normals. The cube extends from \( x = -1 \) to \( x = 1 \), \( y = -1 \) to \( y = 1 \), and \( z = -1 \) to \( z = 1 \).

Step 2: Recall the Flux Formula

The flux of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) across a closed surface \( \sigma \) is given by \( \Phi = \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) is the outward unit normal.

Step 3: Problem (a) Calculate the Flux for \( \mathbf{F}(x, y, z)=x \mathbf{i} \)

Since the vector field has only the \( \mathbf{i} \) component, the contribution of \( \mathbf{F} \) to the flux is non-zero only on the surfaces perpendicular to the x-axis. Calculate flux through \( x = 1 \) and \( x = -1 \): 1. For \( x = 1 \): \( \Phi_{x=1} = \int_{-1}^{1} \int_{-1}^{1} 1 \, dy \, dz = 4 \).2. For \( x = -1 \): \( \Phi_{x=-1} = \int_{-1}^{1} \int_{-1}^{1} (-1) \, dy \, dz = -4 \).Flux for (a): \( \Phi = 4 + (-4) = 0 \).

Step 4: Problem (b) Calculate the Flux for \( \mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \)

This vector field has components in all directions, so compute flux through all six faces:1. \( x = 1 \): \( \Phi_{x=1} = \int_{-1}^{1} \int_{-1}^{1} 1 \, dy \, dz = 4 \).2. \( x = -1 \): \( \Phi_{x=-1} = -4 \).3. \( y = 1 \): \( \Phi_{y=1} = \int_{-1}^{1} \int_{-1}^{1} 1 \, dx \, dz = 4 \).4. \( y = -1 \): \( \Phi_{y=-1} = -4 \).5. \( z = 1 \): \( \Phi_{z=1} = 4 \).6. \( z = -1 \): \( \Phi_{z=-1} = -4 \).Flux for (b): \( \Phi = 4 + (-4) + 4 + (-4) + 4 + (-4) = 0 \).

Step 5: Problem (c) Calculate the Flux for \( \mathbf{F}(x, y, z)=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k} \)

This vector field has squared components, compute flux through all six faces:1. \( x = 1 \): \( \Phi_{x=1} = \int_{-1}^{1} \int_{-1}^{1} 1^2 \, dy \, dz = 4 \).2. \( x = -1 \): \( \Phi_{x=-1} = 4 \).3. \( y = 1 \): \( \Phi_{y=1} = 4 \).4. \( y = -1 \): \( \Phi_{y=-1} = 4 \).5. \( z = 1 \): \( \Phi_{z=1} = 4 \).6. \( z = -1 \): \( \Phi_{z=-1} = 4 \).Flux for (c): \( \Phi = 4 + 4 + 4 + 4 + 4 + 4 = 24 \).

Key Concepts

Surface Integrals

A surface integral is an important concept when dealing with vector fields and surfaces. Imagine you're trying to measure the "flow" of a field through a surface. The surface integral helps you quantify this flow. For our problem, we're dealing with a cube, and the surface integral will measure how much of the vector field passes through its faces.

In mathematical terms, you express the surface integral of a vector field \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) over a surface \( \sigma \) as \( \Phi = \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS \). Here, \( \mathbf{n} \) is the unit normal vector that points outwards on the surface.

This concept is very useful as it allows us to understand and calculate the amount of vector field, such as fluid flow or electromagnetic field, that crosses a given surface. It's essential when you're asked to find the flux of vector fields across surfaces like spheres, cubes, and other geometrical shapes.

Vector Fields

Vector fields are functions that assign a vector to every point in space. They are often used to represent physical quantities like velocity fields of a fluid or force fields. For example, wind speed across the surface of the earth at each point can be represented as a vector field. Each vector in the field shows both the direction and magnitude of the wind where it points.

In the exercise provided, there are three different vector fields given:

• In part (a), \( \mathbf{F}(x, y, z) = x \mathbf{i} \) represents a field pointing along the x-axis with magnitude proportional to x, uniform across the y and z directions.


• In part (b), \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) assigns a vector to each point in space where the vector points outwards from the origin with magnitude increasing linearly with distance.


• In part (c), \( \mathbf{F}(x, y, z) = x^2 \mathbf{i} + y^2 \mathbf{j} + z^2 \mathbf{k} \) increases the magnitude according to the square of each coordinate, causing a stronger field further from the origin.

Understanding the nature of these vector fields helps predict how they will interact with the cube's surfaces, which is essential in calculating the flux.

Gauss's Theorem

Gauss's theorem, also known as the Divergence Theorem, is a fundamental principle connecting the flux through a closed surface to a volume integral over the region inside. It states that the total flux of a vector field out of a closed surface is equal to the integral of the divergence of the field over the volume enclosed by the surface.

Mathematically, this is represented as \( \iint_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{V} (abla \cdot \mathbf{F}) \, dV \), where \( \sigma \) is the closed surface, and \( V \) is the volume it encloses. \( abla \cdot \mathbf{F} \) is the divergence of the vector field. This theorem simplifies the problem of calculating the flux through a complicated surface by potentially reducing it to a volume integral.

In the original exercise, applying Gauss's theorem to vector fields and closed surfaces like the cube can greatly aid in understanding why, for problems (a) and (b), the flux results in zero. This is because the divergence in those scenarios sums to zero over the cube's volume.