Problem 38 - Thomas Calculus

Question

The flow of a gas with a density of \(\delta=0.3 \mathrm{kg} / \mathrm{m}^{2}\) over the closed curve \(\mathbf{r}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}, 0 \leq t \leq 2 \pi,\) is given by the vector field \(\mathbf{F}=\delta \mathbf{v},\) where \(\mathbf{v}=x^{2} \mathbf{i}-y \mathbf{j}\) is a velocity field measured in meters per second. Find the flux of \(\mathbf{F}\) across the curve \(\mathbf{r}(t) .\)

Step-by-Step Solution

Answer

The flux of \( \mathbf{F} \) across the curve is zero.

Step 1: Understand the Concept of Flux

Flux is a measure of the flow of a field through a surface or a curve. In this problem, we need to find the flux of the vector field \( \mathbf{F} = \delta \mathbf{v} \) across the closed curve defined by \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \). The curve represents a circle parameterized by \( t \) from \( 0 \) to \( 2\pi \).

Step 2: Setup of the Flux Integral

The flux of the vector field \( \mathbf{F} \) across a curve \( C \) is given by the line integral \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds \), where \( \mathbf{n} \) is the unit normal vector to the curve \( C \) and \( ds \) is the differential arc length.

Step 3: Parameterize the Unit Circle

The curve \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \) is a parameterization of the unit circle centered at the origin. The differential arc length \( ds \) can be expressed as \( ds = |\mathbf{r}'(t)| \, dt \). Compute \( \mathbf{r}'(t) = (-\sin t) \mathbf{i} + (\cos t) \mathbf{j} \), so \(|\mathbf{r}'(t)| = \sqrt{(-\sin t)^2 + (\cos t)^2} = 1\). Thus, \( ds = dt \).

Step 4: Compute the Velocity Vector Field \( \mathbf{v} \)

The velocity field is given as \( \mathbf{v} = x^2 \mathbf{i} - y \mathbf{j} \). Substitute \( x = \cos t \) and \( y = \sin t \) from the parameterization, obtaining \( \mathbf{v} = (\cos^2 t) \mathbf{i} - (\sin t) \mathbf{j} \).

Step 5: Calculate the Vector Field \( \mathbf{F} \)

\( \mathbf{F} = \delta \mathbf{v} \) with \( \delta = 0.3 \). Substituting \( \mathbf{v} \) from Step 4, we get \( \mathbf{F} = 0.3(\cos^2 t \mathbf{i} - \sin t \mathbf{j}) \).

Step 6: Find the Outward Unit Normal \( \mathbf{n} \)

The outward unit normal for a counterclockwise oriented curve \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \) can be \( \mathbf{n} = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \), the same as the parameterization because it is a unit circle.

Step 7: Compute the Dot Product \( \mathbf{F} \cdot \mathbf{n} \)

Compute \( \mathbf{F} \cdot \mathbf{n} = 0.3 (\cos^2 t \mathbf{i} - \sin t \mathbf{j}) \cdot (\cos t \mathbf{i} + \sin t \mathbf{j}) = 0.3 (\cos^3 t - \sin^2 t \cos t) \).

Step 8: Evaluate the Line Integral for Flux

The flux \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds \) becomes \( \int_0^{2\pi} 0.3 (\cos^3 t - \sin^2 t \cos t) \, dt \). Simplify and calculate this integral. Note that due to symmetry and the nature of trigonometric integrals over a full period, the evaluation yields zero due to the periodic nature of sine and cosine being symmetric about zero.

Key Concepts

Vector Calculus

Vector calculus is an essential area of mathematics that deals with vector fields and operations on them. It provides tools to analyze and understand geometrical and physical properties in multidimensional space.
In this problem, we focus on the vector field \( \mathbf{F} = \delta \mathbf{v} \), where \( \delta \) is a constant density, and \( \mathbf{v} \) is a velocity field. A vector field is a function that assigns a vector to each point in space.

• In our scenario, imagine each point in space having a small arrow that represents the velocity of gas at that location.
• Understanding the flow of these vectors through a curve helps us understand the gas flow across a boundary.

Vector calculus uses rules and operations such as differentiation and integration in multiple dimensions.
This branch of calculus allows us to describe how vector fields like velocity change and interact in space, ultimately helping us calculate things like flux, which is crucial in fluid dynamics and electromagnetism.

Line Integrals

Line integrals are a fundamental concept in vector calculus, used to measure the cumulative effect of a vector field along a curve.
In the context of calculating flux like in our example, a line integral measures the total flow through a curve.
This is represented by the integral \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds \), where \( \mathbf{n} \) is the unit normal vector and \( ds \) represents the arc length.

• This integral calculates the sum of vector field contributions along every differential part of a curve \( C \).
• The dot product \( \mathbf{F} \cdot \mathbf{n} \) determines how much the vector field penetrates through the curve, resembling a projection.

Line integrals help us to quantify how vector fields like electric, magnetic, or fluid fields pass through a given path in space.
Evaluating them requires careful consideration of how the field moves and interacts with the geometry of the surface or curve in question.

Parameterization

Parameterization is a method of describing curves, surfaces, and solids using parameters. It simplifies evaluating difficult integrals over complex shapes, like our unit circle.
Given \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} \), it describes the path of a point traversing around a unit circle in terms of \( t \), a parameter.
In this particular example:

• Each value of \( t \) defines a specific location on the circle.
• \( t \) runs from \( 0 \) to \( 2\pi \), completing a full circle.

Using parameterization, we transform problems into manageable integrals with limits aligning with our parametrization.
This approach is incredibly useful in vector fields, particularly when calculating complex integrals, like those involving arc lengths or line integrals.

Velocity Fields

Velocity fields represent the speed and direction at each point in a region of space, fundamental in fluid dynamics.
In our problem, the velocity field \( \mathbf{v} = x^2 \mathbf{i} - y \mathbf{j} \) defines how the gas particles move, determining the vector field \( \mathbf{F} = \delta \mathbf{v} \).

• Each point in space has a specific velocity given by \( x \) and \( y \) coordinates, derived from parameters like \( \cos t \) and \( \sin t \).
• The velocity defines the direction and magnitude of gas flow.

Understanding how velocity varies across a region is essential for tasks like calculating flux, as it is the interaction of velocity with the surface that determines the net flow.
These fields let us model and predict the movement of fluid and other substances, highlighting airflow, water currents, and more.