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Chapter 17: Problem 19

Use the Divergence Theorem to compute the -net outward flux of the following fields across the given surface \(S .\) \(\mathbf{F}=\langle x, 2 y, z\rangle ; S\) is the boundary of the tetrahedron in the first octant formed by the plane \(x+y+z=1\)

Short Answer

Expert verified
Answer: The net outward flux of the vector field 饾悈 = 鉄▁, 2y, z鉄� across the surface S of the tetrahedron is 4/3.

Step by step solution

01

Compute the Divergence of Vector Field

To compute the divergence of the vector field \(\mathbf{F}=\langle x, 2 y, z\rangle\), we apply the divergence operator to obtain: $$ \nabla \cdot \mathbf{F} = \frac{\partial x}{\partial x} + \frac{\partial (2y)}{\partial y} + \frac{\partial z}{\partial z} = 1 + 2 + 1 = 4. $$
02

Set up the Volume Integral using the Divergence Theorem

According to the Divergence Theorem, the net outward flux of a vector field across a closed surface is equal to the volume integral of the divergence of that vector field inside the region encompassed by the surface. Thus, we have: $$ \text{Flux} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV = \iiint_V 4\, dV. $$
03

Convert the Volume Integral to the Tetrahedron's Coordinate System

We are given that the surface \(S\) is the boundary of the tetrahedron in the first octant formed by the plane \(x+y+z=1\). This region can be described in the following way: \(0 \le x \le 1\), \(0 \le y \le 1-x\), and \(0 \le z \le 1-x-y\). So, the volume integral becomes: $$ \text{Flux} = \int_{0}^1 \int_{0}^{1-x} \int_{0}^{1-x-y} 4 \,dz\, dy\, dx. $$
04

Compute the Volume Integral

Next, evaluate the iterated triple integral: $$ \text{Flux} = \int_{0}^1 \int_{0}^{1-x} 4\left[\int_{0}^{1-x-y} \,dz\right]\, dy\, dx = \\ \int_{0}^1 \int_{0}^{1-x} 4(1-x-y) \,dy\, dx = \\ \int_{0}^1 \left[4\left(\int_{0}^{1-x} (1-x-y) \,dy\right)\right] dx. $$ First, we integrate with respect to \(y\) and simplify: $$ \int_{0}^1 \left[4\left(\int_{0}^{1-x} (1-x-y) \,dy\right)\right] dx = \\ \int_{0}^1 \left[4\left[ (1-x)y -\frac{1}{2}y^2 \right]_0^{1-x}\right] dx = \\ \int_{0}^1 4\left[(1-x)^2 - \frac{1}{2}(1-x)^2 \right] dx. $$ Now, we integrate with respect to \(x\): $$ \text{Flux} = 4\left[\int_{0}^1 \frac{1}{2}(1-x)^2 dx\right] = \\ 4\left[\frac{1}{2}\int_{0}^1 (1-2x+x^2) dx\right] = \\ 4\left[\frac{1}{2}\left[x - x^2 +\frac{1}{3}x^3\right]_0^1\right] = \\ 4\left[\frac{1}{2}\left(1 - 1 + \frac{1}{3}\right)\right] = \frac{4}{3}. $$ Thus, the net outward flux of the vector field \(\mathbf{F} = \langle x, 2y, z \rangle\) across the surface \(S\) of the tetrahedron is \(\frac{4}{3}\).

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Key Concepts

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

Net Outward Flux
The term net outward flux refers to the total amount of a vector field that flows out of a closed surface. Think of it as imagining air particles moving outwardly from a balloon鈥檚 surface; the combined effect of these particles rushing out represents the net outward flux.

When dealing with fields such as electric fields, fluid velocity fields, or any spatial vector quantity, the concept of net outward flux becomes important. It provides a measure of how much of that field 鈥榚scapes鈥� through a boundary. In the exercise, we are dealing with a vector field \( \mathbf{F} \), and to understand the net outward flux, we looked at the flow across the surface of a tetrahedron.

