Problem 16 - Calculus Early Transcendentals

Question

Find the flux of the vector field \(\mathbf{F}\) across \(\sigma\) in the direction of positive orientation. \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} ; \sigma\) is the portion of the sphere $$ \mathbf{r}(u, v)=2 \sin u \cos v \mathbf{i}+2 \sin u \sin v \mathbf{j}+2 \cos u \mathbf{k} $$ with \(0 \leq u \leq \pi / 3,0 \leq v \leq 2 \pi\)

Step-by-Step Solution

Answer

The flux of the vector field across the surface is \( \frac{64\pi\sqrt{3}}{3} \).

Step 1: Identify the vector field and surface

The given vector field is \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). The surface \( \sigma \) is part of a sphere parameterized by \( \mathbf{r}(u, v) = 2 \sin u \cos v \mathbf{i} + 2 \sin u \sin v \mathbf{j} + 2 \cos u \mathbf{k} \), where \( 0 \leq u \leq \pi / 3 \) and \( 0 \leq v \leq 2\pi \).

Step 2: Find the normal vector

Compute the normal vector to the surface \( \sigma \) by finding the cross product of the partial derivatives of \( \mathbf{r}(u, v) \). Let \( \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} \) and \( \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} \). Then compute \( \mathbf{r}_u \times \mathbf{r}_v \).

Step 3: Compute partial derivatives

Calculate \( \mathbf{r}_u = \frac{\partial \mathbf{r}}{\partial u} = 2 \cos u \cos v \mathbf{i} + 2 \cos u \sin v \mathbf{j} - 2 \sin u \mathbf{k} \) and \( \mathbf{r}_v = \frac{\partial \mathbf{r}}{\partial v} = -2 \sin u \sin v \mathbf{i} + 2 \sin u \cos v \mathbf{j} \).

Step 4: Evaluate the cross product

Evaluate \( \mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 \cos u \cos v & 2 \cos u \sin v & -2 \sin u \ -2 \sin u \sin v & 2 \sin u \cos v & 0 \end{vmatrix} \), and find the components of the resulting vector.

Step 5: Convert to a simple expression

After evaluating the determinant, the cross product \( \mathbf{r}_u \times \mathbf{r}_v = 4 \sin^2 u \cos v \mathbf{i} + 4 \sin^2 u \sin v \mathbf{j} + 4 \sin u \cos u \mathbf{k} \). This normal vector is already outward, given the orientation of the surface.

Step 6: Setup the flux integral

The flux \( \Phi \) through the surface \( \sigma \) is given by \( \Phi = \int\int_\sigma \mathbf{F} \cdot d\mathbf{S} \), where \( d\mathbf{S} = (\mathbf{r}_u \times \mathbf{r}_v) \, du \, dv \). Compute \( \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) \).

Step 7: Compute \( \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) \)

Given \( \mathbf{F}(x, y, z) = 2 \sin u \cos v \mathbf{i} + 2 \sin u \sin v \mathbf{j} + 2 \cos u \mathbf{k} \), substitute into the dot product with the normal vector and simplify: \( \mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v) = 16 \sin^3 u + 8 \sin u \cos u \cos u \).

Step 8: Integrate over the parameter domain

Compute the integral \( \Phi = \int_{0}^{2\pi} \int_{0}^{\pi/3} (16 \sin^3 u + 8 \sin u \cos^2 u) \, du \, dv \). Integrate with respect to \( u \) first, followed by \( v \).

Step 9: Calculate the antiderivatives and evaluate

Integrate the function \( 16 \sin^3 u + 8 \sin u \cos^2 u \) with respect to \( u \), then substitute the bounds \( 0 \) and \( \pi/3 \). The antiderivative involves the use of substitution and trigonometric identities.

Step 10: Complete the double integral with respect to \( v \)

Once you have the value from the \( u \) integration, multiply by the simple range for \( v \), which is \( \int_{0}^{2\pi} dv = 2\pi \).

Step 11: Final flux computation

Combine the results from the earlier integrations to find the total flux across the surface. Compute: \( \Phi = 2\pi \left( \frac{32\sqrt{3}}{3} \right) \), which simplifies to \( \frac{64\pi\sqrt{3}}{3} \).

