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Chapter 5: Problem 56

For the following exercises, the heat flow vector field for conducting objects i \(\mathbf{F}=-k \nabla T\), where \(T(x, y, z)\) is the temperature in the object and \(k>0\) is a constant that depends on the material. Find the outward flux of F across the following surfaces \(S\) for the given temperature distributions and assume \(k=1\) $$ T(x, y, z)=100 e^{-x-y} ; \text { S consists of the faces of cube }|x| \leq 1,|y| \leq 1,|z| \leq 1 $$

Short Answer

Expert verified
The outward flux of \( \mathbf{F} \) across surface \( S \) is 0.

Step by step solution

01

Understand the Vector Field

The heat flow vector field is given by \( \mathbf{F} = -k abla T = -abla T \). Since \( k = 1 \), \( \mathbf{F} = -abla T \). Therefore, we need to find the gradient \( abla T \) of the temperature function \( T(x, y, z) = 100 e^{-x-y} \).
02

Compute the Gradient \( \nabla T \)

The gradient of \( T \) is computed as:\[ abla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right) \]For \( T(x, y, z) = 100 e^{-x-y} \), we have:\[ \frac{\partial T}{\partial x} = -100 e^{-x-y}, \quad \frac{\partial T}{\partial y} = -100 e^{-x-y}, \quad \frac{\partial T}{\partial z} = 0 \]Therefore, \( abla T = (-100 e^{-x-y}, -100 e^{-x-y}, 0) \).
03

Define the Vector Field \( \mathbf{F} \)

Since \( \mathbf{F} = -abla T \), we find:\[ \mathbf{F} = (100 e^{-x-y}, 100 e^{-x-y}, 0) \]
04

Set Up the Integral for Flux

The outward flux through surface \( S \) is given by the surface integral:\[ \iint_S \mathbf{F} \cdot \mathbf{n} \, dS \]where \( \mathbf{n} \) is the outward unit normal vector to surface \( S \).
05

Parameterize Each Face of the Cube

The cube has six faces. For simplicity, we only need to consider contributions for faces where the normal has non-zero components. Calculate flux for faces with consistent changes in \( x \) and \( y \) directions, namely, faces perpendicular to the \( x \) and \( y \) axes.
06

Flux Calculation on \( x = \pm 1 \) Faces

For \( x = 1 \) and \( x = -1 \), the normals are \( \mathbf{n} = (\pm 1, 0, 0) \). Calculate:\[ \text{Flux on } x=1 = \int_{-1}^{1} \int_{-1}^{1} \left(100 e^{-1-y} \right)(1) \ dy \, dz = 200 \sinh(1) \] \[ \text{Flux on } x=-1 = -\int_{-1}^{1} \int_{-1}^{1} \left(100 e^{1-y} \right)(-1) \ dy \, dz = -200 \sinh(1) \]
07

Flux Calculation on \( y = \pm 1 \) Faces

For \( y = 1 \) and \( y = -1 \), the normals are \( \mathbf{n} = (0, \pm 1, 0) \). Calculate:\[ \text{Flux on } y=1 = \int_{-1}^{1} \int_{-1}^{1} \left(100 e^{-x-1} \right)(1) \ dx \, dz = 200 \sinh(1) \] \[ \text{Flux on } y=-1 = -\int_{-1}^{1} \int_{-1}^{1} \left(100 e^{-x+1} \right)(-1) \ dx \, dz = -200 \sinh(1) \]
08

Sum Contributions from Relevant Cube Faces

Combine all faces' contributions. Other four faces do not contribute as they either cancel out or the flux integrals evaluate to zero (as in \( z \) direction contributions are zero for symmetric assembly in \( z \)). Total flux is zero since the contributions from opposite faces cancel each other.

