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

Step-by-Step Solution

Answer

The flux is \(\frac{\pi a^5}{32}\).

Step 1: Understand the Problem

We need to find the flux of the vector field \(\mathbf{F} = z\mathbf{k}\) 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 2: Parametrize the Surface

We can parametrize the sphere using spherical coordinates: let \(x = a \sin \theta \cos \phi\), \(y = a \sin \theta \sin \phi\), and \(z = a \cos \theta\). The parameters \(\theta\) and \(\phi\) range from \(0\) to \(\frac{\pi}{2}\) for the sphere in the first octant.

Step 3: Find the Normal Vector

To find the normal vector, we compute the cross product of the partial derivatives of the parametrization with respect to \(\theta\) and \(\phi\):\[\frac{\partial \mathbf{r}}{\partial \theta} = (a \cos \theta \cos \phi, a \cos \theta \sin \phi, -a \sin \theta)\]\[\frac{\partial \mathbf{r}}{\partial \phi} = (-a \sin \theta \sin \phi, a \sin \theta \cos \phi, 0)\]The normal vector \(\mathbf{n}\) is:\[\mathbf{n} = \frac{\partial \mathbf{r}}{\partial \theta} \times \frac{\partial \mathbf{r}}{\partial \phi}\]

Step 4: Compute the Cross Product

Perform the cross product:\[\mathbf{n} = (a \cos \theta \cos \phi, a \cos \theta \sin \phi, -a \sin \theta) \times (-a \sin \theta \sin \phi, a \sin \theta \cos \phi, 0)\]Compute each component:\(\mathbf{i}: (0 - (-a^2 \sin^2 \theta \cos \phi)) = a^2 \sin^2 \theta \cos \phi\)\(\mathbf{j}: (0 - a^2 \sin^2 \theta \sin \phi) = -a^2 \sin^2 \theta \sin \phi\)\(\mathbf{k}: (a^2 \sin \theta \cos \theta \cos^2 \phi + a^2 \sin \theta \cos \theta \sin^2 \phi) = a^2 \sin \theta \cos \theta\)Thus, \(\mathbf{n} = (a^2 \sin^2 \theta \cos \phi, -a^2 \sin^2 \theta \sin \phi, a^2 \sin \theta \cos \theta)\).

Step 5: Evaluate the Flux Integral

Compute the dot product \(\mathbf{F} \cdot \mathbf{n}\):\[ \mathbf{F} \cdot \mathbf{n} = z \cdot a^2 \sin \theta \cos \theta = a^3 \cos^2 \theta \sin \theta\]The surface element is \(d\sigma = |\mathbf{n}| d\theta d\phi = a^2 \sin \theta d\theta d\phi\).The flux integral becomes:\[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \ d\sigma = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} a^3 \cos^2 \theta \sin \theta \cdot a^2 \sin \theta \, d\theta \, d\phi \]

Step 6: Solve the Integral

Calculate the integral:\[ = a^5 \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \cos^2 \theta \sin^2 \theta \, d\theta \, d\phi \]The integral \(\int_{0}^{\frac{\pi}{2}} \cos^2 \theta \sin^2 \theta \, d\theta\) can be evaluated by substitution or using a trigonometric identity. Let \(u = \cos \theta\) then the integral simplifies to \(\frac{1}{8}\). Thus, the flux is:\[ a^5 \cdot \frac{\pi}{4} \cdot \frac{1}{8} \].

Step 7: Final Computation

Compute the final value:\[ \frac{\pi a^5}{32}\].

Step 8: Conclusion

The flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} \, d \sigma\) across the specified portion of the sphere is \(\frac{\pi a^5}{32}\).

Key Concepts

Spherical Coordinates

Spherical coordinates are a system of coordinates used to describe the position of a point in three-dimensional space. It is particularly useful for dealing with spheres and spherical surfaces, as it simplifies the equations and calculations involved. Typically, a point in space is expressed in spherical coordinates by three quantities: the radius \( r \), the polar angle \( \theta \), and the azimuthal angle \( \phi \).

• \( r \): The distance from the origin to the point.
• \( \theta \): The angle between the positive z-axis and the line connecting the origin and the point.
• \( \phi \): The angle in the x-y plane from the positive x-axis.

In our problem, the sphere's surface can be parametrized in spherical coordinates using:
\[x = a \sin \theta \cos \phi, \quad y = a \sin \theta \sin \phi, \quad z = a \cos \theta\]
This representation is very useful for integrating over spherical surfaces because it streamlines the integration limits and expressions.

Parametrization of Surfaces

Parametrization of surfaces involves expressing the surface in terms of parameters, typically denoted as \( u \) and \( v \). This method is essential in vector calculus for transforming three-dimensional surfaces into a form that is more manageable for mathematical operations like integration. When dealing with complex surfaces such as spheres, parametrization allows you to use more suitable coordinate systems like spherical or cylindrical coordinates rather than Cartesian coordinates.

In our exercise, the surface of the sphere is parametrized by the angles \( \theta \) and \( \phi \), which are the variables in spherical coordinates. This is important as it provides a systematic way to describe every point on the sphere using these angles, adhering to:
\[ \mathbf{r} (\theta, \phi) = (a \sin \theta \cos \phi, a \sin \theta \sin \phi, a \cos \theta)\]
Understanding how to parametrize surfaces allows one to find the normal vector to the surface, which is crucial when calculating quantities like flux across a surface.

Vector Calculus

Vector calculus extends calculus to vector fields, supporting essential operations like differentiation and integration of vector-valued functions. A vector field assigns a vector to every point in space and can represent various physical quantities, like force fields or fluid velocity fields. In the context of this problem, we are dealing with the vector field \( \mathbf{F} = z \mathbf{k} \), where \( \mathbf{k} \) is the unit vector along the z-axis.

This particular exercise requires calculating the flux across a surface, where vector calculus concepts like dot products and cross products are invaluable. Specifically, the cross product is used to find a normal vector to a surface defined by a parametric equation.
Once you have the normal vector, the dot product between this vector and the vector field gives a scalar field that can be integrated over the surface to find the flux.

Surface Integral

A surface integral extends the concept of an integral to integration over surfaces, which can be either scalar fields or vector fields. In vector calculus, surface integrals are essential for computing fluid flow across a surface or the electric field through a surface. For the vector field \( \mathbf{F} = z \mathbf{k} \), the integral we are evaluating computes the flux through a specified section of a sphere.

The procedure involves:

• Calculating the dot product \( \mathbf{F} \cdot \mathbf{n} \), where \( \mathbf{n} \) is the normal vector to the surface.
• Finding the differential surface area element \( d\sigma \), a measure of a small area on the surface, often expressed as \( |\mathbf{n}| d\theta d\phi \) in spherical coordinates.

The integral is then performed over the parameter range that describes the surface. In this exercise, the integral results in:
\[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d \sigma = \frac{\pi a^5}{32}\]
The result represents the flux across the specified portion of the sphere.