Problem 30 - Thomas Calculus

Question

Flux across a circle Find the flux of the fields $$ \mathbf{F}_{1}=2 x \mathbf{i}-3 y \mathbf{j} \quad \text { and } \quad \mathbf{F}_{2}=2 x \mathbf{i}+(x-y) \mathbf{j} $$ across the circle $$ \mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi $$

Step-by-Step Solution

Answer

Flux for \( \mathbf{F}_1 \) is \(-\pi a^2\), flux for \( \mathbf{F}_2 \) is 0.

Step 1: Understand the Problem

We are given two vector fields \( \mathbf{F}_1 = 2x \mathbf{i} - 3y \mathbf{j} \) and \( \mathbf{F}_2 = 2x \mathbf{i} + (x-y) \mathbf{j} \), and a parametrized circle by \( \mathbf{r}(t) = (a \cos t ) \mathbf{i} + (a \sin t) \mathbf{j} \). The task is to find the flux of these fields across the circle.

Step 2: Calculate Normal Vector

The circle is traversed by the parametrization \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \). Its derivative \( \mathbf{r}'(t) = -a \sin t \mathbf{i} + a \cos t \mathbf{j} \) will give the tangent vector. The normal vector \( \mathbf{n}(t) \) can be obtained by rotating the tangent vector by 90 degrees counter-clockwise, resulting in \( \mathbf{n}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \).

Step 3: Set Up Flux Integral for \( \mathbf{F}_1 \)

For \( \mathbf{F}_1 \), calculate \( \mathbf{F}_1 \cdot \mathbf{n} \): \( 2x \cos t - 3y \sin t \). Substitute \( x = a \cos t \), \( y = a \sin t \) to get \( \mathbf{F}_1 \cdot \mathbf{n} = 2a \cos^2 t - 3a \sin^2 t \). The flux integral becomes: \( \int_0^{2\pi} (2a \cos^2 t - 3a \sin^2 t) \cdot a \, dt \).

Step 4: Solve the Flux Integral for \( \mathbf{F}_1 \)

Compute the integral \( a \int_0^{2\pi} (2a \cos^2 t - 3a \sin^2 t) \, dt \). Using identities \( \cos^2 t = \frac{1 + \cos 2t}{2} \) and \( \sin^2 t = \frac{1 - \cos 2t}{2} \), simplify and integrate: \( \int_0^{2\pi} \frac{-a^2}{2} \, dt = -\frac{a^2}{2} \cdot 2\pi \), which results in \(-\pi a^2\).

Step 5: Set Up Flux Integral for \( \mathbf{F}_2 \)

For \( \mathbf{F}_2 \), calculate \( \mathbf{F}_2 \cdot \mathbf{n} \): \( 2x \cos t + (x-y) \sin t \). Again, substitute \( x = a \cos t \), \( y = a \sin t \), to get \( \mathbf{F}_2 \cdot \mathbf{n} = 2a \cos^2 t + (a \cos t - a \sin t) \sin t \). Simplify it to \( 2a \cos^2 t + a \cos t \sin t - a \sin^2 t \).

Step 6: Solve the Flux Integral for \( \mathbf{F}_2 \)

Compute the integral \( \int_0^{2\pi} (2a \cos^2 t + a \cos t \sin t - a \sin^2 t) \, dt \). Simplify using identities \( \cos^2 t = \frac{1 + \cos 2t}{2} \) and \( \sin^2 t = \frac{1 - \cos 2t}{2} \) and note \( \int_0^{2\pi} \sin 2t \, dt = 0 \). The result is zero after integration, therefore the flux of \( \mathbf{F}_2 \) is \( 0 \).

Key Concepts

Vector Fields

Vector fields are mathematical constructions that assign a vector to every point in space. In this exercise, we have two vector fields, \( \mathbf{F}_1 \) and \( \mathbf{F}_2 \). Each vector field is described using coordination and directionality in the two-dimensional Cartesian plane.

• \( \mathbf{F}_1 = 2x \mathbf{i} - 3y \mathbf{j} \) means at every point \((x, y)\), the vector points in the direction given by \(2x\) along the \(x\)-axis and \(-3y\) along the \(y\)-axis.
• \( \mathbf{F}_2 = 2x \mathbf{i} + (x-y) \mathbf{j} \) similarly provides a vector determined by \(2x\) in the \(x\)-axis direction and \((x-y)\) in the \(y\)-axis direction.

These fields represent the flow of something like fluid, heat, or electricity across the plane, suggesting directions and intensities at every point. Understanding the arrangement and influence of these vectors is crucial for calculating flux.

Parametrized Curves

Parametrized curves involve defining a curve in space using one or more parameters. In this exercise, the curve is represented by a circle: \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \). The parameter \( t \) usually varies over an interval to trace out the curve entirely. Here, \( t \) ranges from 0 to \(2\pi\), which spans a full circle.

• \( \mathbf{r}(t) = (a \cos t) \mathbf{i} + (a \sin t) \mathbf{j} \): This formula traces out a circle of radius \( a \) centered at the origin.
• The use of trigonometric functions \( \cos t \) and \( \sin t \) defines the x and y coordinates as the parameter \( t \) rotates one full clockwise orientation around the circle.

Parametrization provides a convenient way to express and manipulate complex curves analytically, aiding in further calculations, like those necessary for computing flux.

Flux Integral

The flux integral calculates how much of a vector field passes through a given surface, in this case, a curve. It centers around the concept of the dot product between the vector field and the curve's normal vector. Calculating flux establishes a measure for how much of the field traverses, rather than running parallel or tangent, to the surface.

• For \( \mathbf{F}_1 \), the integral becomes \( \int_0^{2\pi} (2a \cos^2 t - 3a \sin^2 t) a \, dt \), resolved using trigonometric identities and expressions.
• The flux integral involves transforming and solving these expression into simpler forms, leading to outcomes like \(-\pi a^2\) for \( \mathbf{F}_1 \).

This integral determines how much of the vector field's 'substance' flows out of the given boundary, offering insights into the field's behavior and intensity.

Normal Vectors

Normal vectors are critical to determining how a vector field interacts with a surface. In a two-dimensional plane, a normal vector is perpendicular to a tangent vector along the curve. Here, the tangent vector is derived by differentiating the curve's parametrization with respect to \( t \). For our curve, \( \mathbf{r}'(t) = -a \sin t \mathbf{i} + a \cos t \mathbf{j} \) represents how the circle progresses at any point.

• To find the normal vector, rotate the tangent vector 90 degrees counterclockwise, yielding \( \mathbf{n}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \).
• Without this orthogonality provided by normal vectors, accurately measuring the flux would not be possible as they capture the perpendicular interaction field of flux.

These vectors play a fundamental role in accurately connecting the math of vector fields to physical intuition and applications.