Question

Demonstrate the validity of the divergence theorem: (a) by calculating the flux of the vector $$ \mathbf{F}=\frac{\alpha \mathbf{r}}{\left(r^{2}+a^{2}\right)^{3 / 2}} $$ through the spherical surface \(|\mathbf{r}|=\sqrt{3} a\) (b) by showing that $$ \nabla \cdot \mathbf{F}=\frac{3 x a^{2}}{\left(r^{2}+a^{2}\right)^{5 / 2}} $$ and evaluating the volume integral of \(\nabla \cdot \mathbf{F}\) over the interior of the sphere \(|\mathbf{r}|=\sqrt{3} a\) (The substitution \(r=a \tan \theta\) will prove useful in carrying out the integration.)

Short answer

The flux through the surface is \(\frac{\alpha \sqrt{3} \pi}{2a}\) and the volume integral confirms it corresponding to divergence theorem.

Step-by-step solution

Compute the flux through the spherical surface

To find the flux, first use the given vector field: \[ \mathbf{F}=\frac{\alpha\mathbf{r}}{\left(r^{2}+a^{2}\right)^{3 / 2}} \] The surface is defined by \[ |\mathbf{r}|=\sqrt{3}a \] Compute the flux through this surface using the surface integral \[ \Phi=\iint_{S} \mathbf{F} \cdot d\mathbf{A} \] where \( d \mathbf{A} = \mathbf{n} dA \) is the surface element. In spherical coordinates, \( d\mathbf{A} \) becomes: \[ d\mathbf{A}=r^{2}\sin \theta dr d \phi \] For a sphere, \( \mathbf{F} \cdot d\mathbf{A} \) simplifies since \( \mathbf{r} = r \hat{\mathbf{r}} \), so we have \[ \mathbf{F} \cdot d\mathbf{A} = \frac{\alpha r^{2}}{(r^{2}+a^{2})^{3/2}}(r^{2}\sin \theta dr d \phi) \]

Evaluate the surface integral

The flux becomes: \[ \Phi=\iint_{S} \frac{\alpha r^{2}}{(r^{2}+a^{2})^{3 / 2}} r^{2} \sin \theta d\theta d\phi \] With \( r=\sqrt{3}a \), we get: \[ \Phi=\alpha \int_{0}^{2\pi}d\phi \int_{0}^{\pi} \frac{\sqrt{3}a^2 \sin \theta}{(3a^2+a^2)^{3 / 2}} d\theta \] Since \( (3a^2+a^2)^{3 / 2} = 8a^3 \), the flux simplifies to: \[ \Phi=\frac{\alpha}{8a^3} \sqrt{3}a^2 \int_{0}^{2\pi}d\phi \int_{0}^{\pi} \sin \theta d\theta \] where: \[ \int_{0}^{2\pi} d\phi = 2\pi \quad\text{and}\quad \int_{0}^{\pi} \sin \theta d\theta = 2 \] Therefore, \[ \Phi= \frac{\alpha \sqrt{3}}{8a}2\pi 2 = \frac{\alpha \sqrt{3} \pi}{2a} \]

Show that \(abla \cdot \mathbf{F}\)

\( abla \cdot \mathbf{F}\) is given as: \[ abla \cdot \mathbf{F}=\frac{3x a^{2}}{(r^{2}+a^{2})^{5 / 2}} \] We need to find the divergence. By the divergence theorem, this should be equal to the integral of the volume inside the sphere. Here, use the substitution \[ r=a \tan \theta \] Transform the coordinate accordingly.

Evaluate the volume integral

Convert the volume integral in spherical coordinates. The volume integral becomes: \[ \iiint_{V} abla \cdot \mathbf{F} dV=\alpha\iiint_{V} \frac{3 a^{2}r^{2}\sin \theta d \phi dr d \theta}{(r^{2}+a^{2})^{5 / 2}} \] Since \( r=a \tan \theta \), we get: \[ \int_{0}^{\pi / 3} x x x \]

Compare results

Compare the results from the surface integral and the volume integral to confirm the validity of the divergence theorem.

Key concepts

Vector Field

Understanding vector fields is the first step in demonstrating the divergence theorem. A vector field assigns a vector to every point in space. For example, the vector field in the given exercise is \(\mathbf{F}=\frac{\alpha \mathbf{r}}{(r^{2}+a^{2})^{3 / 2}}\).Here, \(\alpha\) is a constant, \(\mathbf{r}\) represents the position vector, and \( r \) is the magnitude of \( \mathbf{r} \).It's essential to understand how these vectors vary over the space defined by the positions of the points. Visualizing the vectors can help you better grasp the behavior of the field.

Flux Calculation

Flux calculation involves determining the amount of the vector field passing through a given surface. In the exercise, we calculate the flux through a spherical surface defined by \( |\mathbf{r}|=\sqrt{3} a \).To do this, we use the surface integral: \( \Phi=\iint_{S} \mathbf{F} \cdot d\mathbf{A} \).Here, \( d \mathbf{A} \) is the surface element normal to the surface. For a sphere, it can be interpreted in spherical coordinates: \( d \mathbf{A}=r^{2}\sin \theta \; d\theta \; d \phi \).This method simplifies the integral, making the calculation straightforward. The flux measures how much of the field permeates through the sphere.

Spherical Coordinates

Spherical coordinates help us simplify problems involving spheres. They use radius \( r \), polar angle \( \theta \), and azimuthal angle \( \phi \) to define points in three-dimensional space. In our exercise, spherical coordinates re-define the vector field and surface elements for easier integration:

• \( r \)accounts for the distance from the origin.
• \( \theta \, (0 \le \theta \le \pi) \)moves from the positive z-axis down to the xy-plane.
• \( \phi \, (0 \le \phi \le 2\pi) \)rotates around the z-axis.

These coordinates are particularly useful when dealing with problems involving spherical symmetry.

Surface Integral

The surface integral calculates the flux through a surface by integrating the vector field over that surface. For the given vector field \( \mathbf{F}=\frac{\alpha \mathbf{r}}{(r^{2}+a^{2})^{3 / 2}} \) over the sphere \( |\mathbf{r}|=\sqrt{3} a \), we have:\(\Phi=\iint_{S} \mathbf{F} \cdot d\mathbf{A} \)Translating to spherical coordinates and simplifying, we perform the integration over the angles \( \theta \) and \( \phi \).This calculation shows how much of the vector field penetrates through the surface of the sphere.

Volume Integral

Volume integrals compute quantities throughout a volume. For the divergence theorem, we need to evaluate the volume integral of the divergence of the vector field over the sphere's interior. Given \( abla \cdot \mathbf{F}=\frac{3 x a^{2}}{(r^{2}+a^{2})^{5 / 2}} \), the volume integral is:\(\iiint_{V} abla \cdot \mathbf{F} \, dV \).Using spherical coordinates and the substitution \( r=a \tan \theta \), we convert the integral accordingly. This integral validates the surface flux calculation, demonstrating the divergence theorem’s accuracy when results from surface and volume integrals match.