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Magazine Chapter 16: Problem 30
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 $$ \mathbf{r}(t)=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi $$
Short Answer
Expert verified
Flux for \( \mathbf{F}_1 \) is \(-\pi a^2\), flux for \( \mathbf{F}_2 \) is 0.
Step by step solution
01
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.
02
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 \( \mathbf{n}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} \).
03
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: \( \int_0^{2\pi} (2a \cos^2 t - 3a \sin^2 t) \cdot a \, dt \).
04
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 \(-\pi a^2\).
05
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 \( 2a \cos^2 t + a \cos t \sin t - a \sin^2 t \).
06
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 \( 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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 \( \mathbf{F}_1 \).
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.
