Problem 13 - Thomas Calculus

Question

In Exercises \(5-14,\) use Green's Theorem to find the counterclockwise circulation and outward flux field \(\mathbf{F}\) and curve \(C .\) $$ \begin{array}{l}{\mathbf{F}=\left(x+e^{x} \sin y\right) \mathbf{i}+\left(x+e^{x} \cos y\right) \mathbf{j}} \\ {C : \text { The right- hand loop of the lemniscate } r^{2}=\cos 2 \theta}\end{array}$$

Step-by-Step Solution

Answer

The circulation of field \(\mathbf{F}\) around curve \(C\) is \(\frac{\pi}{4}\).

Step 1: Identify the Problem

We need to find the counterclockwise circulation and outward flux of the vector field \(\mathbf{F} = (x + e^x \sin y) \mathbf{i} + (x + e^x \cos y) \mathbf{j}\) along the curve \(C\), which is the right-hand loop of the lemniscate \(r^2 = \cos 2\theta\).

Step 2: Set up Green's Theorem

Green's Theorem relates a line integral around a simple, closed curve \(C\) to a double integral over the region \(R\) it encloses. It states \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \,dA \] where \(\mathbf{F} = M\mathbf{i} + N\mathbf{j}\), with \(M = x + e^x \sin y\) and \(N = x + e^x \cos y\).

Step 3: Calculate Partial Derivatives

Compute the partial derivatives required for Green's Theorem: \(\frac{\partial N}{\partial x} = 1 + e^x \cos y\) and \(\frac{\partial M}{\partial y} = e^x \cos y\).

Step 4: Evaluate the Integrand

Substitute the partial derivatives into the Green's Theorem formula: \(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = (1 + e^x \cos y) - e^x \cos y = 1\).

Step 5: Compute the Double Integral

Set up and compute the double integral over the region \(R\) covered by the right-hand loop of the lemniscate. The integral becomes \[ \iint_{R} 1 \, dA = \text{Area of } R\] which is the area of the right-hand loop of the lemniscate. Utilize symmetry or polar coordinates for simplification.

Step 6: Simplify Using Polar Coordinates

Transform into polar coordinates, where \(r^2 = \cos 2\theta\) becomes \(r = \sqrt{\cos 2\theta}\). The limits for \(\theta\) are from \(-\frac{\pi}{4}\) to \(\frac{\pi}{4}\). The integral becomes \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{\sqrt{\cos 2\theta}} r \, dr \, d\theta \].

Step 7: Evaluate the Double Integral

Solve the integral \[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{\sqrt{\cos 2\theta}} r \, dr \, d\theta = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2} (\cos 2\theta) d\theta \]. Integrate with respect to \(\theta\) to find the area of the region.

Step 8: Final Area Calculation

Evaluate the integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2} \cos 2\theta \, d\theta \), which results in the area being \(\frac{1}{2}[\frac{1}{2} \sin 2\theta]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{1}{2}\). Thus, the total area \( R = \frac{\pi}{4} \).

Step 9: Conclusion

Using Green's Theorem, the counterclockwise circulation of the vector field \(\mathbf{F}\) around curve \(C\) is equal to the area of the region \(R\), which is \(\frac{\pi}{4}\).

Key Concepts

Understanding Vector Fields

Vector fields are a fundamental concept in vector calculus and are incredibly useful in visualizing physical phenomena. Imagine them as fields that assign a vector to each point in a subset of space. These vectors can represent anything from wind speed and direction in meteorology to gravitational or electromagnetic forces in physics.

In a mathematical context, a vector field in the plane is typically expressed as \( \mathbf{F} = M(x, y)\mathbf{i} + N(x, y)\mathbf{j} \), where \( M \) and \( N \) are functions of the coordinates \( x \) and \( y \). The field given in our exercise, \( \mathbf{F} = (x + e^x \sin y) \mathbf{i} + (x + e^x \cos y) \mathbf{j} \), is a perfect example illustrating how vectors vary across space due to the inclusion of both \( x \) and \( y \) in its components.

• \( M(x, y) = x + e^x \sin y \)
• \( N(x, y) = x + e^x \cos y \)

The interaction of these components at each point in space provides a detailed description of the vector field, which can be further analyzed for behaviors such as circulation and flux using Green's theorem.

Exploring Line Integrals

A line integral extends the concept of a regular integral by integrating functions over a curve rather than a simple interval. This type of integral is crucial in calculating work done by a force field and can provide insights into the net flow of a field along a path.

Line integrals can be challenging because they require evaluating a vector field along a given curve \( C \). In simple terms, if you have a curve \( C \) and a vector field \( \mathbf{F} \), the line integral calculates the accumulation of the vector field’s effect along the curve. Mathematically, it is written as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), where \( d\mathbf{r} \) represents a small segment of the curve.

• The integral accounts for both the magnitude and direction of the vector field as experienced along the path.
• This concept is central to physics, particularly in the computation of work and circulation.

In the context of Green's Theorem, the line integral around a closed curve guides us to equate it to a double integral over the enclosed region, simplifying complex vector calculations.

Using Polar Coordinates

Polar coordinates, an alternative to the traditional Cartesian system, are essential for analyzing problems with circular symmetry. They are expressed as \( (r, \theta) \), where \( r \) is the distance from a point to the origin (the radius), and \( \theta \) is the angle measured from the positive x-axis.

Converting to polar coordinates often simplifies the evaluation of integrals, especially those involving circular or radial symmetry. For example, in the solution, the equation \( r^2 = \cos 2\theta \) describes the lemniscate loop. The transformation into polar coordinates allows the problem to be defined in terms of \( r \) and \( \theta \), dramatically simplifying integration because:

• The radial part \( r \) is isolated, making it easier to compute radial limits.
• Angles provide straightforward bounds for the integrals.

In this exercise, expressing the lemniscate in polar coordinates ensures that we effectively compute the area of complex regions, as demonstrated by our transformed integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \int_{0}^{\sqrt{\cos 2\theta}} r \, dr \, d\theta \). This approach leverages the simplicity of polar integration to achieve precise, efficient solutions for vector field applications.