Problem 170 - Calculus Volume 3

Question

[T] Evaluate Green's theorem using a computer algebra system to evaluate the integral \(\int_{C} x e^{y} d x+e^{x} d y\) where \(C\) is the circle given by \(x^{2}+y^{2}=4\) and is oriented in the counterclockwise direction.

Step-by-Step Solution

Answer

The integral around the circle using Green's theorem equals zero.

Step 1: Green's Theorem Setup

Green's Theorem relates a line integral around a closed curve \(C\) to a double integral over the region \(R\) it encloses. The theorem states that for \(\oint_{C} (P \, dx + Q \, dy) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\). Here, \(P = x e^y\) and \(Q = e^x\).

Step 2: Compute Partial Derivatives

Calculate \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\).\(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(e^x) = e^x\)\(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x e^y) = x e^y\)

Step 3: Set Up Double Integral

The double integral to evaluate over the region \(R\) is \ \(\iint_{R} (e^x - x e^y) \, dA\).Since \(C\) is a circle of radius 2 centered at the origin, we'll use polar coordinates for easier integration. The region \(R\) is \(x^2 + y^2 \leq 4\).

Step 4: Convert to Polar Coordinates

In polar coordinates, \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Thus, \ dA = r \, dr \, d\theta. The integral bounds are \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\).The integrand becomes: \(e^{x} = e^{r \cos\theta}\) and \(x e^{y} = r \cos\theta \, e^{r \sin\theta}\).

Step 5: Evaluate the Double Integral

The double integral becomes: \ \(\int_{0}^{2\pi} \int_{0}^{2} \left(e^{r \cos\theta} - r \cos\theta \, e^{r \sin\theta} \right) \, r \, dr \, d\theta\).Use a computer algebra system to solve this integral.

Key Concepts

Line Integral

A line integral is a type of integral where the function to be integrated is evaluated along a curve. In essence, it sums, over a line or curve, functions with respect to arc length, usually in a vector field. Line integrals are crucial in physics and engineering for calculating quantities like work done by a force field:

• In this exercise, we have the line integral represented by \( \int_{C} x e^{y} \, dx + e^{x} \, dy \) along a circle \( C \).
• The curve \( C \) is described as \( x^2 + y^2 = 4 \), which signifies a circle of radius 2 with a counterclockwise direction.

Green's Theorem helps us convert this line integral into a more manageable form by relating it to a double integral.

Polar Coordinates

Polar coordinates offer a useful way to represent points in a plane by expressing them through a distance from the origin (radius \( r \)) and an angle (\( \theta \)) measured from the positive x-axis. This coordinate system is especially handy for circular regions:

• In this exercise, the region of interest is a circle, which makes polar coordinates ideal.
• We convert Cartesian points \((x, y)\) to polar by using \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\).

For the double integral, the area element \( dA \) becomes \( r \, dr \, d\theta \), simplifying integration over circular regions and aiding in solving the integral defined by Green's Theorem.

Double Integral

Double integrals allow us to calculate the volume under a surface within a specified region in the xy-plane. They are extensions of single integrals to two dimensions and are integral (no pun intended) to applying Green's Theorem:

• In our problem, the double integral is \( \iint_{R} (e^{x} - x e^{y}) \, dA \) over the circular region \( R \).
• This changes our problem from calculating a complex line integral manually to evaluating a double integral using Green's theorem, often involving polar coordinates for ease.

In this case, we use bounds from the circle’s radius, transforming integration limits into \(0 \leq r \leq 2\) and \(0 \leq \theta \leq 2\pi\). This transformation makes evaluating the volume under the relevant function possible.

Partial Derivatives

Partial derivatives involve taking the derivative of a function with respect to one variable while keeping other variables constant. This concept is particularly relevant when dealing with functions of multiple variables:

• In the Green's Theorem framework, calculating partial derivatives of \( P \) and \( Q \) is crucial:
• We find \( \frac{\partial Q}{\partial x} \) from \( Q = e^{x} \), resulting in \( e^{x} \).
• Similarly, for \( P = x e^{y} \), \( \frac{\partial P}{\partial y} \) equals \( x e^{y} \).

These partial derivatives are integral to transitioning from a line integral to a double integral using Green's Theorem, allowing us to evaluate complex integrals effectively. Their calculation provides the difference component in the theorem, enabling us to assess the double integral and ultimately solve the exercise.