Problem 233 - Advanced Calculus

Question

Prove Green's Theorem in the plane if the region \(\mathrm{R}\) is representable in both of the forms $$ \mathrm{a} \leq \mathrm{x} \leq \mathrm{b}, \mathrm{f}_{1}(\mathrm{x}) \leq \mathrm{y} \leq \mathrm{f}_{2}(\mathrm{x}) $$ $$ \mathrm{C} \leq \mathrm{y} \leq \mathrm{D}, \mathrm{g}_{1}(\mathrm{y}) \leq \mathrm{x} \leq \mathrm{g}_{2}(\mathrm{y}) $$ as in Fig. 1 . Let \(\mathrm{R}\) be the region in \(\mathrm{R}^{2}\) and let \(\mathrm{C}\) be the curve bounding \(\mathrm{R}\) as given in the figure. \(\mathrm{C}\) is oriented in such a way that \(\mathrm{R}\) is always to the left of \(\mathrm{C}\). Let \(\mathrm{P}(\mathrm{x}, \mathrm{y}), \mathrm{Q}(\mathrm{x}, \mathrm{y})\) be real-valued differentiable functions defined on \(\mathrm{R}\). Then Green's Theorem in the plane states that $$ \oint_{C} P d x+Q d y=\iint_{R}[(\partial Q / \partial x)-(\partial P / \partial y)] d x d y . $$

Step-by-Step Solution

Answer

To prove Green's Theorem for the given region R, we divided the region into horizontal parts divided by the curve y = h(x), then found the individual line integrals of each part of the boundary curve C. After summing these line integrals, we showed that it is equal to the double integral over the region R: $$ \oint_{C} P dx + Q dy = \iint_{R}[(\partial Q / \partial x)-(\partial P / \partial y)] dx dy $$

Step 1: Dividing the region

We will divide the region R into two parts horizontally as shown in the figure, where the division is at y = h(x). Then C is the boundary of the combined regions and can be represented as a curve over the top part (C1), a curve over the bottom part (C2), and two vertical sides (C3 and C4).

Step 2: Finding the line integral

We will find the line integral over each part of the curve and sum them up: $$ \oint_{C} P dx + Q dy = \oint_{C1} P dx + Q dy + \oint_{C2} P dx + Q dy + \oint_{C3} P dx + Q dy + \oint_{C4} P dx + Q dy $$

Step 3: Evaluating the line integral over C1 and C2

To evaluate the line integral over C1 and C2, we will parameterize the curves with respect to x, as follows: $$ C1: x=x, \; f_1(x) \le y \le h(x), \; a \le x \le b \\ C2: x=x, \; h(x) \le y \le f_2(x), \; a \le x \le b \\ $$ Now we can evaluate the line integrals over C1 and C2: $$ \oint_{C1} P dx + Q dy = \int_{a}^{b} \left[P(x,h(x)) - P(x,f_1(x))\right] dx + \left[Q(x,h(x))-Q(x,f_1(x))\right] \frac{dh}{dx} dx \\ \oint_{C2} P dx + Q dy = \int_{a}^{b} \left[P(x,f_2(x))-P(x,h(x))\right] dx + \left[Q(x,f_2(x))-Q(x,h(x))\right] \frac{dh}{dx} dx $$

Step 4: Evaluating the line integral over C3 and C4

To evaluate the line integral over C3 and C4, we will parameterize the curves with respect to y, as follows: $$ C3: y=y, \; g_1(y) \le x \le a, \; C \le y \le D \\ C4: y=y, \; b \le x \le g_2(y), \; C \le y \le D \\ $$ Now we can evaluate the line integrals over C3 and C4: $$ \oint_{C3} P dx + Q dy = \int_{C}^{D} \left[P(a,y)-P(g_1(y),y)\right] \frac{dg_1}{dy} dy + \left[Q(a,y)-Q(g_1(y),y)\right] dy \\ \oint_{C4} P dx + Q dy = \int_{C}^{D} \left[P(g_2(y),y)-P(b,y)\right] \frac{dg_2}{dy} dy + \left[Q(g_2(y),y)-Q(b,y)\right] dy $$

