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Chapter 5: Problem 53

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 the curve is 20.

Step by step solution

01

Understand Flux Across a Curve

The flux of a vector field across a curve is given by \[ \text{flux} = \int_C \mathbf{F} \cdot \mathbf{n} \, ds \]where \( \mathbf{n} \) is a unit normal vector to the curve \( C \), and \( ds \) denotes the differential element of arc length along the curve.
02

Determine the Vector Field and Curve

We are given the vector field \( \mathbf{F} = 5 \mathbf{j} \), which is a constant vector pointing in the positive y-direction, and the curve \( C \) defined by \( y = 0, 0 \leq x \leq 4 \). This implies \( C \) is a straight horizontal line segment on the x-axis from \( x = 0 \) to \( x = 4 \).
03

Compute the Unit Normal Vector

For the curve along the x-axis, the outward normal vector is along the y-direction. Since \( \mathbf{F} \) is also in the y-direction, we find the unit normal vector to be \( \mathbf{n} = \mathbf{j} \). Hence, the dot product \( \mathbf{F} \cdot \mathbf{n} = (5 \mathbf{j}) \cdot \mathbf{j} = 5 \).
04

Calculate the Arc Length Differential

Since the curve is straight on the x-axis from \( x=0 \) to \( x=4 \), the differential element of arc length \( ds \, \) is simply \( dx \), where \( \int_C \, ds = \int_0^4 dx \).
05

Compute the Flux Integral

The flux across the curve \( C \) can now be calculated:\[ \text{flux} = \int_0^4 5 \, dx = 5 \int_0^4 \, dx = 5 \times [x]_0^4 = 5 \times (4 - 0) = 20. \]
06

Conclusion

The flux of the vector field \( \mathbf{F} = 5 \mathbf{j} \) across the curve \( C \) is 20.

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Key Concepts

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

Vector Fields
A vector field is essentially a function that assigns a vector to every point in space. Think of it like a field of arrows pointing in various directions throughout a region. Each arrow represents a vector that has both magnitude (length) and direction. In our specific example, the vector field is given as \( \mathbf{F} = 5 \mathbf{j} \). This defines a constant vector field, meaning every point in the field has the same vector: an arrow pointing vertically upwards with a length of 5.
  • **Components**: Vector fields like \( \mathbf{F} \) are often expressed in components, usually denoted \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) which represent unit vectors along the x, y, and z axes respectively.
  • **Uniform Field**: Since \( \mathbf{F} = 5 \mathbf{j} \) has no \( \mathbf{i} \) or \( \mathbf{k} \) components, it is entirely aligned along the y-axis. This is a simplification since there's no variation in direction or magnitude in the field.
Understanding vector fields is crucial for visualizing how forces, like electric and magnetic fields, or fluid velocities might behave across a region.
Unit Normal Vector
The unit normal vector is a crucial element when working with line integrals and flux across curves. A normal vector is perpendicular to a surface or curve. To convert it into a unit vector, it needs to have a magnitude (or length) of 1.
  • **Finding the normal**: For the curve along the x-axis, a normal vector would naturally point in the y-direction. In our problem, since our curve is horizontal, the unit normal vector \( \mathbf{n} \) is simply \( \mathbf{j} \).
  • **Importance in flux**: When calculating flux, the unit normal vector determines the direction through which the vector field crosses the curve. The dot product of the vector field and the normal vector is a measure of this flux. In our scenario, the dot product \( \mathbf{F} \cdot \mathbf{n} = 5 \) is straightforward because the vectors are aligned.
The understanding of unit normal vectors helps in modeling and solving real-world problems where orientations relative to surfaces or curves are important.
Line Integrals
Line integrals are like integrals in calculus but instead of integrating over a straight line or region, you integrate over a curve. They are vital in fields like physics and engineering for calculating quantities like work done by a force or flux.
  • **The setup**: For a line integral concerning flux, we use \( \int_C \mathbf{F} \cdot \mathbf{n} \, ds \). Here, \( ds \) represents the tiny segments along the curve, akin to how \( dx \) is used in regular integrals. In our case, because the curve \( C \) is a simple straight line along the x-axis, \( ds \) simplifies to \( dx \).
  • **Computing the integral**: With \( \mathbf{F} \cdot \mathbf{n} = 5 \) and \( ds = dx \), the integral \( \int_0^4 5 \, dx = 20 \) succinctly computes the total flux across the curve.
Grasping line integrals allows one to solve problems involving flows and fields around variously shaped paths.

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Most popular questions from this chapter

For the following exercises, describe each vector field by drawing some of its vectors. $$ \mathbf{F}(x, y, z)=2 x \mathbf{i}-2 y \mathbf{j}-2 z \mathbf{k} $$

Use the divergence theorem to evaluate \(\iint_{S} \mathbf{F} \cdot d S .\)Use the divergence theorem to compute the value of the flux integral over the unit sphere with \(\mathbf{F}(x, y, z)=3 z \mathbf{i}+2 y \mathbf{j}+2 x \mathbf{k}\)

For the following exercises, find the gradient vector field of each function f. $$ f(x, y, z)=x \cos \left(\frac{y}{z}\right) $$

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Evaluate surface integral \(\iint_{S} z d \mathrm{~A}\), where \(\mathrm{S}\) is surface \(z=\sqrt{x^{2}+y^{2}}, 0 \leq z \leq 2\).
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