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Chapter 6: Problem 89

For the following exercises, find the flux. Let \(\mathbf{F}=5 \mathbf{j}\) and let \(C\) be curve \(y=0,0 \leq x \leq 4\) Find the flux across \(C .\)

Short Answer

Expert verified
The flux across curve \( C \) is 20.

Step by step solution

01

Understand the Problem

The vector field given is \( \mathbf{F} = 5 \mathbf{j} \), which means the field is constant in the positive \( y \)-direction with a magnitude of 5 at all points. The curve \( C \) is a line along \( y=0 \) from \( x=0 \) to \( x=4 \). We are asked to find the flux across this curve \( C \).
02

Identify Relevant Concepts

Flux through a curve is calculated using the line integral of the vector field \( \mathbf{F} \) across the curve \( C \). The formula for the flux \( \Phi \) is given by \( \Phi = \int_{C} \mathbf{F} \cdot \mathbf{n} \, ds \), where \( \mathbf{n} \) is the unit normal vector to the line and \( ds \) is the differential arc length along the curve.
03

Determine the Unit Normal Vector

Since the curve \( C \) is along the x-axis (\( y = 0 \)), the unit normal vector \( \mathbf{n} \) is simply \( \mathbf{i} \) or \( -\mathbf{i} \), perpendicular to the curve. However, to cross the curve from the inside to outside (conventional for positive flux), take \( \mathbf{n} = \mathbf{j} \).
04

Set Up the Line Integral

Substitute \( \mathbf{F} = 5 \mathbf{j} \) and \( \mathbf{n} = \mathbf{j} \) into the flux formula: \( \Phi = \int_{0}^{4} (5\mathbf{j}) \cdot \mathbf{j} \, dx \). Since \( \mathbf{j} \cdot \mathbf{j} = 1 \), it simplifies to \( \Phi = \int_{0}^{4} 5 \, dx \).
05

Evaluate the Integral

Evaluate the definite integral \( \Phi = \int_{0}^{4} 5 \, dx = 5x \mid_{0}^{4} \). This results in \( \Phi = 5(4) - 5(0) = 20 \).
06

Conclusion

The flux of the vector field \( \mathbf{F} = 5 \mathbf{j} \) across the curve \( C \) is \( 20 \). This value represents the total amount of field crossing the curve from one side to the other.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Field
In mathematics and physics, a vector field is a construction in which each point in space is associated with a vector. Think of it as assigning a direction and magnitude (or "strength") to every point in a region.
For example, a vector field could describe the influence of a magnetic force in a given space or the flow of a fluid.
  • In our exercise, the vector field is given as \( \mathbf{F} = 5 \mathbf{j} \). This indicates a constant field in the \( y \)-direction with a magnitude of 5.
  • Since this vector field is constant, it does not vary with position, making it simpler to calculate aspects like flux.
Understanding a vector field is crucial because it provides context for problems involving force, flow, and more in geometrical spaces. Recognizing the constant nature of the vector field in our problem helps streamline the calculation process.
Line Integral
A line integral, in its simplest form, is a method to sum up values along a curve. Instead of summing up just numbers, you account for varying functions along a line segment or curve.
In the context of a vector field, a line integral can help us understand the "total effect" of a vector field along a path or boundary.
  • For our exercise, the line integral helps calculate the flux across the curve \( C \).
  • Mathematically, this is expressed through the formula \( \Phi = \int_{C} \mathbf{F} \cdot \mathbf{n} \, ds \), where \( \mathbf{F} \) is the vector field, and \( \mathbf{n} \) is the unit normal vector.
The line integral is critical as it transitions the vector field's effects from abstract vectors into measurable quantities over defined paths.
Unit Normal Vector
A unit normal vector is a vector of length 1 (unit) that is perpendicular (normal) to a given surface or curve. It helps in delineating the direction for calculations across that boundary.
In flux calculations, the unit normal vector \( \mathbf{n} \) is particularly significant because it helps determine the orientation from inside to outside.
  • For the exercise, the curve \( C \) lies along the \( x \)-axis, so a logical unit normal vector is \( \mathbf{j} \).
  • This perpendicular orientation along the \( y \)-axis aligns perfectly with our vector field \( \mathbf{F} \), simplifying our calculations significantly.
Having a clear understanding of the unit normal vector's role allows us to establish the proper reference direction for the line integral, ensuring accurate computation of the flux.
Definite Integral
The definite integral calculates the total accumulation of values over a given interval. It's a cornerstone of calculus that allows us to find areas, volumes, and many other quantities that aggregate information.
In our exercise, evaluating a definite integral follows from setting up the line integral. Once we simplify the expression, we calculate the total effect along the entire curve or segment.
  • Here, the definite integral \( \Phi = \int_{0}^{4} 5 \, dx \) results in a simple multiplication, yielding the value of 20.
  • The definite integral's result represents the total flux across the curve \( C \).
Understanding definite integrals in this context is essential because it translates our mathematical expressions into physical quantities, such as the total field effect passing across a boundary.

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