Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 16: Problem 18
Although they are not defined on all of space \(R^{3},\) the fields associated with Exercises \(18-22\) are conservative. Find a potential function for each field and evaluate the integrals as in Example \(6 .\) $$\int_{(0,2,1)}^{(1, \pi / 2,2)} 2 \cos y d x+\left(\frac{1}{y}-2 x \sin y\right) d y+\frac{1}{z} d z$$
Short Answer
Expert verified
The value of the integral is \(\ln\left(\frac{\pi}{2}\right)\).
Step by step solution
01
Identify the vector field
The vector field given by the integral is \[\vec{F}(x, y, z) = \langle 2\cos y, \frac{1}{y} - 2x\sin y, \frac{1}{z} \rangle\]This is the field for which we need to find a potential function.
02
Compute partial derivatives for potential function
Assume the potential function \(f(x, y, z)\) such that \[\frac{\partial f}{\partial x} = 2\cos y, \quad \frac{\partial f}{\partial y} = \frac{1}{y} - 2x \sin y, \quad \frac{\partial f}{\partial z} = \frac{1}{z}\]
03
Integrate with respect to x to find f(x,y,z)
Integrating \(\frac{\partial f}{\partial x} = 2\cos y\) with respect to \(x\), we get:\[f(x, y, z) = 2x \cos y + g(y, z)\]where \(g(y, z)\) is an arbitrary function of \(y\) and \(z\).
04
Differentiate with respect to y and integrate
Differentiate the derived expression:\[\frac{\partial f}{\partial y} = -2x \sin y + \frac{\partial g}{\partial y}\]Set this equal to the given partial derivative \(\frac{1}{y} - 2x \sin y\):\[-2x \sin y + \frac{\partial g}{\partial y} = \frac{1}{y} - 2x \sin y\]Thus, \(\frac{\partial g}{\partial y} = \frac{1}{y}\).Integrate with respect to \(y\) to find:\[g(y, z) = \ln |y| + h(z)\]
05
Differentiate with respect to z and integrate
Differentiate the updated expression:\[\frac{\partial f}{\partial z} = \frac{dh}{dz}\]Set this equal to the given partial derivative \(\frac{1}{z}\):\[\frac{dh}{dz} = \frac{1}{z}\]Integrate with respect to \(z\):\[h(z) = \ln |z| + C\]Combine results to get the potential function:\[f(x, y, z) = 2x \cos y + \ln |y| + \ln |z| + C\]
06
Evaluate integral using potential function
We now evaluate \(f(x,y,z)\) from \((0,2,1)\) to \((1,\pi/2,2)\) using the formula:\[\int_C \vec{F} \cdot d\vec{r} = f(1, \pi/2, 2) - f(0, 2, 1)\]Calculate:- \( f(1, \pi/2, 2) = 2 \cdot 1 \cdot \cos(\pi/2) + \ln| \pi/2 | + \ln| 2 | = \ln(\pi/2) + \ln(2) = \ln(\pi) \)- \( f(0, 2, 1) = 2 \cdot 0 \cdot \cos(2) + \ln| 2 | + \ln| 1 | = \ln(2) \)Therefore, the integral evaluates to:\[ \ln(\pi) - \ln(2) = \ln \left( \frac{\pi}{2} \right) \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservative Field
A conservative vector field is a field where the line integral is path-independent, which means the total work done in moving along a path from one point to another is the same regardless of the path taken. This has significant implications when computing integrals because it simplifies work to evaluating using a potential function instead of along a specific path.
- Conservativity implies a certain simplicity in computation, allowing integration from a field鈥檚 start to end points directly through potential differences.
- In mathematical terms, a vector field \( \vec{F} \) is conservative if it can be expressed as the gradient of some scalar potential function \( f(x, y, z) \), i.e., \( \vec{F} = abla f \).
Potential Function
The potential function, often denoted \( f(x, y, z) \), is a scalar function whose gradient equals the vector field \( \vec{F} \). This means \( abla f = \vec{F} \). If a potential function exists, it simplifies calculations significantly, especially in physics or engineering tasks, by providing a compact representation of force fields.
- The process to find a potential function involves integration of the vector field components.
- This can be seen in the steps where individual components are integrated to get terms of \( f \), and leftover functions of the other variables are incorporated progressively.
Line Integral
A line integral of a vector field calculates the total effect (like work or energy) along a specified path or curve. It is denoted as \( \int_C \vec{F} \cdot d\vec{r} \), where \( \vec{F} \) is your vector field and \( d\vec{r} \) is a differential element of the path over which the integral is being evaluated.
- In conservative fields, the integral simplifies as path independence implies it equals the potential difference between endpoints.
- This offers elegant shortcuts: calculation need not traverse the whole path, only the start and end points鈥� function values are necessary.
Partial Derivatives
Partial derivatives denote rate of change concerning one variable, keeping others constant, and are notated \( \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \), etc. These derivatives are crucial in multi-variable calculus.
- The potential function \( f(x, y, z) \) is found by computing partial derivatives matching each vector field component.
- For example, \( \frac{\partial f}{\partial x} = F_x \) ties directly to integration steps to derive \( f \).
