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Magazine Chapter 18: Problem 2
Use the divergence theorem (18.26) to find \(\iint_{s} F \cdot \mathbf{n} d S\) \(\mathbf{F}=y^{3} e^{z} \mathbf{i}-x y \mathbf{j}+x \arctan y \mathbf{k}\) \(S\) is the surface of the region bounded by the coordinate planes and the plane \(x+y+z=1\).
Short Answer
Expert verified
The flux \( \iint_{S} \mathbf{F} \cdot \mathbf{n} dS \) is \(-\frac{1}{96}\).
Step by step solution
01
Understand the Divergence Theorem
The Divergence Theorem relates the flux of a vector field through a closed surface to the divergence of the vector field in the volume it encloses. Mathematically, it states: \[ \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_{V} abla \cdot \mathbf{F} \, dV \], where \(S\) is the closed surface and \(V\) is the volume inside \(S\).
02
Calculate the Divergence of \( \mathbf{F} \)
Find the divergence of \( \mathbf{F} = y^3e^z \mathbf{i} - xy \mathbf{j} + x \arctan y \mathbf{k} \). The divergence is given by \( abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(y^3e^z) + \frac{\partial}{\partial y}(-xy) + \frac{\partial}{\partial z}(x \arctan y) \). Calculating each term separately, we find: \( \frac{\partial}{\partial x}(y^3e^z) = 0 \), \( \frac{\partial}{\partial y}(-xy) = -x \), and \( \frac{\partial}{\partial z}(x \arctan y) = 0 \). Therefore, \( abla \cdot \mathbf{F} = 0 - x + 0 = -x \).
03
Define the Volume \( V \) and Set Up the Integral
The volume \( V \) is bounded by the coordinate planes and the plane \( x + y + z = 1 \). This can be expressed as the region \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1-x \), and \( 0 \leq z \leq 1-x-y \). Set up the triple integral for the divergence: \[ \iiint_{V} abla \cdot \mathbf{F} \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} (-x) \, dz \, dy \, dx \].
04
Evaluate the Integral
Evaluate the triple integral: First integrate with respect to \( z \), \[ \int_{0}^{1-x-y} (-x) \, dz = -x(1-x-y) \]. Then integrate with respect to \( y \), \[ \int_{0}^{1-x} -x(1-x-y) \, dy = -x[(1-x)y - \frac{1}{2}y^2]_{0}^{1-x} = -x\left[(1-x)(1-x) - \frac{1}{2}(1-x)^2\right] = -x\left[\frac{1}{2}(1-x)^2\right] \]. Finally, integrate with respect to \( x \), \[ \int_{0}^{1} \frac{-x(1-x)^2}{2} \, dx = -\frac{1}{2} \int_{0}^{1} (x - 2x^2 + x^3) \, dx = -\frac{1}{2} \left[\frac{x^2}{2} - \frac{2x^3}{3} + \frac{x^4}{4}\right]_{0}^{1} = -\frac{1}{2} \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4}\right) = -\frac{1}{96}. \]
05
Conclude with the Result
According to the Divergence Theorem, the flux of the vector field \( \mathbf{F} \) through the surface \( S \) is given by the integral over the volume \( V \) we just calculated, which equals \(-\frac{1}{96}.\)
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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 assigns a vector to each point in space. It can be thought of as a collection of vectors showing the direction and magnitude of some force throughout a region. Understanding a vector field is crucial in visualizing how a quantity changes from point to point. In the given problem, the vector field is represented by \( \mathbf{F} = y^3e^z \mathbf{i} - xy \mathbf{j} + x \arctan y \mathbf{k} \). This expression breaks down into three components:
By analyzing these components separately, we can infer how the vector field behaves in space.
- \( y^3e^z \mathbf{i} \) represents the component in the x-direction.
- \( -xy \mathbf{j} \) represents the component in the y-direction.
- \( x \arctan y \mathbf{k} \) represents the component in the z-direction.
By analyzing these components separately, we can infer how the vector field behaves in space.
Flux
Flux measures the flow of a vector field through a surface. It can be visualized as the total "amount" of the field which passes through the surface. In mathematical terms, the flux is given by the surface integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{n} \) is a unit normal to the surface.
The divergence theorem provides a powerful way to compute the flux by converting this surface integral into a volume integral. This conversion often simplifies the calculation significantly. In the given exercise, we use the divergence theorem to find the flux through the surface \( S \), which is bounded by the coordinate planes and the plane \( x + y + z = 1 \).
The divergence theorem provides a powerful way to compute the flux by converting this surface integral into a volume integral. This conversion often simplifies the calculation significantly. In the given exercise, we use the divergence theorem to find the flux through the surface \( S \), which is bounded by the coordinate planes and the plane \( x + y + z = 1 \).
Triple Integral
A triple integral is an extension of the concept of a single or double integral. It computes the integral over a three-dimensional region, providing a way to accumulate quantities throughout a volume.
To apply the divergence theorem, the triple integral becomes particularly useful. In our example, we set up a triple integral over the volume \( V \) enclosed by the surface \( S \). This volume is defined by the bounds \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1-x \), and \( 0 \leq z \leq 1-x-y \).
By evaluating this triple integral, we determine the total contribution from the divergence of \( \mathbf{F} \) across the entire volume, ultimately determining the flux through the surface.
To apply the divergence theorem, the triple integral becomes particularly useful. In our example, we set up a triple integral over the volume \( V \) enclosed by the surface \( S \). This volume is defined by the bounds \( 0 \leq x \leq 1 \), \( 0 \leq y \leq 1-x \), and \( 0 \leq z \leq 1-x-y \).
By evaluating this triple integral, we determine the total contribution from the divergence of \( \mathbf{F} \) across the entire volume, ultimately determining the flux through the surface.
Coordinate Planes
The coordinate planes are the three planes that divide the three-dimensional space: the xy-plane, yz-plane, and xz-plane. Each of these planes can be described as a situation where one of the Cartesian coordinates is zero.
- The xy-plane is where \( z = 0 \).
- The yz-plane is where \( x = 0 \).
- The xz-plane is where \( y = 0 \).
