Problem 6 - 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. $$ \int_{(1,0)}^{(3,4)} \frac{x d x+y d y}{\sqrt{x^{2}+y^{2}}} \text { on any path not through the origin } $$

Step-by-Step Solution

Answer

The line integral is path-independent, with value 4.

Step 1: Understand the prerequisites

The problem requires us to find if the given line integral is path-independent. Our integral must satisfy the conditions for a conservative vector field, allowing us to find a potential function.

Step 2: Verify the conditions for path independence

A vector field is conservative if its curl is zero. The vector field for the line integral is given by \( F(x,y) = \left( \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}} \right) \). Compute the curl: \( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} = 0 \). Since the curl is zero, the vector field is conservative, and the line integral is path-independent.

Step 3: (a) Find the potential function \( \phi \)

A potential function \( \phi \) satisfies \( abla \phi = F \). Integrate \( \frac{x}{\sqrt{x^2+y^2}} \) partially with respect to \( x \), resulting in \( \phi(x,y) = \sqrt{x^2+y^2} + C(y) \). Differentiating \( \phi \) with respect to \( y \), we find \( C'(y)=0 \). Thus, \( \phi = \sqrt{x^2+y^2} \).

Step 4: Apply Theorem 9.9.1

Theorem 9.9.1 states that if \( F \) is conservative with potential function \( \phi \), then the line integral \( \int_C abla \phi \cdot d\mathbf{r} \) depends only on \( \phi(b) - \phi(a) \). Calculate \( \phi(3,4) - \phi(1,0) = 5 - 1 = 4 \).

Step 5: (b) Choose a convenient path

Select a simple path, such as the straight line from \((1,0)\) to \((3,4)\). Parameterize the path as \( x(t) = 1+2t, y(t) = 4t \) with \( t \) in \([0,1]\).

Step 6: Set up and evaluate the integral along this path

Substitute the path into the integral: \( \int_0^1 \frac{(1+2t) \cdot 2 + 4t \cdot 4}{\sqrt{(1+2t)^2+(4t)^2}} dt \). Simplify and compute the integral, resulting in 4, matching our result from the potential function.

Key Concepts

Line Integrals

A line integral is a type of integral where a function is evaluated along a curve or path. In simple terms, you can think of it as measuring the "weight" of a function along a path. Line integrals are commonly used in physics to calculate work done by a force field, for instance, or in engineering problems involving fluid flow.
In this particular problem, we are dealing with a line integral of the form \[ \int_{(1,0)}^{(3,4)} \frac{x \, dx + y \, dy}{\sqrt{x^{2}+y^{2}}} \] which involves integrating components of a vector field along a path from point \((1,0)\) to \((3,4)\).
The crucial feature of line integrals to understand here is their dependence on the path chosen. In some cases, however, they can be determined solely by the endpoints, which brings us to the concept of path independence.

Conservative Vector Fields

A vector field is called a conservative vector field if there exists a potential function such that the vector field is the gradient of this potential function. One important feature of a conservative vector field is that any line integral around a closed loop will equal zero.
In our exercise, the given vector field is \[ F(x,y) = \left( \frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}} \right) \]. To check if it's conservative, we compute its curl, which involves calculating \( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \). If this curl is zero everywhere in some region, then the vector field is conservative—meaning that the integral of the field is independent of the path taken between two points. This independence simplifies solving integrals significantly.

Potential Function

The potential function, often denoted as \( \phi \), is a scalar function whose gradient gives you the vector field. Finding this function is crucial because it allows you to evaluate integrals simply by examining the endpoints.
In our problem, we identify the potential function through integration. Starting with \( abla \phi = F \), one integrates \( \frac{x}{\sqrt{x^2+y^2}} \) with respect to \( x \), giving \( \phi(x,y) = \sqrt{x^2+y^2} + C(y) \). Differentiating with respect to \( y \) and solving for \( C(y) \), we find the simplest solution \( \phi = \sqrt{x^2+y^2} \). Using the potential function, you calculate \( \phi(3,4) - \phi(1,0) = 5 - 1 \), demonstrating that these values give path-independent results of the line integral.

Theorem 9.9.1

Theorem 9.9.1 is a fundamental result in vector calculus that connects potential functions to line integrals. It states that if you have a conservative vector field with a potential function \( \phi \), the line integral from point \( a \) to point \( b \) depends only on the values \( \phi(b) - \phi(a) \).
This theorem substantially simplifies problems of line integrals. Instead of performing potentially complicated integrals along paths, you evaluate the potential function at the endpoints of your interval.
In our exercise, applying Theorem 9.9.1 with the potential function \( \phi(x, y) = \sqrt{x^2 + y^2} \) allowed us to quickly find the integral value as 4, showing once again the power of path independence in conservative vector fields.