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Key Concepts

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

Flux Calculation
The concept of flux in vector calculus is vital for understanding how a field behaves through a surface. Specifically, in this exercise, we are dealing with the outward flux of a heat flow vector field denoted as \( \mathbf{F} \). This field represents the flow of heat through a conducting object. To know how much of the vector field penetrates through a surface, we must calculate the surface integral of \( \mathbf{F} \) across a surface \( S \), using the formula:\[\iint_S \mathbf{F} \cdot \mathbf{n} \, dS\]Here, \( \mathbf{n} \) is the outward unit normal vector at each point of the surface.
  • The "dot product" \( \mathbf{F} \cdot \mathbf{n} \) measures how much of the vector field passes through the surface perpendicularly.
  • The surface integral sums this quantity over the entire surface \( S \).
In our exercise, the surface \( S \) is a cube, made up of six faces. To calculate the total flux, we assess each face individually, noting the symmetry that allows for simplifications.
Gradient Vector
The gradient vector, denoted as \( abla T \), is essential for understanding how temperature changes within the object. In this problem, the temperature \( T \) is given by a particular function of \( x \), \( y \), and \( z \): \( T(x, y, z) = 100 e^{-x-y} \).
  • The gradient is a vector that points in the direction of the greatest increase of the function.
  • It is determined by partial derivatives of \( T \) with respect to each variable.
Mathematically, the gradient is computed as:\[abla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z} \right)\]Each component tells us the rate and direction of change of temperature along the respective axis. In our exercise:
  • \( \frac{\partial T}{\partial x} = -100 e^{-x-y} \)
  • \( \frac{\partial T}{\partial y} = -100 e^{-x-y} \)
  • \( \frac{\partial T}{\partial z} = 0 \)
Thus, the gradient vector \( abla T \) is \( (-100 e^{-x-y}, -100 e^{-x-y}, 0) \), which directly influences the formulation of the vector field \( \mathbf{F} = -abla T \).
Heat Flow
Heat flow, represented by the vector field \( \mathbf{F} \), is the main subject of our calculation. This flow is described by the equation \( \mathbf{F} = -k abla T \), where \( k \) is a constant depending on the material. It quantifies how heat moves within an object and radiates outward through its surfaces.
  • The negative sign in the equation signifies that heat flows from areas of higher temperature to lower temperature, opposite the direction of the gradient.
  • In our problem, we have \( k = 1 \), simplifying \( \mathbf{F} = -abla T \).
Understanding heat flow is crucial for calculating the amount of energy being transferred across a surface. The calculation of the flux gives us insight into this transfer, allowing for predictions and analysis of thermal behaviors in various materials.This process of determining the outward flux, assisted by knowing \( abla T \), gives us a comprehensive picture of how the temperature distribution within an object effects the heat flow across its boundary.

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Most popular questions from this chapter

A lamina has the shape of a portion of sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) that lies within cone \(z=\sqrt{x^{2}+y^{2}}\). Let \(S\) be the spherical shell centered at the origin with radius \(a\), and let \(C\) be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the \(z\) -axis. Suppose the vertex angle of the cone is \(\phi_{0}\), with \(0 \leq \phi_{0}<\frac{\pi}{2} .\) Determine the mass of that portion of the shape enclosed in the intersection of \(S\) and \(C .\) Assume \(\delta(x, y, z)=x^{2} y^{2} z .\)

Find the divergence of \(\mathrm{F}\) at the given point. $$ \mathbf{F}(x, y, z)=e^{x} \sin y \mathbf{i}-e^{x} \cos y \mathbf{j} \text { at }(0,0,3) $$

For the following exercises, describe each vector field by drawing some of its vectors. $$ \mathbf{F}(x, y)=y \mathbf{i}+\sin x \mathbf{j} $$

For the following exercises, evaluate \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d s\) for vector field \(F\), where \(\mathbf{N}\) is an outward normal vector to surface \(S\). \(\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}\), and \(S\) is hemisphere \(z=\sqrt{1-x^{2}-y^{2}} .\)

For the following exercises, Fourier's law of heat transfer states that the heat flow vector \(\mathrm{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T\), which means that heat energy flows hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of joules per meter per second-kelvin or watts per meter-kelvin. A temperature function for region \(D\) is given. Use the divergence theorem to find net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{N} d S=-k \iint_{S} \nabla T \cdot \mathbf{N} d S\) across the boundary \(S\) of \(D\), where \(k=1$$T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}} ; D\) is the sphere of radius a centered at the origin.
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