Key Concepts

Surface Parameterization

In vector calculus, surface parameterization is an essential technique used to describe surfaces in space. By representing a surface using a parameterization, we transform it into functions of two variables, typically denoted by \(u\) and \(v\). This is useful when dealing with complex surfaces, as it translates a 3D problem into a more manageable 2D problem.
For example, consider a portion of a sphere parameterized by the equations:

• \(x = 2 \sin u \cos v\)
• \(y = 2 \sin u \sin v\)
• \(z = 2 \cos u\)

These equations map a grid of points in the \(uv\)-plane to points on the sphere. The parameters \(u\) and \(v\) range over specified intervals, here \(0 \leq u \leq \pi/3\) and \(0 \leq v \leq 2\pi\). This means \(u\) influences the latitude, and \(v\) affects the longitude.By using surface parameterization, we simplify integrating over complex shapes and facilitate the computation of physical properties such as area and flux.

Cross Product

The cross product is a mathematical operation that finds a vector perpendicular to two other vectors. This operation is particularly helpful in physics and engineering for finding normal vectors to surfaces.
Given a parameterized surface \(\mathbf{r}(u, v)\), we start by computing its partial derivatives: \(\mathbf{r}_u\) and \(\mathbf{r}_v\). These vectors represent tangent vectors to the surface at each point. For our sphere,

• \(\mathbf{r}_u = 2 \cos u \cos v \mathbf{i} + 2 \cos u \sin v \mathbf{j} - 2 \sin u \mathbf{k}\).
• \(\mathbf{r}_v = -2 \sin u \sin v \mathbf{i} + 2 \sin u \cos v \mathbf{j}\).

To find a normal vector, we compute the cross product: \(\mathbf{r}_u \times \mathbf{r}_v\). The result is another vector perpendicular to the surface, providing a tool to describe its orientation.The determinant method assists in the cross product calculation, involving vector components. The resultant normal vector might come with scaffolded components like \(4 \sin^2 u \cos v \mathbf{i} + 4 \sin^2 u \sin v \mathbf{j} + 4 \sin u \cos u \mathbf{k}\), offering a route to evaluate the flux across a surface.

Double Integral

In calculus, the double integral is an extension of the concept of an integral to functions of two variables. This concept is often used when determining the net flow, or flux, through a surface in a vector field.
Integrating a vector field over a surface typically involves evaluating the double integral of the dot product \(\mathbf{F} \cdot d\mathbf{S}\). Here, the differential area \(d\mathbf{S}\) is expressed using the cross product of the parameterization: \(\mathbf{r}_u \times \mathbf{r}_v\) as a small patch on the surface.
Once you have \(\mathbf{F} \cdot (\mathbf{r}_u \times \mathbf{r}_v)\), the integral becomes straightforward, separated into two parts: integration over \(u\) first, then \(v\):

• \(\int_{0}^{2\pi} \int_{0}^{\pi/3} (16 \sin^3 u + 8 \sin u \cos^2 u) \, du \, dv\).

The inner integral with \(u\) usually accounts for more complexity, as seen in trigonometric expressions like \(\sin^3 u\) and \(\cos^2 u\). These need decomposition into simpler integral forms or identities, making way for thorough evaluation.

Antiderivative Calculation

The concept of finding an antiderivative is crucial in evaluating definite integrals, which appears in the assessment of a vector field's flux via double integrals. Here, antiderivatives help quantify areas under curves within their respective function range.
The integration process starts by computing the antiderivative of an expression like \(16 \sin^3 u + 8 \sin u \cos^2 u\). This often calls for using substitution methods or trigonometric identities. For instance, applying identities such as \(\cos^2 u = 1 - \sin^2 u\) can simplify terms for easier integration.
Once you find the antiderivative with respect to \(u\), substitute the upper and lower bounds from the parameter limits, \(0\) and \(\pi/3\). This continuous process simplifies the expression to reveal its contribution over the prescribed domain.
Finally, once the \(u\) integral completes, the resulting expression is multiplied by the integral with respect to \(v\), usually straightforward, spanning \(0\) to \(2\pi\). This process, collectively, computes the flux over the defined surface segment in the vector field, represented here neatly as \(\frac{64\pi\sqrt{3}}{3}\).