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Chapter 19: Problem 22

Find the flux of \(\vec{H}=2 \vec{i}+3 \vec{j}+5 \vec{k}\) through the surface \(S.\) \(S\) is the disk of radius 1 in the plane \(x=2\) oriented in the positive \(x\) -direction.

Short Answer

Expert verified
The flux through the surface is \( 2\pi \).

Step by step solution

01

Understand the Problem

We need to find the flux of the vector field \( \vec{H}=2 \vec{i}+3 \vec{j}+5 \vec{k} \) through the given surface \( S \). \( S \) is a disk of radius 1 situated in the plane \( x = 2 \), and it is oriented in the positive \( x \)-direction.
02

Determine the Unit Normal Vector

Since the surface is perpendicular to the \( x \)-axis and oriented positively along the \( x \)-axis, the unit normal vector to the surface \( S \) is \( \vec{n} = \vec{i} = (1,0,0) \).
03

Find the Area of the Disk

The disk has a radius of 1. The area \( A \) of the disk is given by \( A = \pi r^2 = \pi \times 1^2 = \pi \).
04

Calculate the Dot Product of Vector Field with Normal

Substitute the vector field \( \vec{H} = (2,3,5) \) and the normal vector \( \vec{n} = (1,0,0) \). The dot product is \( \vec{H} \cdot \vec{n} = (2,3,5) \cdot (1,0,0) = 2 \).
05

Compute the Flux

The flux \( \Phi \) through the surface \( S \) is obtained by multiplying the dot product by the area of the surface. Thus, \( \Phi = \vec{H} \cdot \vec{n} \times A = 2 \times \pi = 2\pi \).

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Key Concepts

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

Surface Integrals
Imagine you have a surface, like a disk or a balloon, and you want to understand how a vector field interacts with this surface. This is where surface integrals come in handy. A surface integral helps us calculate quantities like flux across a surface. To compute a surface integral, we sum up all the contributions of tiny pieces of a surface influenced by the vector field. It's like breaking down the surface into small squares, checking how the vector field pushes or pulls perpendicularly on each square, and then adding these effects. Here's a helpful way to think about it:
  • The vector field gives you the forces or flows at every point.
  • The surface, such as our disk, provides the area where these forces act.
  • The interaction is analyzed by piece-wise contributions, finally summed to get the total flux through the surface.
Surface integrals are powerful because they translate the interaction of vector fields with surfaces into mathematical expressions that can be calculated.
Oriented Surfaces
Understanding the concept of oriented surfaces is crucial when dealing with flux. When we say that a surface is oriented, it means we've chosen a direction that is considered 'positive'. This direction is represented by the normal vector pointing outwards. In the problem we are solving, the disk is located in the plane where x=2 and is oriented in such a way that the positive x-direction is the one that matters. Think of it like a badge pinned on your shirt: It could face either in or out, and which way it faces changes how things affect it. For calculating flux, this directionality is essential:
  • It ensures consistency; everyone agrees on what "positive" means for the calculation.
  • It impacts results, as orienting the surface differently could mathematically flip the sign of flux values.
  • It informs computing the correct unit normal vector, a necessary component in flux calculations.
Unit Normal Vector
The unit normal vector is your guide for understanding how the vector field intersects the surface. It's a vector that stands perpendicular (or "normal") to the surface and has a length of one (hence "unit"). In many cases, finding the correct unit normal vector depends on the configuration and orientation of your surface. For our disk in the plane x=2, oriented positively along the x-axis, the unit normal vector is simple and straightforward鈥攁 pleasant \[ \vec{n} = (1, 0, 0)\]Here鈥檚 why the unit normal vector is valuable:
  • It helps compute how much of the vector field passes through the surface. This is done using the dot product.
  • It ensures calculations are precise by using the \(\text{"unit" length of } 1\) which keeps the math tidy.
  • Its direction guarantees alignment with the orientation of the surface, enhancing accurate flux assessment.
The choice of the direction for the normal vector is linked directly with the surface's orientation, affirming its vital role in surface integrals and flux calculations.

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Most popular questions from this chapter

Calculate the flux of the vector field through the surface. \(\vec{F}=x \vec{i}+2 y \vec{j}+3 z \vec{k}\) and \(S\) is a square of side 2 in the plane \(y=3,\) oriented in the positive \(y\) -direction.

Involve electric fields. Electric charge produces a vector field \(\vec{E},\) called the electric field, which represents the force on a unit positive charge placed at the point. Two positive or two negative charges repel one another, whereas two charges of opposite sign attract one another. The divergence of \(\vec{E}\) is proportional to the density of the electric charge (that is, the charge per unit volume), with a positive constant of proportionality. The electric field at the point \(\vec{r}\) as a result of a point charge at the origin is \(\vec{E}(\vec{r})=k \vec{r} /\|\vec{r}\|^{3}\) (a) Calculate div \(\vec{E}\) for \(\vec{r} \neq \overrightarrow{0}\) (b) Calculate the limit suggested by the geometric definition of \(\operatorname{div} \vec{E}\) at the point (0,0,0) (c) Explain what your answers mean in terms of charge density.

Suppose that \(\vec{E}\) is a uniform electric field on 3 -space, so \(\vec{E}(x, y, z)=a \vec{\imath}+b \vec{\jmath}+c \vec{k},\) for all points \((x, y, z)\) where \(a, b, c\) are constants. Show, with the aid of symmetry, that the flux of \(\overrightarrow{\boldsymbol{E}}\) through each of the following closed surfaces \(S\) is zero: (a) \(S\) is the cube bounded by the planes \(x=\pm 1\) \(y=\pm 1,\) and \(z=\pm 1.\) (b) \(S\) is the sphere \(x^{2}+y^{2}+z^{2}=1.\) (c) \(S\) is the cylinder bounded by \(x^{2}+y^{2}=1, z=0\) and \(z=2.\)

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.

Find the area of the surface \(z=f(x, y)\) over the region \(R\) in the \(x y\) -plane. \(f(x, y)=50+5 x-y, R:\) circle of radius 3 centered at the origin
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