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Chapter 14: Problem 10

Why is the upward flux of a vertical vector field \(\mathbf{F}=\langle 0,0,1\rangle\) across a surface equal to the area of the projection of the surface in the \(x y\) -plane?

Short Answer

Expert verified
Answer: The upward flux of the given vector field \(\mathbf{F}=\langle 0,0,1\rangle\) across a surface is equal to the area of the projection of the surface in the \(xy\)-plane because the vector field points straight up without any horizontal components. This causes the flux to be only in the upward direction, and the area of the projection of the surface onto the \(xy\)-plane is precisely the amount of "space" available for the field to flow through the surface in the positive \(z\)-direction.

Step by step solution

01

Understanding the Concept of Flux

Flux is a measure of how much of a certain quantity passes through a surface per unit time. In the context of a vector field, flux describes how much of the field "flows" through a surface. Mathematically, flux is described as the surface integral of the vector field over a given surface, which can be calculated using the formula: $$\text{Flux} = \iint_{S} \mathbf{F} \cdot d\mathbf{S}$$
02

Given Vector Field

In this problem, we are given a vertical vector field \(\mathbf{F}=\langle 0,0,1\rangle\). This means that the field points straight upwards without any horizontal components. The only direction the field "flows" through a surface is in the positive \(z\)-direction.
03

Calculating the Flux

To calculate the flux, we need to take the surface integral of the vector field over the surface. In this case, since the vector field only has a \(z\)-component, we can simplify the integral by just considering the \(z\)-component of the surface normal vector. Let \(\mathbf{N}\) be the unit normal vector of the surface S. Thus, the surface integral becomes: $$\text{Flux} = \iint_{S} (1)\cdot( \mathbf{N} \cdot \langle 0,0,1\rangle)dS = \iint_{S} N_z dS$$
04

Comparing to the Projected Area

Now let's consider the projection of the surface onto the \(xy\)-plane. Since the field points straight up, we can see the "flow" of the field through the surface is equal to the area of the surface projected onto the \(xy\)-plane. In other words, the flux is equal to the projected area because the vector field \(\mathbf{F}=\langle 0,0,1\rangle\) causes the flux to be only in the upward direction, and the area of the projection of the surface onto the \(xy\)-plane is precisely the amount of "space" available for the field to flow through the surface in the positive \(z\)-direction. Putting everything together, we have shown that the upward flux of the given vector field \(\mathbf{F}=\langle 0,0,1\rangle\) across a surface is indeed equal to the area of the projection of the surface in the \(xy\)-plane.

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