Problem 21 - 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 \(\quad \mathbf{F}=z \mathbf{k}\) across the portion of the sphere \(x^{2}+y^{2}+\) \(z^{2}=a^{2}\) in the first octant in the direction away from the origin

Step-by-Step Solution

Answer

The flux is \( \frac{a^3 \pi}{2} \)."

Step 1: Understand the Problem

We need to calculate the flux of the vector field \( \mathbf{F} = z \mathbf{k} \) across a quarter of a sphere in the first octant. The sphere is defined by \( x^2 + y^2 + z^2 = a^2 \) and the surface normal is pointing away from the origin.

Step 2: Parametrize the Surface

We describe the quarter of the sphere using spherical coordinates. Set \( x = \rho \sin \theta \cos \phi \), \( y = \rho \sin \theta \sin \phi \), \( z = \rho \cos \theta \) with \( \rho = a \), \( 0 \leq \theta \leq \frac{\pi}{2} \), \( 0 \leq \phi \leq \frac{\pi}{2} \). The parameterization of the surface is \( \mathbf{r}(\theta, \phi) = (a \sin \theta \cos \phi, a \sin \theta \sin \phi, a \cos \theta) \).

Step 3: Compute the Normal Vector

The normal vector to the surface is found by taking the cross product of partial derivatives of \( \mathbf{r}(\theta, \phi) \):\[ \mathbf{n} = \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \theta} \ \frac{\partial x}{\partial \phi} & \frac{\partial y}{\partial \phi} & \frac{\partial z}{\partial \phi} \end{vmatrix} \]Calculated, this yields \( \mathbf{n} = a^2 \sin \theta (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) \).

Step 4: Compute the Flux Integral

Substitute the parameterizations into the flux integral \( \iint_S \mathbf{F} \cdot \mathbf{n} \, d\sigma \):\[ \iint_S z \mathbf{k} \cdot \mathbf{n} \, d\sigma = \iint_D a \cos \theta \cdot a^2 \sin^2 \theta \, d\theta \, d\phi \]This simplifies to: \[ a^3 \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \cos \theta \sin^2 \theta \, d\theta \, d\phi \].

Step 5: Integrate and Solve

First, integrate with respect to \( \theta \):\[ \int_0^{\frac{\pi}{2}} \cos \theta \sin^2 \theta \, d\theta = \int_0^{\frac{\pi}{2}} \cos \theta (1 - \cos^2 \theta) \, d\theta = \int_0^{\frac{\pi}{2}} (\cos \theta - \cos^3 \theta) \, d\theta \]Compute this integral: \[ \left[ \sin \theta - \frac{\sin \theta \cos^2 \theta}{3} \right]_0^{\frac{\pi}{2}} = 1 - 0 = 1 \]Then integrate with respect to \( \phi \): \[ \int_0^{\frac{\pi}{2}} 1 \, d\phi = \frac{\pi}{2} \]Finally, multiply the results: \[ a^3 \cdot 1 \cdot \frac{\pi}{2} = \frac{a^3 \pi}{2} \].

Step 6: Final Step: Interpret the Result

The flux of the vector field \( \mathbf{F} = z \mathbf{k} \) across the portion of the sphere in the first octant, with the normal vector pointing away from the origin, is \( \frac{a^3 \pi}{2} \).

Key Concepts

Parametrization

Parametrization is a technique used to represent a surface in terms of two or more parameters. This method allows us to express points on a surface using functions of these parameters. In our case, we are dealing with a portion of a sphere in the first octant. To parametrize the surface, we use angles from spherical coordinates: \( \theta \) and \( \phi \). These angles help specify points on the sphere's surface without directly referencing \( x, y, \) or \( z \) coordinates. By doing so, parametrization simplifies integrating over complex shapes.

Using spherical coordinates, the surface of the sphere can be represented as

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

This transformation is useful because it turns the surface into a parameter space defined by \( \theta \) and \( \phi \). Parametrization is crucial for setting up integrals over surfaces and computing quantities like flux.

Sphere

A sphere is a three-dimensional surface where all points are equidistant from a central point, known as the center. This distance is called the radius. In this exercise, the sphere is defined by the equation \( x^2 + y^2 + z^2 = a^2 \), where \( a \) is the radius. This equation shows that every point on the sphere is exactly \( a \) units away from its center.

For a complete sphere, the parameters \( \theta \) (angle from the positive \( z \)-axis) and \( \phi \) (angle from the positive \( x \)-axis in the \( xy \)-plane) range from 0 to \( \pi \) and 0 to \( 2\pi \), respectively. However, our focus is only on a quarter of the sphere in the first octant. This region is of particular interest because it includes only points where \( x, y, \) and \( z \) are positive, split by the three coordinate planes.

Vector Field

A vector field assigns a vector to each point in space. In our problem, the vector field is given by \( \mathbf{F} = z \mathbf{k} \). This means at every point on the surface, the vector is directed along the \( z \)-axis, scaled by the \( z \)-coordinate.

• This vector field could represent a physical phenomenon, like fluid flow or electromagnetic field, where the strength varies with \( z \).
• The flux through the surface is the total flow of the vector field through the surface.

In our problem, the flux integral \( \iint_S \mathbf{F} \cdot \mathbf{n} \, d\sigma \) measures how much of this flow passes through the surface of the sphere in the specified region, with \( \mathbf{n} \) being the surface normal pointing away from the origin.

Spherical Coordinates

Spherical coordinates are a system to express three-dimensional vectors using three numbers: a radial distance and two angles. This system is particularly useful for dealing with problems involving spheres and circular shapes. The coordinates \( (\rho, \theta, \phi) \) correspond to the distance from the origin, the angle from the positive \( z \)-axis (inclination), and the azimuthal angle in the \( xy \)-plane from the positive \( x \)-axis, respectively.

For our problem:

• \( \rho = a \), as the points lie on a sphere with radius \( a \).
• \( 0 \leq \theta \leq \frac{\pi}{2} \), representing the angle down from the \( z \)-axis, ensuring we stay in the first octant.
• \( 0 \leq \phi \leq \frac{\pi}{2} \), measuring the rotation around the \( z \)-axis, also keeping us in the first octant.

This coordinate system simplifies our surface equations and the integration process, allowing for easier computation of quantities such as the flux through the sphere.