Problem 180 - Calculus Volume 3

Question

Use Green's theorem to evaluate line integral \(\oint_{C} e^{2 x} \sin 2 y d x+e^{2 x} \cos 2 y d y, \quad\) where \(C\) is ellipse \(9(x-1)^{2}+4(y-3)^{2}=36\) oriented counterclockwise.

Step-by-Step Solution

Answer

The line integral evaluates to 0.

Step 1: Identify the Vector Field

The given line integral is \( \oint_{C} e^{2x} \sin 2y \, dx + e^{2x} \cos 2y \, dy \). In vector form, the field is \( \mathbf{F} = (e^{2x} \sin 2y, e^{2x} \cos 2y) \).

Step 2: Apply Green's Theorem

According to Green's theorem, \( \oint_{C} M \, dx + N \, dy = \int\int_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \), where \( D \) is the region enclosed by \( C \). Here, \( M = e^{2x} \sin 2y \) and \( N = e^{2x} \cos 2y \).

Step 3: Compute Partial Derivatives

Find \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \). \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(e^{2x} \cos 2y) = 2e^{2x} \cos 2y \). \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^{2x} \sin 2y) = 2e^{2x} \cos 2y \).

Step 4: Set Up the Double Integral

Substitute the partial derivatives into Green's theorem equation: \( \int\int_{D} \left( 2e^{2x} \cos 2y - 2e^{2x} \cos 2y \right) \, dA = \int\int_{D} 0 \, dA = 0 \).

Step 5: Evaluate the Integral

Since the integrand is 0, the area integral over any region \( D \) is also 0. Therefore, \( \oint_{C} e^{2x} \sin 2y \, dx + e^{2x} \cos 2y \, dy = 0 \).

Key Concepts

Line Integrals

Line integrals are a way to integrate functions over a curve or path, as opposed to a standard integral which is over a region or interval. They are especially useful in physics for calculating quantities like work done by a force field or fluid flow along a path. Unlike traditional integrals, which sum values along a straight line interval, line integrals sum values along a curved path, which can be complex.

When dealing with line integrals in the context of vector fields, usually notated as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), the expression "\(\mathbf{F} \cdot d\mathbf{r}\)" represents the dot product of the vector field \(\mathbf{F}\) with an infinitesimal segment of the curve \(d\mathbf{r}\). This encompasses both the direction and magnitude of \(\mathbf{F}\) along the path.

In cases where Green’s theorem applies, line integrals can be transformed into double integrals over a region \(D\) enclosed by the path \(C\). This transformation can simplify calculations significantly, as it can convert a complex curvilinear integral into a potentially simpler area integral.

Vector Fields

Vector fields are mathematical constructs where each point in space is associated with a vector. These fields can represent various physical phenomena, such as gravitational forces, wind velocity, or electric and magnetic fields. Understanding vector fields is crucial for applications in physics and engineering.

In the problem, the vector field is given by \( \mathbf{F} = (e^{2x} \sin 2y, e^{2x} \cos 2y) \) where \( M = e^{2x} \sin 2y \) and \( N = e^{2x} \cos 2y \). This means that for every point \((x, y)\) on the plane, the vector \((e^{2x} \sin 2y, e^{2x} \cos 2y)\) describes both the direction and intensity of the vector field at that point.

Visualizing vector fields can help in intuitively understanding their impact or behavior over a region. For example, if we were to visualize \( \mathbf{F} \), we would see a pattern that changes as \(x\) and \(y\) vary, following the exponential growth determined by \(e^{2x}\) and the oscillation caused by the trigonometric functions.

Partial Derivatives

Partial derivatives are derivatives of functions with respect to one variable while keeping the other variables constant. This concept is a fundamental tool in multivariable calculus, particularly when working with functions involving two or more variables.

In the context of Green's Theorem, partial derivatives help in transforming a line integral into a double integral over a region. For our problem, the functions \(M = e^{2x} \sin 2y\) and \(N = e^{2x} \cos 2y\) require us to find partial derivatives: \(\frac{\partial N}{\partial x}\) and \(\frac{\partial M}{\partial y}\).

• \(\frac{\partial N}{\partial x} = 2e^{2x} \cos 2y\)
  
• \(\frac{\partial M}{\partial y} = 2e^{2x} \cos 2y\)
  

Calculating these derivatives is crucial in verifying the criteria of Green's Theorem, where the difference \(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\) equals zero in this case, proving the result \( \oint_{C} e^{2x} \sin 2y \, dx + e^{2x} \cos 2y \, dy = 0 \). Thus, mastering partial derivatives is essential for handling problems involving multivariable functions and theorems like Green's.