Problem 22 - Thomas Calculus

Question

In Exercises \(19-28,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. Sphere \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) across the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) in the direction away from the origin

Step-by-Step Solution

Answer

The flux is \( 4\pi a^{3} \).

Step 1: Understanding the Problem

We need to find the flux of the vector field \( \mathbf{F} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) across the sphere \( x^{2} + y^{2} + z^{2} = a^{2} \), specifically in the direction away from the origin.

Step 2: Parameterize the Sphere

The sphere \( x^{2} + y^{2} + z^{2} = a^{2} \) can be parameterized using spherical coordinates: \( x = a \sin \theta \cos \phi \), \( y = a \sin \theta \sin \phi \), and \( z = a \cos \theta \), where \( 0 \leq \phi < 2\pi \) and \( 0 \leq \theta \leq \pi \).

Step 3: Compute the Normal Vector

The normal vector \( \mathbf{n} \) that points outward from the sphere can be derived by taking the cross product of the tangent vectors derived from the parameterization. The outward unit normal will be \( \mathbf{n} = \frac{\mathbf{r}_\theta \times \mathbf{r}_\phi}{\| \mathbf{r}_\theta \times \mathbf{r}_\phi \|} \), and for a unit sphere, this simplifies to \( \mathbf{n} = \frac{1}{a} (x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \).

Step 4: Compute the Flux Integral

The flux integral is \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma = \iint_{S} \mathbf{F} \cdot \mathbf{n} \, a^{2} \sin \theta \, d\theta \, d\phi \). Since \( \mathbf{F} \cdot \mathbf{n} = \frac{x^{2} + y^{2} + z^{2}}{a} = \frac{a^{2}}{a} = a \), this simplifies to \( a \iint_{S} a^{2} \sin \theta \, d\theta \, d\phi = a^{3} \iint_{S} \sin \theta \, d\theta \, d\phi \).

Step 5: Evaluate the Integral

Evaluate the integral: \( a^{3} \int_{0}^{2\pi} \int_{0}^{\pi} \sin \theta \, d\theta \, d\phi \). The inner integral \( \int_{0}^{\pi} \sin \theta \, d\theta \) evaluates to 2. Thus, the flux is \( a^{3} \cdot 2 \int_{0}^{2\pi} 1 \, d\phi = a^{3} \cdot 2 \cdot 2\pi = 4\pi a^{3} \).

Key Concepts

Parameterization of a Sphere

To begin solving problems involving flux integrals across surfaces like a sphere, we first need to understand parameterization. Parameterization involves expressing the surface in terms of two parameters, usually denoted as \( \theta \) and \( \phi \) in spherical coordinates.
This helps in transforming a three-dimensional surface into a more manageable form. For the sphere defined by \( x^2 + y^2 + z^2 = a^2 \), parameterization in spherical coordinates is used. This is expressed as:

• \( x = a \sin \theta \cos \phi \)
• \( y = a \sin \theta \sin \phi \)
• \( z = a \cos \theta \)

Here, \( \theta \) is the polar angle, ranging from \( 0 \) to \( \pi \), and \( \phi \) is the azimuthal angle, ranging from \( 0 \) to \( 2\pi \). This transformation simplifies the calculations involved in evaluating flux and other surface integrals.

Spherical Coordinates

Spherical coordinates are an essential tool when dealing with three-dimensional problems like those involving spheres. Unlike Cartesian coordinates, spherical coordinates use a radius and two angles to describe a point in space.
In this system, a point is represented by three values: the radius \( r \), the polar angle \( \theta \), and the azimuthal angle \( \phi \).

• \( r \) is the distance from the origin to the point, which equals the radius \( a \) for points on the surface of the sphere.
• \( \theta \) is the angle from the positive z-axis to the point.
• \( \phi \) is the angle from the positive x-axis in the x-y plane.

Using these coordinates simplifies many calculations, especially when calculating integrals over spherical surfaces, as it aligns naturally with the symmetry of the problem.

Normal Vector

The normal vector is crucial in computing surface integrals, including the flux integral. In this context, we need a vector that points away from the surface and is perpendicular to it.
To find this for the parameterized sphere, we first calculate the tangent vectors through partial derivatives with respect to \( \theta \) and \( \phi \).
The normal vector \( \mathbf{n} \) is found using the cross product of these tangent vectors. This gives us a vector perpendicular to the surface:
\[\mathbf{n} = \frac{\mathbf{r}_\theta \times \mathbf{r}_\phi}{\|\mathbf{r}_\theta \times \mathbf{r}_\phi\|} \]
For a unit sphere with radius \( a \), this formula simplifies to:
\[\mathbf{n} = \frac{1}{a}(x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \]
The normal vector is scaled to have a magnitude of one, ensuring its direction is outward.

Flux Calculation

Flux calculation involves finding how much of a field is passing through a surface. For our problem, it means integrating the dot product of the vector field \( \mathbf{F} \) and the outward-pointing normal vector \( \mathbf{n} \).
The flux integral is given by:
\[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma \]
In the case of a unit sphere, where \( \mathbf{F} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) and \( \mathbf{n} = \frac{1}{a}(x \mathbf{i} + y \mathbf{j} + z \mathbf{k}) \), we find \( \mathbf{F} \cdot \mathbf{n} = a \).
Thus, the flux through the sphere simplifies to:
\[ a \iint_{S} a^2 \sin \theta \, d\theta \, d\phi = a^3 \int_{0}^{2\pi} \int_{0}^{\pi} \sin \theta \, d\theta \, d\phi \]
Evaluating this integral yields the final result: \( 4\pi a^3 \). This gives the total measure of the field passing through the spherical surface.