Problem 197 - Calculus Volume 3

Question

Use Green's Theorem to evaluate integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}, \quad\) where \(\mathbf{F}(x, y)=\left(x y^{2}\right) \mathbf{i}+x \mathbf{j}, \quad\) and \(C\) is a unit circle oriented in the counterclockwise direction.

Step-by-Step Solution

Answer

The integral evaluates to \( \pi \).

Step 1: Understand Green's Theorem

Green's Theorem relates a line integral around a closed curve \( C \) to a double integral over the region \( D \) bounded by \( C \). The theorem states: \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] for \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \).

Step 2: Identify P and Q from F

Given \( \mathbf{F}(x, y) = (x y^2) \mathbf{i} + x \mathbf{j} \), we identify \( P(x, y) = x y^2 \) and \( Q(x, y) = x \).

Step 3: Calculate Partial Derivatives

Calculate the partial derivatives \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \). We find \( \frac{\partial Q}{\partial x} = 1 \) and \( \frac{\partial P}{\partial y} = 2xy \).

Step 4: Set Up the Double Integral

Substitute the partial derivatives into Green's Theorem, we get: \[ \iint_{D} \left( 1 - 2xy \right) \, dA \]. Here, \( D \) is the unit circle \( x^2 + y^2 \leq 1 \).

Step 5: Convert to Polar Coordinates

Because \( D \) is a circle, it's convenient to use polar coordinates: \( x = r \cos \theta \), \( y = r \sin \theta \), \( dA = r \, dr \, d\theta \), and \( r \) ranges from 0 to 1 and \( \theta \) from 0 to \( 2\pi \).

Step 6: Evaluate the Double Integral

Substitute the polar expressions into the integral: \[ \int_{0}^{2\pi} \int_{0}^{1} \left( 1 - 2r^2 \cos \theta \sin \theta \right) r \, dr \, d\theta \]. Simplify and integrate: \[ \int_{0}^{2\pi} \left. \left( \frac{r^2}{2} - \frac{r^4}{2} \sin 2\theta \right) \right|_{0}^{1} \, d\theta = \int_{0}^{2\pi} \frac{1}{2} \, d\theta = \pi \].

Step 7: Conclusion

The value of the integral \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \) using Green's Theorem is \( \pi \).

Key Concepts

Line Integral

A line integral is a fundamental concept in vector calculus that helps measure the degree of circulation of a vector field along a given path. In simpler terms, it calculates the integral of a vector field along a specified curve. You imagine it as a way to add up tiny segments of work done by a force or field along a path. Here's why it's interesting:

• It can be used to evaluate how a field like force acts over a path.
• It can provide insights into physical scenarios, such as work done by a force field on a particle traveling along a path.

For a vector field \( \mathbf{F} = P\mathbf{i} + Q\mathbf{j} \), the line integral along a curve \( C \) is:\[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} \]Green's Theorem gives us a toolkit where, instead of evaluating this possibly complex line integral directly, we convert it into an equivalent double integral over the region \( D \) enclosed by the curve \( C \). This is often simpler to compute.

Double Integral

A double integral extends the concept of a single integral to compute the volume under a surface in a given region. In the context of Green's Theorem, it transforms a line integral around a closed curve into a double integral over the plane region it encircles. When evaluating the double integral in Green's Theorem, we compute:\[\iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\]Here’s what happens in a practical sense:

• Once partial derivatives are calculated, they are plugged into the equation.
• Now, instead of analyzing the path itself, the focus shifts to examining the region bounded by the path.

This process simplifies calculations, especially in symmetric regions or when the vector field setup is complicated. It also allows transforming real-world physical problems into a more treatable mathematical framework.

Polar Coordinates

Polar coordinates offer an advantageous system, especially when dealing with circular or symmetric regions like the unit circle. By translating Cartesian coordinates \( (x, y) \) into polar coordinates using \( x = r \cos \theta \) and \( y = r \sin \theta \), we can handle circular domains more gracefully. Here's why polar coordinates are effective:

• They simplify integrations over circular areas.
• They inherently account for the radial symmetry of the circle.
• Changing from Cartesian differential area \( dA \) to \( r \, dr \, d\theta \) matches the circle's geometry better.

In the solution to the Green's Theorem problem, polar coordinates were introduced to convert the domain from the Cartesian plane into the circular region, where \( r \) ranges between 0 and 1, while \( \theta \) extends from 0 to \( 2\pi \). This simplification is crucial not just for the ease of integration but also for accurately capturing the essence of circular trajectories.