Step 5: Summing the line integrals

Adding up the line integrals, we get: $$ \oint_{C} P dx + Q dy = \int_{a}^{b}\left[P(x,f_2(x))-P(x,f_1(x))\right] dx + \int_{a}^{b}\left[Q(x,f_2(x))-Q(x,f_1(x))\right] \frac{dh}{dx} dx \\ \;\;\;\;+\ \int_{C}^{D}\left[P(a,y)-P(g_1(y),y)\right] \frac{dg_1}{dy} dy + \int_{C}^{D}\left[Q(a,y)-Q(g_1(y),y)\right] dy \\ \;\;\;\;+\ \int_{C}^{D}\left[P(g_2(y),y)-P(b,y)\right] \frac{dg_2}{dy} dy + \int_{C}^{D}\left[Q(g_2(y),y)-Q(b,y)\right] dy $$

Step 6: Finding the double integral

According to Green's theorem, the double integral is given by: $$ \iint_{R}[(\partial Q / \partial x)-(\partial P / \partial y)] dx dy $$ Since region R satisfies both conditions given in the exercise, we can write the double integral as: $$ \int_{a}^{b} \int_{f_1(x)}^{f_2(x)} \left[(\partial Q / \partial x)-(\partial P / \partial y)\right] dy dx $$

Step 7: Proving Green's Theorem

Now, we need to show that the line integral over the boundary curve C is equal to the double integral over region R: $$ \oint_{C} P dx + Q dy = \iint_{R}[(\partial Q / \partial x)-(\partial P / \partial y)] dx dy $$ By showing that the sum of line integrals in Step 5 is equal to the double integral in Step 6, we prove Green's Theorem for the given region R.

Key Concepts

Line Integral

A line integral is a way to measure a function along a curve. It sums up the values of a function at points along a path, multiplied by small distance segments along that path. Think of it as extending the idea of an integral to functions, not just over intervals on the x-axis, but over curves in a plane. This is crucial when we deal with fields or paths, such as when you want to compute the work done by a force field along a path.

For example, if you have a force field in a plane, the line integral can help you calculate the work done by the field along a curve path from one point to another. The formula for a line integral of a vector field \(\textbf{F} = (P, Q)\) across a curve \(C\) is expressed as:

\[ \oint_C P \, dx + Q \, dy \]

This integral considers both horizontal and vertical components (\(P\) and \(Q\)), representing the amount of the field passing along the x and y directions across the path. In Green's theorem context, evaluating this line integral plays a significant role in understanding how curves and their parameterization can help describe the behavior of a field.

Double Integral

Double integrals allow you to compute the volume under a surface defined above a region in a plane. They are considered over a two-dimensional area and are crucial when working with functions representing surfaces rather than lines.

For a given region \(R\) in the plane, the double integral can be expressed as:

\[ \iint_{R} f(x, y) \, dx \, dy \]

This calculates the sum of tiny slices of space within the region \(R\), building up to represent volumes. In the context of Green's Theorem, double integrals help bridge the relationship between the line integral around a closed curve \(C\) bounding \(R\) and the behavior of a vector field within \(R\).

Green's Theorem specifically uses the double integral to relate the circulation around the curve \(C\) to what happens inside the region \(R\), making it a critical concept in field analysis.

Parameterization of Curves

Parameterization of curves provides a systematic way to describe curves using a parameter, usually denoted by \(t\). By expressing coordinates of a curve as functions of a parameter, we gain the ability to traverse each point on the curve smoothly, calculating properties like tangents and integrating functions over those paths.

In Green's Theorem problems, parameterizing curves allow for calculating line integrals over complex paths. It's about transforming the evaluation of the line integral according to the configuration of the curve. For example, a curve could be described as a pair of equations:

- \(x = f(t)\)
- \(y = g(t)\)

As \(t\) varies over an interval, these equations trace out a curve in the plane. This system is crucial when working with line integrals, simplifying the computation by transforming the variables, often converting curves into simpler paths for easier integration.

Differentiable Functions

Differentiable functions are those with derivatives at every point in their domain. In basic terms, differentiability means you can draw a tangent at any point, indicating a smooth curve with no sharp edges.

In the context of Green's Theorem, \(P(x, y)\) and \(Q(x, y)\) should be differentiable functions within the region \(R\). This requirement ensures the integrity of the mathematical operations, such as partial derivatives, used in constructing and proving the theorem.

• \( \frac{\partial Q}{\partial x} \): Represents how \(Q\) changes as \(x\) changes.
• \( \frac{\partial P}{\partial y} \): Represents how \(P\) changes as \(y\) changes.

Differentiable functions allow the use of calculus operations, enabling evaluations like finding the exact difference in some plane's properties that is necessary for Green's Theorem to apply. By differentiating \(P\) and \(Q\), you can determine aspects of the field's behavior over and within the region.