Problem 173 - Calculus Volume 3

Question

Evaluate \(\int \frac{x d x+y d y}{C x^{2}+y^{2}},\) where \(C\) is any piecewise, smooth simple closed curve enclosing the origin, traversed counterclockwise.

Step-by-Step Solution

Answer

The integral evaluates to zero.

Step 1: Understanding the Problem

We need to evaluate the integral \( \int \frac{x dx + y dy}{Cx^2 + y^2} \) where \( C \) is a closed curve enclosing the origin. As it's a simple closed curve enclosing the origin, we can consider converting this problem to polar coordinates.

Step 2: Convert to Polar Coordinates

In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). Hence, the expressions modify to \( x^2 + y^2 = r^2 \), \( dx = -r \sin \theta d \theta + \cos \theta dr \), and \( dy = \cos \theta d \theta + \sin \theta dr \). Substituting them into the integral terms.

Step 3: Evaluate the Integral using Polar Coordinates

The numerator becomes \( x dx + y dy = r \cos \theta (-r \sin \theta d\theta + \cos \theta dr) + r \sin \theta (\cos \theta d\theta + \sin \theta dr) \), which simplifies to \( r dr \). The denominator becomes \( Cx^2 + y^2 = Cr^2 + r^2 = (C + 1)r^2 \). Thus, the integral simplifies to \( \int \frac{r dr}{(C+1)r^2} \), simplifying further to \( \frac{1}{C+1} \int \frac{dr}{r} \).

Step 4: Integrate and Identify the Integration Bounds

The integral \( \int \frac{dr}{r} \) evaluates to \( \ln |r| \). Since \( r \) runs around a path enclosing the origin, the integration bounds are effectively a full circle, which cancels out as it covers a full loop in polar coordinates.

Step 5: Analyze Logarithmic Properties

Due to the nature of the logarithm \( \ln |r| \), and the fact that the path is a closed loop enclosing the origin, the boundary evaluation cancels out, leaving the net integral for the entire path equal to zero.

Key Concepts

Contour Integration

In complex analysis, contour integration is a technique used to evaluate certain types of integrals in the complex plane. It involves integrating a complex function along a prescribed path or contour.
These contours can be open or closed, but for this exercise, we'll focus on closed contours.
This method is particularly useful when dealing with complex functions that have singularities.

• Closed contours are loops with no endpoint; they start and end at the same point.
• Integrals over closed contours often reveal interesting properties because of these loops.

Using contour integration, we can efficiently evaluate integrals like the one in the exercise by considering properties of the function and the curve.

Polar Coordinates

Converting to polar coordinates can greatly simplify calculations in problems involving circles or curves around the origin.
In polar coordinates, any point in the plane is determined by a distance from the origin, denoted as \( r \), and an angle \( \theta \).

• The relationships between Cartesian and polar coordinates: \( x = r \cos \theta \) and \( y = r \sin \theta \).
• Change in a curve using polar involves modifying differentials: \( dx = -r \sin \theta d \theta + \cos \theta dr \) and \( dy = \cos \theta d\theta + \sin \theta dr \).

This transformation lets us handle problems concerning circular or rotating systems with ease. For integrating along a closed path as in this exercise, polar coordinates simplify the integrals due to the radial symmetry.

Closed Curve

A closed curve in our context is a path in the complex plane that begins and ends at the same point, forming a loop.
It's important in our problem because such curves enclose an area, and they typically relate to integrals yielding zero, akin to Green's theorem context in real analysis.

• Simple closed curves don’t intersect themselves.
• This property is fundamental to applying integration theorems in complex analysis.

For our integral, the understanding that it traverses a simple closed curve enclosing the origin helps simplify calculations profoundly, often leading to intuitive results due to the symmetry and loops involved.

Cauchy's Theorem

Cauchy's Theorem is a cornerstone of complex analysis and is pivotal in evaluating contour integrals around closed curves.
It states that if a function is analytic inside and on some closed contour, then the integral of that function along the curve is zero.

• The theorem applies to functions without singularities inside the contour.
• It greatly simplifies calculations as it can sometimes negate the need for detailed computation.

In our exercise, although the integral represents a real-valued function rather than complex, the concepts surrounding Cauchy's theorem, notably the behavior of integrals over closed paths, suggests why the result of the integral is zero, given the closed loop's characteristics.