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Chapter 19: Problem 35

Calculate the flux of the vector field through the surface. \(\vec{F}=2 \vec{\imath}+3 \vec{\jmath}\) through the square of side \(\pi\) in the \(x y-\) plane, oriented upward.

Short Answer

Expert verified
The flux through the surface is 0.

Step by step solution

01

Identify the Surface

The problem states that the vector field passes through a square of side \( \pi \) in the \( xy \)-plane. This square is centered at the origin and lies flat on the \( xy \)-plane with vertices at \(( \pm \pi/2, \pm \pi/2, 0)\). The upward orientation implies that the normal vector to the surface is \( \hat{k} \).
02

Determine the Surface Normal Vector

Since the square lies in the \(xy\)-plane and is oriented upward, the outward normal vector \(\hat{n}\) is \(\hat{k}\). This is because the positive \(z\)-axis points upwards and the surface normal points in the positive \(z\) direction.
03

Evaluate the Flux Integral

The flux of \( \vec{F} \) across a surface \( \Sigma \) with normal vector \( \hat{n} \) is given by \( \iint_{\Sigma} \vec{F} \cdot \hat{n} \, dS \). Here, \(\vec{F} = 2 \hat{i} + 3 \hat{j}\), \(\hat{n} = \hat{k}\), and the area element \(dS = dx \, dy\). Since \(\vec{F} \cdot \hat{n} = 0 \) because \(\hat{i}\) and \(\hat{j}\) have no \(\hat{k}\) component, the integral evaluates to 0.
04

Integrate Over the Surface

We integrate \( \vec{F} \cdot \hat{n} \) as follows: \[ \iint_{\Sigma} (2 \hat{i} + 3 \hat{j}) \cdot \hat{k} \, dx \, dy = \iint_{\Sigma} 0 \, dx \, dy = 0. \] Therefore, the flux through the surface is \(0\).

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Key Concepts

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

Vector Field
A vector field is basically a function that assigns a vector to every point in space. Imagine it like a vast mesh or fabric, where each thread represents a direction and magnitude of force. In mathematics, vector fields are often used to represent things that have both a direction and a force, such as wind blowing over a landscape or water currents in a river.
For the given exercise, the vector field is \( \vec{F} = 2 \hat{i} + 3 \hat{j} \). This means that at every point in the field, the vector points in the direction of the positive x-axis two units and the positive y-axis three units. It's like saying that there's a constant wind blowing diagonally, with a particular intensity, over the grid of our square.
Surface Integral
When we talk about surface integrals, we are discussing a way to integrate over a surface to calculate something like the total force through that surface. It's akin to spreading a net over a surface, and figuring out how much of this vector field 'flows' through it.
In the exercise, the surface integral involves integrating over the square in the xy-plane. This square has side length \( \pi \) and is centered at the origin. To compute the surface integral, you evaluate the vector field over this square, making sure to consider how it interacts with the surface's normal vector.
Normal Vector
The normal vector is pivotal in defining the orientation of a surface. It stands perpendicular to the surface and helps us understand the direction of the flow in the flux calculation.
  • Imagine a hand extending a thumb; the thumb is like a normal vector, standing out at a right angle to your palm.
For a surface lying in the xy-plane, the normal vector points along the z-axis. Since the exercise specifies an upward orientation, the normal vector for our square is \( \hat{k} \), which means it points upwards, perpendicular to our square surface.
The role of the normal vector is essential when calculating flux because it helps determine the direction and amount of vector field that genuinely 'pierces' the surface.
Flux Integral
A flux integral calculates how much of a vector field "flows" through a given surface. It is given by the formula \( \iint_{\Sigma} \vec{F} \cdot \hat{n} \, dS \), which captures how the field interacts with the surface's orientation and area.
Here, \( \vec{F} \) is our vector field, and \( \hat{n} \) is the normal vector to the surface. The dot product \( \vec{F} \cdot \hat{n} \) tells us how much of \( \vec{F} \) is aligned with \( \hat{n} \). If \( \vec{F} \) and \( \hat{n} \) are perpendicular, this dot product equals zero, indicating no flux flows through.
  • In the exercise, since both \( \vec{i} \) and \( \vec{j} \) components of \( \vec{F} \) are perpendicular to \( \hat{k} \), the flux integral evaluates to zero, implying no net flow through the square.
Thus, the result of the flux integral is zero, suggesting that the overall flow does not penetrate or exit the surface.

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Most popular questions from this chapter

Calculate the flux of the vector field through the surface. \(\vec{F}=z \vec{k}\) through a square of side \(\sqrt{14}\) in a horizontal plane 2 units below the \(x y\) -plane and oriented downward.

Let \(S\) be the cube with side length \(2,\) faces parallel to the coordinate planes, and centered at the origin. (a) Calculate the total flux of the constant vector field \(\vec{v}=-\vec{i}+2 \vec{j}+\vec{k}\) out of \(S\) by computing the flux through each face separately. (b) Calculate the flux out of \(S\) for any constant vector field \(\vec{v}=a \vec{i}+b \vec{j}+c \vec{k}\) (c) Explain why the answers to parts (a) and (b) make sense.

Are the statements true or false? Give reasons for your answer. \(\operatorname{div} \vec{F}\) is a scalar whose value can vary from point to point.

A vector field, \(\vec{v},\) in the plane is a point source at the origin if its direction is away from the origin at every point, its magnitude depends only on the distance from the origin, and its divergence is zero away from the origin. (a) Explain why a point source at the origin must be of the form \(\vec{v}=\left(f\left(x^{2}+y^{2}\right)\right)(x \vec{i}+y \vec{j})\) for some positive function \(f\) (b) Show that \(\vec{v}=K\left(x^{2}+y^{2}\right)^{-1}(x \vec{i}+y \vec{j})\) is a point source at the origin if \(K>0\) (c) What is the magnitude \(\|\vec{v}\|\) of the source in part (b) as a function of the distance from its center? (d) Sketch the vector field \(\vec{v}=\left(x^{2}+y^{2}\right)^{-1}(x \vec{i}+y \vec{j})\) (e) Show that \(\phi=\frac{K}{2} \log \left(x^{2}+y^{2}\right)\) is a potential function for the source in part (b).

An infinitely long straight wire lying along the \(z\) -axis carries an electric current \(I\) flowing in the \(\vec{k}\) direction. Ampere's Law in magnetostatics says that the current gives rise to a magnetic field \(\vec{B}\) given by $$\vec{B}(x, y, z)=\frac{I}{2 \pi} \frac{-y \vec{i}+x \vec{j}}{x^{2}+y^{2}}$$
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