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Chapter 4: Problem 8

Show that for any constants \(a, b\) and any closed simple curve \(C, \oint_{C} a d x+b d y=0\)

Short Answer

Expert verified
\( \oint_{C} a dx + b dy = 0 \) because the integral can be expressed as zero when applying Green's Theorem.

Step by step solution

01

Understand the Integral

The integral given is a line integral of the form \(\frac{dF}{dx} = a\) and \(\frac{dF}{dy} = b\). This is identified as a linear combination of differential forms dx and dy along a closed simple curve C.
02

Express the Integral in Terms of a Potential Function

Identify a function F(x, y) such that \(dF = a dx + b dy\). By integrating, we get \(F(x, y) = ax + by + C\), where C is a constant.
03

Apply Green's Theorem

Green's Theorem relates a line integral around a simple closed curve C to a double integral over the region D enclosed by C: \(\begin{aligned}\oint_{C} M dx + N dy = \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA\the d{aligned}\)
04

Match Terms with Green's Theorem

In our case, \(M = a \) and \(N = b \). Compute the partial derivatives: \( \frac{\partial N}{\partial x} = 0\), \(\frac{\partial M}{\partial y} = 0 \). Substituting these into Green's Theorem results in: \(\iint_{D} \left( 0 - 0 \right) dA = 0\).
05

Conclude the Proof

Since the double integral is zero, the corresponding line integral must also be zero. Therefore, \(\begin{aligned}\oint_{C} adx + b dy = 0\thed{aligned}\)

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Key Concepts

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

Green's Theorem
Green's Theorem is a fundamental result in vector calculus that connects the line integral around a simple closed curve to a double integral over the region it encloses. This theorem is particularly useful for transforming complex line integrals into more manageable double integrals.

Green's Theorem states that for a positively oriented, simple closed curve C enclosing a region D, and functions M(x, y) and N(x, y) with continuous partial derivatives, we have:
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Most popular questions from this chapter

For Exercises \(12-23\), prove the given formula \((r=\|\mathbf{r}\|\) is the length of the position vector field \(\mathbf{r}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k})\). $$ \operatorname{div}(f \mathbf{F})=f \operatorname{div} \mathbf{F}+\mathbf{F} \cdot \nabla f $$

Suppose that \(\Delta u=0\) (i.e. \(u\) is harmonic) over \(\mathbb{R}^{3}\), Define the normal derivative \(\frac{\partial u}{\partial n}\) of \(u\) over a closed surface \(\Sigma\) with outward unit normal vector \(\mathbf{n}\) by \(\frac{\partial u}{\partial n}=D_{n} u=\mathbf{n} \cdot \nabla u,\) Show that \(\iint \frac{\partial u}{\partial n} d \sigma=0 .\) (Hint: Use Green's second identity.)

For Exercises \(1-4,\) use Green's Theorem to evaluate the given line integral around the curve \(C,\) traversed counterclockwise. $$ \oint_{C}\left(x^{2}-y^{2}\right) d x+2 x y d y ; C \text { is the boundary of } R=\left\\{(x, y): 0 \leq x \leq 1,2 x^{2} \leq y \leq 2 x\right\\} $$

For Exercises \(1-6,\) find the Laplacian of the function \(f(x, y, z)\) in Cartesian coordinates. $$ f(x, y, z)=e^{-x^{2}-y^{2}-z^{2}} $$

Evaluate the surface integral \(\iint \mathbf{f} \cdot d \boldsymbol{\sigma},\) where \(\mathbf{f}(x, y, z)=x^{2} \mathbf{i}+x y \mathbf{j}+z \mathbf{k}\) and \(\Sigma\) is the part of the plane \(6 x+3 y+2 z=6\) with \(x \geq 0, y \geq 0,\) and \(z \geq 0,\) with the outward unit normal \(n\) pointing in the positive \(z\) direction.
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