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Let the piecewise smooth, simple closed curve \(C\) be the boundary of a region \(S\) in the \(x y\) -plane. Modify the argument in Example 2 to show that $$ A(S)=\oint_{C}(-y) d x=\oint_{C} x d y $$

Short Answer

Expert verified
Both integrals represent the area of region \(S\) using Green's Theorem.

Step by step solution

01

Understanding the Problem

We need to show that the area of a region \( S \) can be expressed as \( A(S) = \oint_{C}(-y) dx = \oint_{C} x dy \), where \( C \) is the boundary of \( S \).
02

Recall Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the region \( S \) bounded by \( C \). It states: \[ \oint_{C} (L \, dx + M \, dy) = \iint_{S} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA. \]
03

Apply Green's Theorem to xdy

For \( \oint_{C} x \, dy \), set \( L = 0 \) and \( M = x \). According to Green's Theorem, we have: \[ \iint_{S} \left( \frac{\partial x}{\partial x} - \frac{\partial 0}{\partial y} \right) dA = \iint_{S} 1 \, dA = A(S). \]
04

Apply Green's Theorem to -ydx

For \( \oint_{C} -y \, dx \), set \( L = -y \) and \( M = 0 \). According to Green's Theorem, we have: \[ \iint_{S} \left( \frac{\partial 0}{\partial x} - \frac{\partial (-y)}{\partial y} \right) dA = \iint_{S} 1 \, dA = A(S). \]
05

Conclude the Solution

Both \( \oint_{C} x \, dy \) and \( \oint_{C} -y \, dx \) equal \( A(S) \) by Green's Theorem, which shows that \( A(S) = \oint_{C}(-y) dx = \oint_{C} x dy \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Line Integrals
Line integrals are crucial when dealing with vector fields and curves, especially in understanding Green's Theorem. A line integral along a curve essentially adds up all values of a function along that curve. For example, if we have a curve \( C \) and a function defined on this curve, the line integral \( \oint_{C} f(x, y) \, ds \) sums up all the small contributions of \( f(x, y) \) along \( C \).
This concept helps in understanding flow along a path or modeling various physical quantities like work or energy transfer. Instead of the whole area, line integrals focus on movement along the boundary, taking into account both the magnitude of the function and the path's direction.
  • When evaluating line integrals \( \oint_{C} x \, dy \) or \( \oint_{C} -y \, dx \), you can feel the connection between the curve and the region it encircles.
  • Green's Theorem facilitates this connection by relating these line integrals to double integrals, thus transforming boundary problems into domain problems.
Demystifying Double Integrals
Double integrals help in calculating a wide range of quantities like area, volume, and more across a two-dimensional region. When you use a double integral over a region \( S \), it is similar to stacking up several tiny areas to get an overall total. Each tiny point in the region \( S \) contributes a tiny piece to this sum.
The formula \( \iint_{S} \, dA \) is essentially adding up all the infinitesimal areas \( dA \) over region \( S \). This sum gives us the total area, which is crucial in understanding Green's Theorem application. In this context,
  • Each increment of area in the plane is considered, and all these small pieces eventually sum up to provide a comprehensive measure.
  • Through integrating, we step beyond mere length along curves to include whole regions, relating the effects observed along the boundary to the entirety of the space they bound.
A double integral forms the core of equivalence used in expressing areas via line integrals, as revealed by applying Green's Theorem.
Area Calculation Using Green's Theorem
Green's Theorem beautifully relates line integrals to double integrals, providing a pivotal method for area calculation. Specifically, if you have a closed curve \( C \) that forms the boundary of a region \( S \), Green's Theorem can transform challenging problems into solvable ones by flipping between boundary and region insights.
To compute the area of \( S \) using line integrals like \( A(S) = \oint_{C} (-y) \, dx \) or \( \oint_{C} x \, dy \), we apply Green's theorem. By setting up the correct functions\(, L = -y \) and \( M = 0 \) for one case, and \( L = 0 \) and \( M = x \) for the other,
  • Both setups yield the double integral \( \iint_{S} \, dA \), representing the area of \( S \).
  • This verification of area by boundary analysis brilliantly showcases Green's Theorem's ability to bind external path integration with internal region evaluation.
In essence, this theorem establishes a critical bridge between knowing a shape by its perimeter and understanding the extensive space it encompasses, greatly simplifying area calculations for complex regions.

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Most popular questions from this chapter

Plot the parametric surface over the indicated domain. \(\mathbf{r}(u, v)=u \mathbf{i}+3 v \mathbf{j}+\left(4-u^{2}-v^{2}\right) \mathbf{k} ; 0 \leq u \leq 2\), \(0 \leq v \leq 1\).

Let $$\mathbf{F}=\frac{y}{x^{2}+y^{2}} \mathbf{i}-\frac{x}{x^{2}+y^{2}} \mathbf{j}=M \mathbf{i}+N \mathbf{j}$$ (a) Show that \(\partial N / \partial x=\partial M / \partial y\). (b) Show, by using the parametrization \(x=\cos t, y=\sin t,\) that \(\oint_{C} M d x+N d y=-2 \pi,\) where \(C\) is the unit circle. (c) Why doesn't this contradict Green's Theorem?

This problem shows the need for simple connectedness in the "if" statement of Theorem C. Let \(\mathbf{F}=(y \mathbf{i}-x \mathbf{j}) /\left(x^{2}+y^{2}\right)\) on the set \(D=\left\\{(x, y): x^{2}+y^{2} \neq 0\right\\}\). Show each of the following. (a) The condition \(\partial M / \partial y=\partial N / \partial x\) holds on \(D\). (b) \(\mathbf{F}\) is not conservative on \(D\). Hint: To establish part (b), show that \(\int_{C} \mathbf{F} \cdot d \mathbf{r}=-2 \pi\) where \(C\) is the circle with parametric equations \(x=\cos t, y=\sin t\) \(0 \leq t \leq 2 \pi\)

Use Stokes's Theorem to calculate \(\oint_{C} \mathbf{F} \cdot \mathbf{T} d s\) \(\mathbf{F}=2 z \mathbf{i}+x \mathbf{j}+3 y \mathbf{k} ; C\) is the ellipse that is the intersection of the plane \(z=x\) and the cylinder \(x^{2}+y^{2}=4,\) oriented clockwise as viewed from above.

Consider the velocity field \(\mathbf{v}(x, y, z)=-\omega y \mathbf{i}+\omega x \mathbf{j}\), \(\omega>0\) (see Example 2 and Figure 1 ). Note that \(\mathbf{v}\) is perpendicular to \(x \mathbf{i}+y \mathbf{j}\) and that \(\|\mathbf{v}\|=\omega \sqrt{x^{2}+y^{2}}\). Thus, \(\mathbf{v}\) describes a fluid that is rotating (like a solid) about the \(z\) -axis with constant angular velocity \(\omega .\) Show that \(\operatorname{div} \mathbf{v}=0\) and \(\operatorname{curl} \mathbf{v}=2 \omega \mathbf{k}\).

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