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Magazine Chapter 13: Problem 45
The temperature at the point \((x, y, z)\) in a substance with conductivity \(K=6.5\) is \(u(x, y, z)=2 y^{2}+2 z^{2} .\) Find the rate of heat flow inward across the cylindrical surface \(y^{2}+z^{2}=6,0 \leqslant x \leqslant 4\).
Short Answer
Expert verified
The rate of heat flow is \(-1248\pi\).
Step by step solution
01
Understanding the Problem
We need to find the rate of heat flow across a cylindrical surface defined by \(y^{2}+z^{2}=6\) and bounded between \(x=0\) and \(x=4\). This requires using the formula for heat flow through a surface using the temperature function and thermal conductivity.
02
Setting Up the Heat Flow Formula
The formula for heat flow \( \,\Phi\, \) is given by \( \Phi = -K \int_S abla u \cdot \mathbf{n} \, dS \), where \(K\) is the thermal conductivity, \(abla u\) is the gradient of temperature \(u\), and \(\mathbf{n}\) is the unit normal to the surface \(S\).
03
Computing the Gradient \(\nabla u\)
The temperature function is \(u(x, y, z) = 2y^2 + 2z^2\). Compute the gradient: \(abla u = (0, 4y, 4z)\), because the partial derivatives are \(\frac{\partial u}{\partial x} = 0\), \(\frac{\partial u}{\partial y} = 4y\), \(\frac{\partial u}{\partial z} = 4z\).
04
Determining the Normal Vector \(\mathbf{n}\)
The unit normal vector \(\mathbf{n}\) to the cylindrical surface \(y^2 + z^2 = 6\) is \((0, \frac{y}{\sqrt{6}}, \frac{z}{\sqrt{6}})\) because it is normal to the radial direction \((0, y, z)\) scaled to unit length.
05
Calculating \(\nabla u \cdot \mathbf{n}\)
Substitute the expressions from Steps 3 and 4: \( abla u \cdot \mathbf{n} = (0, 4y, 4z) \cdot (0, \frac{y}{\sqrt{6}}, \frac{z}{\sqrt{6}}) = \frac{4y^2}{\sqrt{6}} + \frac{4z^2}{\sqrt{6}}\). Simplifying gives \( abla u \cdot \mathbf{n} = \frac{4(y^2 + z^2)}{\sqrt{6}} = \frac{24}{\sqrt{6}}\).
06
Evaluating the Surface Integral
The surface \(S\) is a cylinder with base circles defined by \(y^2 + z^2 = 6\) and height \(4\). Thus, \(dS\) is the lateral surface area of the cylinder, \(dS = \sqrt{6} \, dx \, d\theta\), using polar coordinates. The integral becomes \(-K \int_{x=0}^{4} \int_{\theta=0}^{2\pi} \frac{24}{\sqrt{6}} \times \sqrt{6} \, dx \, d\theta\).
07
Solving the Integral
Integrate over \(x\) and \(\theta\): \(-6.5 \int_{0}^{4} \int_{0}^{2\pi} 24 \, dx \, d\theta = -6.5 \times 24 \times 4 \times 2\pi = -6.5 \times 192 \pi \).
08
Final Calculation
Compute the result: \(-6.5 \times 192 \pi = -1248\pi\). The negative sign indicates the direction of heat flow.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient of Temperature
In calculus, the gradient of a scalar field represents the rate and direction of change in the field. For temperature, the gradient tells us how temperature changes spatially in a medium. In this example, the temperature function is given by \( u(x, y, z) = 2y^2 + 2z^2 \).
The gradient, noted as \( abla u \), is calculated by taking the partial derivatives with respect to each variable: \( x, y, \) and \( z \). This serves to identify how the temperature alters in each direction:
The gradient, noted as \( abla u \), is calculated by taking the partial derivatives with respect to each variable: \( x, y, \) and \( z \). This serves to identify how the temperature alters in each direction:
- The partial derivative with respect to \( x \) is zero since the temperature function does not change with \( x \).
- With respect to \( y \), it is \( 4y \), meaning the temperature changes four times faster in the \( y \) direction.
- Similarly, the partial derivative with respect to \( z \) is \( 4z \), indicating a similar rate of change in the \( z \) direction.
Thermal Conductivity
Thermal conductivity is a property of materials that signifies their ability to conduct heat. In equations, it is denoted by the symbol \( K \), and it significantly influences the rate of heat flow in a substance. For the given problem, the constant value is \( K = 6.5 \).
It acts as a scaling factor in the heat flow formula:
It acts as a scaling factor in the heat flow formula:
- Incorporated into the formula as the term \( -K \int_S abla u \cdot \mathbf{n} \, dS \), where \( K \) adjusts the strength of the heat flow modeled by the gradient term \( abla u \) across the surface \( S \).
- Materials with high thermal conductivity will transfer heat more rapidly than those with lower values.
Surface Integral
A surface integral is a powerful mathematical tool used for calculating the total magnitude of a field passing through a surface. In the context of heat flow, this is where its utility shines. It helps to integrate the product of the temperature gradient and the normal vector over a surface, like our cylinder in the given example.
For the cylinder, defined by \( y^2 + z^2 = 6 \) from \( x = 0 \) to \( x = 4 \), the integral setup involves:
For the cylinder, defined by \( y^2 + z^2 = 6 \) from \( x = 0 \) to \( x = 4 \), the integral setup involves:
- The area element \( dS \), representing a tiny section of the cylinder's surface, is calculated using cylindrical coordinates as \( \sqrt{6} \, dx \, d\theta \).
- The integral then runs over this lateral surface area to evaluate the total heat flow, expressed ultimately as \( \int_{x=0}^{4} \int_{\theta=0}^{2\pi} \frac{24}{\sqrt{6}} \times \sqrt{6} \, dx \, d\theta \).
- This combined with the conductivity \( K \) gives the final physical result.
