Problem 29 - Calculus Early Transcendentals

Question

Let \(\mathbf{F}=\|\mathbf{r}\|^{k} \mathbf{r},\) where \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) and \(k\) is a constant. (Note that if \(k=-3,\) this is an inverse-square field.) Let \(\sigma\) be the sphere of radius \(a\) centered at the origin and oriented by the outward normal \(\mathbf{n}=\mathbf{r} /\|\mathbf{r}\|=\mathbf{r} / a\). (a) Find the flux of \(\mathbf{F}\) across \(\sigma\) without performing any integrations. I Hint: The surface area of a sphere of radius \(\left.a \text { is } 4 \pi a^{2} .\right]\) (b) For what value of \(k\) is the flux independent of the radius of the sphere?

Step-by-Step Solution

Answer

(a) 4\pi if k = -3. (b) k = -3.

Step 1: Understanding the Vector Field

The vector field is given by \( \mathbf{F} = \| \mathbf{r} \|^k \mathbf{r} \), where \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). In this context, \( \| \mathbf{r} \| = \sqrt{x^2 + y^2 + z^2} \) represents the magnitude of the vector \( \mathbf{r} \).

Step 2: Relating the Field and Surface

The sphere \( \sigma \) of radius \( a \) has an outward normal \( \mathbf{n} = \mathbf{r} / \| \mathbf{r} \| = \mathbf{r} / a \). Hence, the unit outward normal vector is proportional to \( \mathbf{r} \). The flux of \( \mathbf{F} \) through the surface \( \sigma \) can be deduced by considering how \( \mathbf{F} \) and \( \mathbf{n} \) are related.

Step 3: Applying the Divergence Theorem

According to the Divergence Theorem, the flux of a field \( \mathbf{F} \) through a closed surface \( \sigma \) is equal to the volume integral of \( abla \cdot \mathbf{F} \) over the region inside \( \sigma \). First, find \( abla \cdot \mathbf{F} \).

Step 4: Calculating the Divergence

Compute \( abla \cdot \mathbf{F} \) where \( \mathbf{F} = \| \mathbf{r} \|^k \mathbf{r} \). Begin with the observation that \( abla \cdot (f \cdot \mathbf{r}) = abla f \cdot \mathbf{r} + f abla \cdot \mathbf{r} \). Here \( abla \cdot \mathbf{r} = 3 \), and \( f = \| \mathbf{r} \|^k \), so \( abla f = k \| \mathbf{r} \|^{k-2} \mathbf{r} \). Substitute these results to get \( abla \cdot \mathbf{F} = k \| \mathbf{r} \|^{k-2} \| \mathbf{r} \|^2 + 3 \| \mathbf{r} \|^k = (k+3) \| \mathbf{r} \|^k \).

Step 5: Ensuring Independence from Radius

For the flux to be independent of \( a \), the divergence \( abla \cdot \mathbf{F} \) must give a simple scale by volume. Thus, require \( (k+3) = 0 \), leading to \( k = -3 \).

Step 6: Evaluate the Flux

The flux through the sphere \( \sigma \) is computed by \( \int_\sigma \mathbf{F} \cdot \mathbf{n} \, dA = 4 \pi a^2 \mathbf{F} \cdot \mathbf{n} \). Substituting \( \mathbf{F} = \| \mathbf{r} \|^{-3} \mathbf{r} = a^{-3} a \mathbf{r} = a^{-2} \mathbf{r} \), the flux becomes \( 4\pi a^2 a^{-2} a^{-1} a^2 = 4\pi \), constant for \( k = -3 \).

Step 7: Concluding Independence

Thus, if \( k = -3 \), the flux is \( 4\pi \), which is indeed independent of radius \( a \).

Key Concepts

Vector Field

Vector fields are an integral part of understanding how multidimensional systems behave. They are a way to assign a vector to every point in a space, effectively describing how a particular quantity, like a force or a velocity, varies in that space. In our context, the vector field is given as \( \mathbf{F} = \| \mathbf{r} \|^k \mathbf{r} \), where \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). This expression tells us that the field intensities depend on both the position in space and the parameter \( k \).

Understanding the basics of a vector field helps in visualizing how the vector quantity (e.g., force) at each point can be described by a combination of the field's direction and magnitude. Here, \( \| \mathbf{r} \| \) represents the magnitude of the initial vector \( \mathbf{r} \), which scales the direction given by \( \mathbf{r} \) itself. This setup is crucial when analyzing how the vector field interacts with a surface like a sphere.

Divergence Theorem

The Divergence Theorem is a powerful tool that helps relate the flux of a vector field through a closed surface to the divergence of that field over the volume contained by the surface. The theorem states that the total flux through a closed surface \( \sigma \) (the sphere in our case) is equal to the integral of the divergence of \( \mathbf{F} \) over the volume inside \( \sigma \). Mathematically, this is expressed as:

• \( \int_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dA = \int_V abla \cdot \mathbf{F} \, dV \)

In our scenario, we apply this theorem to avoid performing complex surface integrals by instead calculating the divergence \( abla \cdot \mathbf{F} \). For the vector field \( \mathbf{F} = \| \mathbf{r} \|^k \mathbf{r} \), computing divergence involves using formulas for the divergence of a scalar field. This way, we can confirm if the flux remains constant across different sphere radii by checking for which \( k \) the divergence scales correctly.

Sphere Surface Area

Understanding the geometric properties of spheres is crucial when dealing with problems related to flux and vector fields. The sphere's surface area is given by the formula \( 4 \pi a^2 \), where \( a \) is the radius. This formula is pivotal in computing the flux of a vector field through the sphere, as it directly determines the extent of the area across which the field lines intersect.

In the given exercise, knowing the sphere's surface area allows us to calculate the flux through it without performing complex integrations. The flux \( \int_{\sigma} \mathbf{F} \cdot \mathbf{n} \, dA \) simplifies when using symmetry and given results, reducing the complexity of the problem and providing insights into how vector fields interact with surfaces. The hybrid use of known geometric formulas with vector calculus avenues like the Divergence Theorem makes solving such problems manageable.

Inverse-Square Field

An inverse-square field is a specific type of vector field where the field's strength decreases with the square of the distance from the source. This concept often appears in physics, notably in the study of gravitational and electrostatic fields. In terms of our vector field \( \mathbf{F} = \| \mathbf{r} \|^k \mathbf{r} \), the inverse-square field is realized when \( k = -3 \).

This setup means the magnitude of the field at any point is inversely proportional to the square of the distance from the origin: \( \| \mathbf{r} \|^{-3} \). This formulation reflects a natural scaling where the influence of the field diminishes rapidly as one moves away from the source. In the exercise at hand, it is shown that for \( k = -3 \), the flux becomes independent of the sphere's radius, providing a classical example of how inverse-square laws manifest in mathematical formulations and align with physical observations.