Problem 35 - Thomas Calculus

Question

Flow integrals Find the flow of the velocity field \(\mathbf{F}=\) \((x+y) \mathbf{i}-\left(x^{2}+y^{2}\right) \mathbf{j}\) along each of the following paths from \((1,0)\) to \((-1,0)\) in the \(x y\) -plane. a. The upper half of the circle \(x^{2}+y^{2}=1\) b. The line segment from \((1,0)\) to \((-1,0)\) c. The line segment from \((1,0)\) to \((0,-1)\) followed by the line segment from \((0,-1)\) to \((-1,0)\)

Step-by-Step Solution

Answer

a. -2, b. 0, c. -2.

Step 1: Express the Velocity Field

The velocity field is given by \( \mathbf{F}=(x+y) \mathbf{i}-(x^{2}+y^{2}) \mathbf{j} \). This means \( P(x,y) = x + y \) and \( Q(x,y) = -(x^2 + y^2) \).

Step 2: Solution for Path a - Upper Half of Circle

The path is the semicircular arc from \((1,0)\) to \((-1,0)\) on \(x^2 + y^2 = 1\). Parameterize this arc with \(x=\cos t\) and \(y=\sin t\), where \(t\) goes from \(0\) to \(\pi\). The flow integral is \[ \int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{0}^{\pi} ((\cos t + \sin t)\cdot (-\sin t) - (\cos^2 t + \sin^2 t)\cdot \cos t) dt = \int_{0}^{\pi} (-\cos t) dt = -2. \]

Step 3: Solution for Path b - Straight line from (1,0) to (-1,0)

The path can be parameterized by \( x = 1 - 2t \) and \( y = 0 \), where \( t \) goes from \(0\) to \(1\). The derivative \( d\mathbf{r} = (-2, 0) \). The flow integral is \[ \int_{0}^{1} ((1-2t) + 0) \cdot (-2) \, dt = \int_{0}^{1} (2 - 4t) \, dt = 0. \]

Step 4: Solution for Path c - Line segments from (1,0) to (0,-1) and (0,-1) to (-1,0)

First segment: from \((1,0)\) to \((0,-1)\) can be parameterized by \( x = 1-t \), \( y = -t \) for \( t \) from \(0\) to \(1\). The flow integral is \[ \int_{0}^{1} ((1-t) + (-t)) (-1, -1) \cdot d\mathbf{r} \approx -1. \]Second segment: from \((0,-1)\) to \((-1,0)\) can be parameterized by \( x = -t \), \( y = -1 + t \) for \( t \) from \(0\) to \(1\). The flow integral is \[ \int_{0}^{1} (-t + (-1+t)) \cdot (-1, 0) \cdot d\mathbf{r} \approx 1. \]Combining both segments gives a total flow of -2.

Key Concepts

Velocity Field

In vector calculus, a velocity field is a vector field that represents the velocity of every point in a fluid flow. The velocity field for a region in space can vary depending on the position, meaning that in different areas, the fluid may flow with different speeds and in different directions.
In the given exercise, the velocity field is represented by \(\mathbf{F} = (x+y) \mathbf{i} - (x^2+y^2) \mathbf{j}\). This can be understood better by breaking it down into its components:

• The component \(P(x, y) = x + y\) describes the flow along the \(x\)-direction.
• The component \(Q(x, y) = -(x^2 + y^2)\) describes the flow along the \(y\)-direction.

These components indicate how the velocity changes as a function of \(x\) and \(y\), which allows us to analyze how objects will move within this field.

Parameterization

Parameterization is a technique used in mathematics to describe a path or a surface using one or more variables. When solving problems involving line integrals, parameterization helps to redefine the path in terms of a single variable, usually \(t\), which represents time or some other progression variable.
In the context of this exercise, the paths are parameterized to calculate the flow of the velocity field along them:

• For a semicircular path represented by \(x^2 + y^2 = 1\), we use \(x = \cos t\) and \(y = \sin t\), with \(t\) ranging from \(0\) to \(\pi\), which traces the upper half of the circle.
• For a straight line from \((1, 0)\) to \((-1, 0)\), parameterization is done by setting \(x = 1 - 2t\) and \(y = 0\), progressing as \(t\) goes from 0 to 1.
• In piecewise paths, each segment is parameterized separately to simplify computation and clarify the change in variables over each segment.

By parameterizing the paths correctly, it becomes much easier to carry out the line integration and study the behavior of the velocity field along different paths.

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and the operations on them. These operations include integration and differentiation of vector fields, which are essential in understanding physical phenomena such as fluid and electromagnetic fields.
Key aspects of vector calculus used in the exercise include:

• Vector Fields: A representation where each point in space is associated with a velocity vector that indicates direction and magnitude.
• Gradients, Divergence, and Curl: Fundamental concepts in vector calculus that help define how vector fields function and change.

In particular, solving a flow integral within a vector field requires understanding how the vector field behaves along a path. It often involves calculating the dot product of the velocity and an infinitesimal segment of the path, which in this context is managed through line integrals.

Line Integrals

Line integrals are a crucial concept in vector calculus. They allow us to evaluate the integral of a function along a curve, which in the case of vectors, means finding how much a vector field contributes along that path.
For the velocity field, a line integral calculates the total flow along a given path. The integral of a vector field \(\mathbf{F}\) along a path \(C\) is expressed as \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\). Here \(d\mathbf{r}\) represents a small segment of the path.
Steps to compute a line integral include:

• Parameterizing the path to express \(d\mathbf{r}\) in terms of a parameter, usual variable \(t\).
• Substituting \(\mathbf{F}\) and \(d\mathbf{r}\) into the integral expression.
• Evaluating the integral over the specified limits of \(t\).

This process provides the measure of how much flow or flux occurs through a specified path, giving insight into properties of the field, such as direction and intensity of flow along different sections.