Problem 21 - Thomas Calculus

Question

a. Show that the outward flux of the position vector field \(\mathbf{F}=\) \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) through a smooth closed surface \(S\) is three times the volume of the region enclosed by the surface. b. Let n be the outward unit normal vector field on \(S .\) Show that it is not possible for \(F\) to be orthogonal to \(n\) at every point of \(S .\)

Step-by-Step Solution

Answer

The flux is three times the volume, and \( \mathbf{F} \) cannot be orthogonal to \( \mathbf{n} \) on \( S \).

Step 1: Recall the Divergence Theorem

The divergence theorem states that for a vector field \( \mathbf{F} \), the outward flux through a closed surface \( S \) is equal to the volume integral of the divergence of \( \mathbf{F} \) over the region \( V \) enclosed by \( S \). Mathematically, it is given by: \[ \int_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \int_{V} abla \cdot \mathbf{F} \, dV \]

Step 2: Find the Divergence of \( \mathbf{F} \)

The vector field \( \mathbf{F} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) has components \( F_1 = x \), \( F_2 = y \), and \( F_3 = z \). The divergence \( abla \cdot \mathbf{F} \) is calculated as: \[ abla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3 \]

Step 3: Calculate the Volume Integral

Substitute the divergence \( abla \cdot \mathbf{F} = 3 \) into the volume integral: \[ \int_{V} abla \cdot \mathbf{F} \, dV = \int_{V} 3 \, dV = 3 \int_{V} \, dV = 3V \] where \( V \) is the volume of the region enclosed by the surface \( S \).

Step 4: Compare with Flux Integral

According to the divergence theorem, the flux integral through the closed surface \( S \) is: \[ \int_{S} \mathbf{F} \cdot \mathbf{n} \, dS = 3V \] This shows that the outward flux of \( \mathbf{F} \) through \( S \) is indeed three times the volume of the region enclosed by \( S \).

Step 5: Analyze Orthogonality Condition

Suppose \( \mathbf{F} \) is orthogonal to \( \mathbf{n} \) everywhere on \( S \). This would mean \( \mathbf{F} \cdot \mathbf{n} = 0 \) for all points on \( S \), implying a zero flux through \( S \). However, we have shown that the flux is \( 3V \). Since \( 3V eq 0 \), \( \mathbf{F} \) cannot be orthogonal to \( \mathbf{n} \) at every point of \( S \).

Key Concepts

Vector Field

A **vector field** is a mathematical construct that assigns a vector to every point in space. This is useful for representing quantities that have both magnitude and direction at each point, such as wind velocity in weather models or the gravitational force in physics.
In the given exercise, the vector field is expressed as \( \mathbf{F} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). This notation means that at any point \((x, y, z)\), the value of \( \mathbf{F} \) is determined by its components along the x, y, and z directions, respectively. Here:

• \( \mathbf{i} \) represents the unit vector in the direction of the x-axis.
• \( \mathbf{j} \) represents the unit vector in the direction of the y-axis.
• \( \mathbf{k} \) represents the unit vector in the direction of the z-axis.

Such vector fields can describe physical phenomena where forces or velocities vary over space, and they are fundamental in applying calculus to multidimensional analysis and physics problems.

Outward Flux

The **outward flux** refers to the amount of a vector field passing through a surface. When we consider a closed surface, like a sphere or a cube, the outward flux is calculated with respect to the surface's outer normals, which are vectors perpendicular to the surface pointing outwards.
The calculation of outward flux through a closed surface \( S \) involves integrating the dot product of the vector field \( \mathbf{F} \) and the outward normal vector \( \mathbf{n} \) over the entire surface. Mathematically, this is expressed as:
\[\int_{S} \mathbf{F} \cdot \mathbf{n} \, dS\]This integral provides a scalar value representing the net rate at which the vector field 'flows' out of the volume bounded by the surface.
In the problem, you determine that this flux equals three times the enclosed volume by using the Divergence Theorem, tying the concept of flux to that of the volume integral.

Closed Surface

A **closed surface** is a surface that completely encloses a volume with no gaps or openings, akin to the surface of a balloon or a closed box. In mathematical terms, a closed surface is one for which the boundary does not exist, meaning it bounds a finite volume within.
Such surfaces are crucial for applying the Divergence Theorem, which connects the flux through this type of surface to the volume integral of the divergence of the vector field inside the surface.
Examples of closed surfaces include:

• Spheres
• Cylinders with capped ends
• Cubes or rectangular prisms

The closed nature of these surfaces is vital when performing calculations using vector fields, as it ensures every point on the surface contributes to the total flux out of the enclosed volume, making the Divergence Theorem applicable.

Volume Integral

A **volume integral** involves integrating a function over a three-dimensional region, usually denoted by \( V \), covered by a closed surface. This technique is essential for calculating quantities like mass, charge, or total flux inside a volume, based on a density function or divergence within.
The volume integral, especially in the context of vector fields and the Divergence Theorem, is given by integrating the divergence of a vector field over the volume \( V \) enclosed by a closed surface \( S \). Mathematically, this is expressed as:
\[\int_{V} abla \cdot \mathbf{F} \, dV\]where \( abla \cdot \mathbf{F} \) represents the divergence of the vector field \( \mathbf{F} \).
This integral accounts for the total 'outflowing' property of the vector field within the enclosed space. In our exercise, the value of this volume integral equates to three times the volume \( V \) due to the constant divergence of 3, thus simplifying the equation to \( 3V \), aligning perfectly with the requirements of the Divergence Theorem.