Problem 4 - Advanced Engineering Mathematics

Question

Show that the given line integral is independent of the path. Evaluate in two ways: (a) Find a potential function \(\phi\) and then use Theorem 9.9.1, and (b) Use any convenient path between the endpoints of the path. $$ \int_{(0,0)}^{(\pi / 2,0)} \cos x \cos y d x+(1-\sin x \sin y) d y $$

Step-by-Step Solution

Answer

The line integral evaluates to 1 and is path independent.

Step 1: Determine the Condition for Path Independence

For the line integral to be independent of the path, the vector field \( \mathbf{F} = \langle \cos x \cos y, 1 - \sin x \sin y \rangle \) must be conservative. This means that there exists a potential function \( \phi(x, y) \) such that \( abla \phi = \mathbf{F} \). For \( \mathbf{F} \) to be conservative, its components must satisfy \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), where \( P(x, y) = \cos x \cos y \) and \( Q(x, y) = 1 - \sin x \sin y \).

Step 2: Verify Curl Condition

Calculate \( \frac{\partial P}{\partial y} = -\cos x \sin y \) and \( \frac{\partial Q}{\partial x} = -\cos x \sin y \). Since \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), the line integral is independent of the path.

Step 3: Find the Potential Function

Since \( \mathbf{F} = abla \phi \), we integrate \( P = \cos x \cos y \) with respect to \( x \) to obtain \( \phi(x, y) = \sin x \cos y + h(y) \), where \( h(y) \) is a function of \( y \) only.

Step 4: Determine Function h(y)

Differentiate \( \phi(x, y) \) with respect to \( y \): \( \frac{\partial \phi}{\partial y} = -\sin x \sin y + h'(y) \). Since \( \frac{\partial \phi}{\partial y} = Q = 1 - \sin x \sin y \), equate terms to find \( h'(y) = 1 \), which integrates to \( h(y) = y + C \). Thus, \( \phi(x, y) = \sin x \cos y + y + C \).

Step 5: Apply Theorem 9.9.1 to Evaluate the Integral

Use Theorem 9.9.1 states that if \( \phi \) is a potential function for \( \mathbf{F} \), then \( \int_{C} \mathbf{F} \cdot d\mathbf{r} = \phi(\pi/2,0) - \phi(0,0) \). Calculate \( \phi(\pi/2, 0) = 1 + 0 + C \) and \( \phi(0, 0) = 0 + 0 + C \). Therefore, the line integral evaluates to \( 1 + C - C = 1 \).

Step 6: Evaluate the Integral Using a Convenient Path

Choose the path consisting of two line segments: from \( (0,0) \) to \( (\pi/2,0) \), the path is straight along the x-axis. Here, \( dy = 0 \), so \( \int_{0}^{\pi/2} \cos x \cos 0 \, dx = \int_{0}^{\pi/2} \cos x \, dx = \left. \sin x \right|_{0}^{\pi/2} = 1 \). The result is consistent with Theorem 9.9.1.

Key Concepts

Conservative Vector Fields

In vector calculus, a vector field is termed conservative if it is path-independent. This means the integral of the vector field between two points is the same for any path connecting them.
To determine if a vector field \( \mathbf{F} \) is conservative, one needs to check if it satisfies the curl condition given by \( abla \times \mathbf{F} = \mathbf{0} \). This is equivalent to verifying that the partial derivatives satisfy the relationship \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), where \( \mathbf{F} = \langle P, Q \rangle \).
If this condition holds, there exists a potential function \( \phi \) such that the vector field is the gradient of this function \( \mathbf{F} = abla \phi \). This characteristic makes conservative vector fields very useful because they simplify computations of line integrals.

Potential Functions

Potential functions are crucial in understanding conservative vector fields. A potential function \( \phi(x, y) \) is a scalar function whose gradient gives the vector field, \( abla \phi = \mathbf{F} \).
To find a potential function, you perform integral operations on the components of the vector field. Starting with one component, integrate it partially to obtain a general form for \( \phi \).

• For example, integrating \( P(x, y) = \cos x \cos y \) with respect to \( x \) gives \( \phi(x, y) = \sin x \cos y + h(y) \), where \( h(y) \) is a function only of \( y \).
• You then determine \( h(y) \) by ensuring the derived function satisfies the remaining part of the vector field. For \( Q(x, y) = 1 - \sin x \sin y \), differentiating \( \phi(x, y) \) with respect to \( y \) and equating with \( Q \) helps find \( h(y) \).

Thus, the potential function uniquely describes the vector field in question, simplifying our computations.

Path Independence

Path independence is a powerful concept in vector calculus that tells us the value of an integral is the same regardless of the path taken. For a vector field to be path independent between two points, it must be conservative.
This makes computations easier as we don't need to consider each specific path between points.

• If a potential function \( \phi \) exists, the evaluation of line integrals boils down to calculating \( \phi \) at the endpoints, thus making the path irrelevant.
• Theorem 9.9.1 leverages this by providing a quick way to determine the value based solely on the potential function: \( \int_{C} \mathbf{F} \cdot d \mathbf{r} = \phi(b) - \phi(a) \).

This results in significant computational advantages, especially for complex or intricate paths.

Line Integrals

Line integrals allow us to integrate functions along a curve or path. They are foundational in understanding the work done by a force field along a path.
The line integral of a vector field \( \mathbf{F} = \langle P, Q \rangle \) over a curve \( C \) is computed as:

• \( \int_{C} P \,dx + Q \,dy \)

For conservative vector fields, line integrals become particularly simple.

• With path independence, you only need to compute the difference in potential functions at the endpoints, greatly simplifying the process.
• Choosing a specific path can sometimes make calculations easier, as shown when integrating along segments where the integral becomes straightforward, like integrating \( \cos x \) from \( 0 \) to \( \pi/2 \).

Thus, the concept of line integrals is invaluable, providing insight and practical tools in fields such as physics and engineering.