Problem 18 - Thomas Calculus

Question

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 \(6 .\) $$\int_{(0,2,1)}^{(1, \pi / 2,2)} 2 \cos y d x+\left(\frac{1}{y}-2 x \sin y\right) d y+\frac{1}{z} d z$$

Step-by-Step Solution

Answer

The value of the integral is \(\ln\left(\frac{\pi}{2}\right)\).

Step 1: 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.

Step 2: Compute partial derivatives for potential function

Assume the potential function \(f(x, y, z)\) such that \[\frac{\partial f}{\partial x} = 2\cos y, \quad \frac{\partial f}{\partial y} = \frac{1}{y} - 2x \sin y, \quad \frac{\partial f}{\partial z} = \frac{1}{z}\]

Step 3: 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 \(z\).

Step 4: 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:\[g(y, z) = \ln |y| + h(z)\]

Step 5: 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:\[f(x, y, z) = 2x \cos y + \ln |y| + \ln |z| + C\]

Step 6: 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:\[ \ln(\pi) - \ln(2) = \ln \left( \frac{\pi}{2} \right) \]

Key Concepts

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’s 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., \( \vec{F} = abla f \).

When dealing with vector fields in physics or engineering, conservativity often represents energy conservation or equilibrium conditions. Identifying a field as conservative lets us use potential energy concepts to ease computations.

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.

When handling conservative fields, calculating a potential function becomes a powerful tool as integrals over any path between two points boil down to a simple evaluation of potential differences.

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’ function values are necessary.

The elegance of the line integral in conservative fields underscores its importance across disciplines, allowing elegant, simplified solutions for what would be laborious path calculations otherwise.

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 \( f \).

Using partial derivatives enables us to tie components of vector fields directly to functions describing their potential energy, and find potential functions through methodical integration of each component separately. They form foundational tools in establishing relations between fields and scalar potentials.