To find the net outward flux using calculus, one typically applies the Divergence Theorem, which connects the flow across the surface with a volume integral of the divergence of \( \mathbf{F} \) 鈥� that鈥檚 essentially a summing up of how much \( \mathbf{F} \) is diverging (or spreading out) at every point within the volume.
Volume Integral
A volume integral represents the total of a function over a three-dimensional space. If you imagine a physical space filled with a quantity - it could be mass, charge, heat, or in our case, the divergence of a field - the volume integral measures how much there is throughout the entire volume.

In the context of the exercise, we compute the volume integral of the divergence of the vector field within the boundaries of a tetrahedron. This process involves an iterated triple integral where the function is integrated over one variable at a time, considering the limits for each variable determined by the tetrahedron鈥檚 geometry.

The process might feel complex, but it boils down to summing up contributions from every minuscule 鈥榗ube鈥� of the volume. The step-by-step solution shows us integrating with respect to \(z\), then \(y\), and finally \(x\), narrowing our three-dimensional problem down to a single number, which represents the total divergence鈥攊.e., the net outward flux鈥攐f our vector field throughout the volume.
Vector Field Divergence
The divergence of a vector field is a measure of how much the field vectors are spreading out from a point. Mathematically, we find this by taking the dot product of the del operator (a vector operator denoted by \( abla \) and pronounced 'nabla') with the vector field in question.

Imagine a group of runners starting from a single point: If they all run away from the point in different directions, they're 'diverging'. The divergence at each point in a vector field represents the 'rate of spreading' at that point, and when you calculate the divergence over a volume, it tells you the net 'source strength' within that volume.

In our exercise, the divergence of the vector field \( \mathbf{F} = \langle x, 2y, z \rangle \) was found to be constant at 4. This means that at every point within our space, the vectors in the field are collectively spreading out at the same rate, contributing uniformly to the net outward flux. The divergence being constant simplifies our calculations, as seen in the volume integral section.

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Most popular questions from this chapter

Conservation of energy Suppose an object with mass \(m\) moves in a region \(R\) in a conservative force field given by \(\mathbf{F}=-\nabla \varphi,\) where \(\varphi\) is a potential function in a region \(R .\) The motion of the object is governed by Newton's Second Law of Motion, \(\mathbf{F}=m \mathbf{a},\) where a is the acceleration. Suppose the object moves from point \(A\) to point \(B\) in \(R\) a. Show that the equation of motion is \(m \frac{d \mathbf{v}}{d t}=-\nabla \varphi\) b. Show that \(\frac{d \mathbf{v}}{d t} \cdot \mathbf{v}=\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v})\) c. Take the dot product of both sides of the equation in part (a) with \(\mathbf{v}(t)=\mathbf{r}^{\prime}(t)\) and integrate along a curve between \(A\) and B. Use part (b) and the fact that \(\mathbf{F}\) is conservative to show that the total energy (kinetic plus potential) \(\frac{1}{2} m|\mathbf{v}|^{2}+\varphi\) is the same at \(A\) and \(B\). Conclude that because \(A\) and \(B\) are arbitrary, energy is conserved in \(R\).

Integration by parts (Gauss' Formula) Recall the Product Rule of Theorem \(17.13: \nabla \cdot(u \mathbf{F})=\nabla u \cdot \mathbf{F}+u(\nabla \cdot \mathbf{F})\) a. Integrate both sides of this identity over a solid region \(D\) with a closed boundary \(S\), and use the Divergence Theorem to prove an integration by parts rule: $$\iiint_{D} u(\nabla \cdot \mathbf{F}) d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \nabla u \cdot \mathbf{F} d V$$ b. Explain the correspondence between this rule and the integration by parts rule for single-variable functions. c. Use integration by parts to evaluate \(\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V\) where \(D\) is the cube in the first octant cut by the planes \(x=1\) \(y=1,\) and \(z=1\)

Find a vector field \(\mathbf{F}\) with the given curl. In each case, is the vector field you found unique? $$\operatorname{curl} \mathbf{F}=\langle 0, z,-y\rangle$$

Find the general formula for the surface area of a cone with height \(h\) and base radius \(a\) (excluding the base).

Prove the following identities. Assume \(\varphi\) is a differentiable scalar- valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla \cdot(\varphi \mathbf{F})=\nabla \varphi \cdot \mathbf{F}+\varphi \nabla \cdot \mathbf{F} \quad \text { (Product Rule) }$$
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