Problem 50 - Calculus Early Transcendentals

Question

Evaluate the integral \(\int_{-C} \frac{x d y-y d x}{x^{2}+y^{2}},\) where \(C\) is the circle \(x^{2}+y^{2}=a^{2}\) traversed counterclockwise.

Step-by-Step Solution

Answer

The integral evaluates to \(-2\pi\).

Step 1: Identify Integration Path and Parameterize

The task is to evaluate the integral \(\int_{-C} \frac{x dy - y dx}{x^2 + y^2}\), where the path \(-C\) is the circle \(x^2 + y^2 = a^2\) traversed counterclockwise. First, establish a parameterization for this path. For a circle of radius \(a\), we use the parameterization \(x = a \cos(t)\) and \(y = a \sin(t)\), where \(t\) ranges from \(0\) to \(2\pi\) when traced counterclockwise. Since the integral is over \(-C\), it implies reversing the direction, which effectively means changing the limits of integration or adding a negative sign to the entire integral later.

Step 2: Express Differential Elements

With the parameterization of \(x\) and \(y\) as \(x = a \cos(t)\) and \(y = a \sin(t)\), compute the differential elements: \(dx = -a \sin(t)dt\) and \(dy = a \cos(t)dt\). These expressions will substitute into the integral.

Step 3: Substitute and Simplify the Integral

Substitute \(x = a \cos(t)\), \(y = a \sin(t)\), \(dx\), and \(dy\) into the integral. The integral becomes: \[\int_0^{2\pi} \frac{a \cos(t) \cdot a \cos(t)dt - a \sin(t) \cdot (-a \sin(t))dt}{a^2 \cos^2(t) + a^2 \sin^2(t)}\].Since \(x^2 + y^2 = a^2\), this simplifies to:\[ a^2 \int_0^{2\pi} \frac{\cos^2(t) + \sin^2(t)}{a^2} dt = \int_0^{2\pi} 1 dt\].This results in : \(\int_0^{2\pi} 1 dt = 2\pi\).

Step 4: Correct for the Reversed Path

Since the integral is actually over \(-C\) (not \(C\), which was traced counter-clockwise), we need to account for the path reversal. Thus, multiply the result from the previous step by \(-1\), giving: \(- 2\pi\). This produces the final result for the integral.

Key Concepts

Parameterization

Parameterization is a technique used to describe a curve in terms of a single variable, typically denoted as \( t \). In the context of a circular path, this makes evaluating integrals much simpler because it transforms a problem involving coordinates \( x \) and \( y \) into a problem involving \( t \). The path of a circle is often parameterized with trigonometric functions because these functions naturally relate to circles. For example, we use \( x = a \cos(t) \) and \( y = a \sin(t) \) to represent a circle of radius \( a \). Here, \( t \) varies from \( 0 \) to \( 2\pi \), covering the whole circle in one counterclockwise loop.This parameterization helps us compute differentials like \( dx \) and \( dy \) easily, enabling straightforward substitution into the integral, simplifying our calculations.

Coordinates Transformation

Coordinates transformation involves changing variables to make problems easier to solve. In our context, it particularly simplifies the calculation of line integrals along curves. Starting from the parameterized equations \( x = a \cos(t) \) and \( y = a \sin(t) \), we differentiate with respect to \( t \) to find \( dx \) and \( dy \). Here, \( dx = -a \sin(t)dt \) and \( dy = a \cos(t)dt \). These transformations allow us to replace \( x \) and \( y \) in our integral with their expressions in terms of \( t \). This converts the integral into a single-variable integral, which is generally much easier to evaluate. In essence, it leverages the nature of circles and trigonometric identities to simplify the problem.

Counterclockwise Traversal

Counterclockwise traversal is a specific direction of moving around a path, which is important in many mathematical contexts such as Green's Theorem or line integrals. Standard orientation for measures like curves is usually counterclockwise, but sometimes problems require us to adjust this.When dealing with line integrals around a closed path, the direction of traversal affects the sign of the integral. In the original problem, the circle is traversing counterclockwise, but the integral is over \(-C\), implying we change direction. To adjust for reversed traversal, we place a negative sign to indicate clockwise movement. This little adjustment significantly impacts the integral result, transforming \( 2\pi \) to \(-2\pi\). This shows how important understanding traversal direction is for the correct calculation of line integrals.