Problem 22 - Thomas Calculus

Question

Use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\begin{array}{l}{\text { Sphere } \quad \mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} \text { across the sphere } x^{2}+y^{2}+} \\\ {z^{2}=a^{2} \text { in the direction away from the origin }}\end{array}\)

Step-by-Step Solution

Answer

The flux across the sphere is \( 4\pi a^2 \).

Step 1: Understand the Problem and Given Function

We need to find the flux of the vector field \( \mathbf{F} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) across the sphere given by the equation \( x^2 + y^2 + z^2 = a^2 \). The sphere is centered at the origin and has a radius \( a \). The flux is in the direction away from the origin.

Step 2: Parametrize the Surface of the Sphere

To parametrize the sphere, we use spherical coordinates. The parametrization is \( \vec{r}(\theta, \phi) = (a \sin\phi \cos\theta, a \sin\phi \sin\theta, a \cos\phi) \), where \( 0 \leq \theta \leq 2\pi \) and \( 0 \leq \phi \leq \pi \).

Step 3: Calculate the Normal Vector

The normal vector to the parametrized surface, \( \mathbf{n} \), can be found by taking the cross product of the partial derivatives of \( \vec{r} \). Calculate \( \frac{\partial \vec{r}}{\partial \theta} \) and \( \frac{\partial \vec{r}}{\partial \phi} \), and then take their cross product.

Step 4: Find \( \mathbf{F} \cdot \mathbf{n} \)

Substitute \( \vec{r}(\theta, \phi) \) into \( \mathbf{F} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} = a \sin\phi \cos\theta \mathbf{i} + a \sin\phi \sin\theta \mathbf{j} + a \cos\phi \mathbf{k} \). Then compute the dot product \( \mathbf{F} \cdot \mathbf{n} \).

Step 5: Set Up the Integral for Flux

The flux integral is \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma \), where \( d\sigma = || \frac{\partial \vec{r}}{\partial \theta} \times \frac{\partial \vec{r}}{\partial \phi} || \, d\theta \, d\phi \). Substitute \( \mathbf{F} \cdot \mathbf{n} \) and \( d\sigma \) into the integral.

Step 6: Evaluate the Flux Integral

Perform the integration by evaluating the double integral over the range \( 0 \leq \theta \leq 2\pi \) and \( 0 \leq \phi \leq \pi \). The integral should simplify to a result thanks to symmetric properties of sphere and vector field.

Step 7: Interpret the Result

Based on symmetry and the previous calculations, the result of the flux integral should equal \( 4\pi a^2 \). This result confirms that the flux outward on a sphere of radius \( a \) for the given vector field is equal to the surface area of the sphere.

Key Concepts

Parametrization

The concept of parametrization is essential when dealing with complex surfaces, as it provides a way to express a surface using parameters that simplify calculations. In this exercise, we are interested in finding the flux of a vector field across a sphere. The sphere's surface can be described using spherical coordinates. Spherical coordinates are especially useful for parametrizing spheres due to their natural fit with the geometry of a sphere.

By using the angles \( \theta \) and \( \phi \), we create a mapping from a parameter space onto the sphere's surface. The parametrization \( \vec{r}(\theta, \phi) = (a \sin\phi \cos\theta, a \sin\phi \sin\theta, a \cos\phi) \) transforms the two parameters into 3D coordinates that lie on the sphere. This process simplifies integration and other calculations on the sphere's surface.

In a nutshell, parametrization transforms a potentially complex shape into a set of more manageable, mathematically tractable expressions that reflect its geometry.

Spherical Coordinates

Spherical coordinates are a system of coordinates used to define positions in three-dimensional space. They are particularly useful for dealing with objects possessing rotational symmetry, such as spheres, because they efficiently describe points in terms of radius and angles rather than rectangular coordinates.

The setup includes:

• \( r \): the radius from the origin to the point, here it's the radius \( a \) of the sphere.
• \( \theta \): the angle in the xy-plane from the positive x-axis.
• \( \phi \): the polar angle from the positive z-axis.

In our scenario with the sphere, these coordinates help not only in defining the location of points on the sphere but also in determining how the parameters play a role in equations for calculations like finding normal vectors or flux.

By using spherical coordinates, we can efficiently convert problems into integrals that mirror the physical, often simpler symmetries inherent to spherical shapes.

Vector Field Integration

Integrating a vector field to find flux involves calculating how much of the field passes through a given surface. Here, it involves taking the dot product of the vector field with the surface's normal vector across the surface.

The process is typically divided into several steps:

• Firstly, substitute the parametrized coordinates into the vector field.
• Next, calculate the normal vector to the surface through cross products of partial derivatives derived from the parametrization.
• Then, compute the dot product of the vector field with the normal vector.

This dot product represents the component of the vector field passing perpendicularly through the surface.

Finally, it leads to setting up an integral that spans the entire surface area, using differentials derived from the parametrization. For the sphere and our vector field, this integration respects the natural symmetry of the scenario, ultimately leading to results like \( 4\pi a^2 \), representing the total flux through the sphere.