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Chapter 15: Problem 39

Show that if \(R\) is a region in the plane bounded by a piecewise smooth, simple closed curve \(C\), then $$ \text { Area of } R=\oint x d y=-\oint y d x $$

Short Answer

Expert verified
Both expressions, \( \oint x \, dy \) and \( -\oint y \, dx \), equal the area of region \( R \).

Step by step solution

01

Understanding the Notation and Context

The curve \( C \) is closed and piecewise smooth, meaning it does not intersect itself and is made up of a finite number of smooth segments. We are to find the area \( A \) of the region \( R \) it encloses using line integrals.
02

Using Green's Theorem

Apply Green's Theorem, which relates the line integral around a simple closed curve to a double integral over the region it encloses. Green's Theorem states: \[ \oint_C (L dx + M dy) = \iint_R \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \, dx \, dy. \] Here, we choose \( L = 0 \) and \( M = x \), yielding: \[ \oint_C x \, dy = \iint_R \left( 1 \right) dx \, dy = \text{Area of } R. \]
03

Calculating the Line Integral \( \oint x \, dy \)

Compute \( \oint x \, dy \), which, by Green's Theorem, gives us the area of the region \( R \). Thus, \( A = \oint_C x \, dy \).
04

Using Green's Theorem Again

Now, consider the line integral \( - \oint y \, dx \). We apply Green's Theorem again with \( L = -y \) and \( M = 0 \), giving: \[ \oint_C (-y) dx = \iint_R \left( 1 \right) dx \, dy = \text{Area of } R. \]
05

Equating Both Expressions for Area

Since both expressions, \( \oint x \, dy \) and \( -\oint y \, dx \), equal the area \( A \), we conclude \( \oint x \, dy = -\oint y \, dx = \text{Area of } R \).

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Key Concepts

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

Line Integrals
Line integrals extend the concept of integration to functions over a curve. They are crucial in fields like physics and engineering, providing a way to sum quantities along a curve. When calculating a line integral,
  • We consider functions depending on position along the curve.
  • Integrate these functions over each segment of the curve.
In mathematics, a line integral along curve \( C \) can be represented as \( \oint_C f(x, y) \, ds \), where \( ds \) is an infinitesimal element of the curve. To gain intuition: Imagine walking a path, where at every point, you collect some value as you move. This process, adding up the collected values, is akin to the line integral. In our context:
  • The expression \( \oint x \, dy \) explores how the "x-component" evolves along the curve.
  • Similarly, \( \oint y \, dx \) provides insights into the "y-component's" journey.
Piecewise Smooth Curve
A piecewise smooth curve is made up of multiple smooth segments joined together. It鈥檚 important in calculus and analysis since not all curves in real-life applications are perfectly smooth. This kind of curve:
  • Is finite in nature, meaning it consists of a finite number of distinct smooth pieces.
  • May change direction sharply at specific points or joins.
Understanding these curves is essential when applying theorems like Green's Theorem, which was used in our exercise. A piecewise smooth boundary is necessary for Green's Theorem's conditions:
  • Closed: The curve forms a complete loop without breaks.
  • Non-intersecting: The curve doesn't cross itself, ensuring a clearly defined enclosed region.
Double Integral
Double integrals are the extension of single-variable integrals but over a two-dimensional area. They help calculate quantities spread over regions in the plane, such as mass or area itself. Here's how they generally work:
  • You pick a region, say \( R \), and a function \( f(x, y) \).
  • The double integral then sums up the function's values over every point in the region, \( \iint_R f(x, y) \, dx \, dy \).
In our case, the double integral finds the area \( A \) of region \( R \). By setting the function to 1 over region \( R \), the double integral directly yields the area, simplifying the process using Green's Theorem. It connects naturally with line integrals to allow computation of two-dimensional quantities by tracing along a boundary curve.
Area of a Region
The area of a region in a plane is typically calculated using geometric or calculus methods. When the region is defined by a curve, especially a "piecewise smooth" closed curve, like in our exercise, Green's Theorem becomes a powerful tool for finding this area through line integrals.The area \( A \) of a region \( R \) can be calculated by:
  • Employing line integral expressions like \( \oint x \, dy \) or \( -\oint y \, dx \).
  • Using Green's Theorem to relate these line integrals to a double integral over \( R \).
So, rather than directly computing the area through traditional methods, you integrate around its boundary, producing \( A = \oint x \, dy = -\oint y \, dx \). This method particularly shines when traditional geometric methods are cumbersome or when the boundary curve defines an odd pattern or region shape.

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Most popular questions from this chapter

Outward flux of a gradient field \(\quad\) Let \(S\) be the surface of the portion of the solid sphere \(x^{2}+y^{2}+z^{2} \leq a^{2}\) that lies in the first octant, and let \(f(x, y, z)=\ln \sqrt{x^{2}+y^{2}+z^{2}} .\) Calculate $$\iint_{S} \nabla f \cdot \mathbf{n} d \sigma$$ \((\nabla f \cdot \mathbf{n} \text { is the derivative of } f \text { in the direction of outward normal } \mathbf{n} .\) )

Conservation of mass \(\quad\) Let \(\mathbf{v}(t, x, y, z)\) be a continuously differentiable vector field over the region \(D\) in space, and let \(p(t, x,\) \(y, z)\) be a continuously differentiable scalar function. The variable t represents the time domain. The Law of Conservation of Mass asserts that $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=-\iint_{S} p \mathbf{v} \cdot \mathbf{n} d \sigma,$$ where \(S\) is the surface enclosing \(D\). a. Give a physical interpretation of the conservation of mass law if \(\mathbf{v}\) is a velocity flow field and \(p\) represents the density of the fluid at point \((x, y, z)\) at time \(t\) b. Use the Divergence Theorem and Leibniz's Rule, $$\frac{d}{d t} \iiint_{D} p(t, x, y, z) d V=\iiint_{D} \frac{\partial p}{\partial t} d V,$$ to show that the Law of Conservation of Mass is equivalent to the continuity equation, $$\nabla \cdot p \mathbf{v}+\frac{\partial p}{\partial t}=0.$$ (In the first term \(\nabla \cdot p \mathbf{v},\) the variable \(t\) is held fixed, and in the second term \(\partial p / \partial t,\) it is assumed that the point \((x, y, z)\) in \(D\) is held fixed.)

Use the surface integral in Stokes' Theorem to calculate the flux of the curl of the field \(\mathbf{F}\) across the surface \(S .\) $$\mathbf{F}=(y-z) \mathbf{i}+(z-x) \mathbf{j}+(x+z) \mathbf{k}$$ \(S: \quad \mathbf{r}(r, \theta)=(r \cos \theta) \mathbf{i}+(r \sin \theta) \mathbf{j}+\left(9-r^{2}\right) \mathbf{k}, 0 \leq r \leq 3, \quad 0 \leq \theta \leq 2 \pi,\) in the direction away from the origin.

In Exercises 29 and \(30 .\) find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. \(\mathbf{F}(x, y, z)=-\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}\) \(S:\) rectangular surface \(z=0, \quad 0 \leq x \leq 2, \quad 0 \leq y \leq 3\) direction k

In Exercises \(19-28,\) use a parametrization to find the flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d \sigma\) across the surface in the specified direction. \(\mathbf{F}=x y \mathbf{i}-z \mathbf{k} \quad\) through the cone \(\quad z=\sqrt{x^{2}+y^{2}}\) \(0 \leq z \leq 1,\) in the direction away from the z-axis
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