Problem 33 - Thomas Calculus

Question

Area as a line integral Show that if \(R\) is a region in the plane bounded by a piecewise smooth, simple closed curve \(C,\) then $$ R=\oint_{C} x d y=-\oint_{C} y d x $$

Step-by-Step Solution

Answer

Use Green's Theorem to verify: both integrals equal the area of region \( R \).

Step 1: Understanding the Problem

We are given a region \( R \) in the plane bounded by a piecewise smooth, simple closed curve \( C \). We need to show that the integral expressions for the area \( \oint_{C} x \, dy \) and \( -\oint_{C} y \, dx \) are equal.

Step 2: Introduction to Green's Theorem

Green's Theorem states that for a simple closed curve \( C \) positively oriented, enclosing a region \( R \), \( \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). We'll use this to relate the line integral to an area integral.

Step 3: Setting the Appropriate Functions

To achieve the desired area representation, let's set \( P = 0 \) and \( Q = x \) for \( \oint_{C} x \, dy \), and \( P = -y \) and \( Q = 0 \) for \( -\oint_{C} y \, dx \).

Step 4: Applying Green's Theorem for \( R = \oint_{C} x \, dy \)

Applying Green’s theorem with \( P = 0 \) and \( Q = x \), we have \( \iint_{R} \left( \frac{\partial x}{\partial x} - \frac{\partial 0}{\partial y} \right) \, dA = \iint_{R} 1 \, dA \), which reduces to the area \( A \) of \( R \).

Step 5: Applying Green's Theorem for \( R = -\oint_{C} y \, dx \)

Similarly, applying Green’s theorem with \( P = -y \) and \( Q = 0 \), we have \( \iint_{R} \left( \frac{\partial 0}{\partial x} - \frac{\partial (-y)}{\partial y} \right) \, dA = \iint_{R} 1 \, dA \), which also reduces to the area \( A \) of \( R \).

Step 6: Concluding Equality of Integrals

Since both approaches using Green's Theorem lead to the same result of \( A \), the original integrals must indeed represent the area of \( R \), verifying that \( \oint_{C} x \, dy = -\oint_{C} y \, dx \).

Key Concepts

Green's Theorem: Bridging Line and Area Integrals

At the heart of the exercise is Green’s Theorem, a fundamental result in vector calculus. This theorem provides the connection between a line integral around a simple closed curve and a double integral over the region it encloses. Simply put, it allows us to relate the circulation around a curve to the flux through the surface it bounds. In mathematical terms, if you have a positively oriented, simple closed curve \( C \) that encompasses a region \( R \), Green's Theorem states that:

• \( \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).

Here, \( P \) and \( Q \) are functions of \( x \) and \( y \). By strategically choosing these functions, Green's Theorem turns the complex problem of determining a line integral into a more manageable area computation.
To find the area contained by \( C \), one sets the functions \( P \) and \( Q \) to be either zero or a simple variable, simplifying the expression significantly. As seen in the original solution, such choices allowed the integrals \( \oint_{C} x \, dy \) and \( -\oint_{C} y \, dx \) to reflect the area of the region \( R \).
This powerful connection helps us understand not just line integrals in isolation, but also their possible interpretations in terms of physical space and areas.

Calculating Area Using Line Integrals

Calculating the area of a region using line integrals might seem counterintuitive at first. However, the use of Green's Theorem shows us that the line integral around the curve \( C \) can equate to computing the area \( A \) of the enclosed region \( R \). Here's the step-by-step idea:

• For \( \oint_{C} x \, dy \), set \( P = 0 \) and \( Q = x \). The term \( \frac{\partial Q}{\partial x} \) simplifies to 1, reducing the area integral \( \iint_{R} 1 \, dA \) to simply the area \( A \).
• For \( -\oint_{C} y \, dx \), set \( P = -y \) and \( Q = 0 \). Similarly, \( \frac{\partial P}{\partial y} \) simplifies to -1, resulting again in the area integral \( \iint_{R} 1 \, dA \), which is \( A \).

Both approaches surprisingly lead to determining area, aligning with the intuitive notion of areas enclosed by curves in geometry. It illustrates how vector calculus allows us to reinterpret classic geometric ideas into new, analytical forms.

Understanding the Simple Closed Curve

A simple closed curve is the boundary of the region we are interested in calculating the area for. But what exactly does "simple closed curve" mean?

• "Simple" implies that the curve does not cross itself. It's much like drawing a line from a point and returning to the same point without lifting your pencil or retracing over any line.
• "Closed" means the curve forms a complete loop with no ends — it defines a distinct inside and outside, enclosing a region.

The condition of the curve being simple and closed is crucial for applying Green’s Theorem. Only when these conditions are satisfied does the transformation between a line integral and an area integral hold true. For practical purposes, this means that any line integral calculated around such a curve will directly relate to the area calculation inside, offering a seamless way to analyze the geometry enclosed by the curve.
The beauty of using simple closed curves lies in their ability to easily demonstrate how intricately connected the properties of a curve and the contained areas are. This serves as a perfect example of how complex problems, when constrained by 'simple' definitions, become much more manageable and understandable.