Problem 9 - Advanced Engineering Mathematics

Question

Use the divergence theorem to find the outward flux \(\iint_{S}(\mathbf{F} \cdot \mathbf{n}) d S\) of the given vector field \(\mathbf{F}\). $$ \begin{aligned} &\mathbf{F}=(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) /\left(x^{2}+y^{2}+z^{2}\right) ; D \text { the region bounded by }\\\ &\text { the concentric spheres } x^{2}+y^{2}+z^{2}=a^{2}, x^{2}+y^{2}+z^{2}=b^{2}\\\ &b>a \end{aligned} $$

Step-by-Step Solution

Answer

The outward flux is zero.

Step 1: Recall the Divergence Theorem

The Divergence Theorem states that for a vector field \( \mathbf{F} \) and a closed surface \( S \) enclosing a region \( D \), \( \iint_{S}(\mathbf{F} \cdot \mathbf{n}) \, dS = \iiint_{D} (abla \cdot \mathbf{F}) \, dV \). Here, \( \mathbf{n} \) is the outward normal vector to \( S \).

Step 2: 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 our vector field \( \mathbf{F} = \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{x^2 + y^2 + z^2} \), calculate each partial derivative to find \( abla \cdot \mathbf{F} \).

Step 3: Simplify the Divergence

Upon simplifying, you find \( abla \cdot \mathbf{F} = 0 \). This is because each term contributes to a total divergence that sums to zero. Specifically, the divergence involves terms that cancel out due to symmetry and the structure of the vector field.

Step 4: Apply the Divergence Theorem

Since \( abla \cdot \mathbf{F} = 0 \), the integral \( \iiint_{D} (abla \cdot \mathbf{F}) \, dV = 0 \). According to the Divergence Theorem, this means the surface integral \( \iint_{S}(\mathbf{F} \cdot \mathbf{n}) \, dS \) also equals zero.

Key Concepts

Vector Field

A vector field is a function that assigns a vector to every point in a subset of space. This can be visualized as a field of arrows, where each arrow has both a direction and a magnitude. In the given exercise, the vector field \( \mathbf{F} \) is defined as \( (x \mathbf{i} + y \mathbf{j} + z \mathbf{k})/(x^2 + y^2 + z^2) \). This type of vector field is sometimes referred to as a radial field, as its direction always points radially outward from the origin.

• Each component of the vector field \( \mathbf{F} \) is dependent on \( x, y, \) and \( z \), highlighting its three-dimensional nature.
• The denominator \( x^2 + y^2 + z^2 \) normalizes the vector, affecting its magnitude without changing its direction.
• Understanding how to calculate the divergence of this vector field is key in using the Divergence Theorem.

The divergence of a vector field gives us a measure of how much the field spreads out or converges at a point. For this vector field, symmetry in the components results in a cancellation that often leads to a zero divergence.

Outward Flux

Outward flux is the flow exiting through a surface. In the context of a vector field, it measures how much \( \mathbf{F} \) flows or passes through a closed surface \( S \). In mathematical terms, this is represented by the surface integral \( \iint_{S}(\mathbf{F} \cdot \mathbf{n}) \, dS \). Here, \( \mathbf{n} \) is the unit normal vector pointing outward from the surface.

• The dot product \( \mathbf{F} \cdot \mathbf{n} \) determines the portion of the vector field flowing perpendicular to the surface.
• If the surface integral results in zero, it implies that the amount of flow entering the surface is equal to the amount exiting, leading to no net flow.
• Using the Divergence Theorem, the integral simplifies to \( \iiint_{D} \, (abla \cdot \mathbf{F}) \, dV \), turning a potentially complex surface computation into a volume calculation.

For the given exercise, the net outward flux is zero due to the divergence of the vector field being zero, demonstrating no net outflow through any portion of the bounding surfaces.

Spherical Coordinates

Spherical coordinates are a system of coordinates that extend polar coordinates into three dimensions using radius, polar angle, and azimuthal angle to locate a point. They can be quite useful when working with spheres or circular symmetry. In contrast to Cartesian coordinates \( (x, y, z) \), spherical coordinates \( (\rho, \theta, \phi) \) describe any point in space as:

• \( \rho \): the radial distance from the origin to the point.
• \( \theta \): the angle from the positive x-axis (in the x-y plane).
• \( \phi \): the angle from the positive z-axis (often called the polar angle).

When solving problems involving spherical shapes, such as the concentric spheres in this exercise, spherical coordinates simplify the integration process.

- The transformation between Cartesian and spherical coordinates is crucial to set up integration, especially in these scenarios.- This setup allows you to exploit the symmetry present in spherical problems, often simplifying the mathematics tremendously.- In the exercise, the region \( D \) is bounded by two spheres described by \( x^2 + y^2 + z^2 = a^2 \) and \( x^2 + y^2 + z^2 = b^2 \), which translate directly into constant \( \rho \) surfaces in spherical coordinates.