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In Exercises \(5-10\) , use Green's Theorem to find the counterclockwise circulation and outward flux for the field \(\mathbf{F}\) and curve \(C .\) \(\mathbf{F}=(x+y) \mathbf{i}-\left(x^{2}+y^{2}\right) \mathbf{j}\) \(C :\) The triangle bounded by \(y=0, x=1,\) and \(y=x\)

Short Answer

Expert verified
Circulation: \(-\frac{7}{6}\); Outward flux: \(\frac{1}{6}\).

Step by step solution

01

Understand Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \(C\) and a double integral over the plane region \(R\) bounded by \(C\). The formula is given by: \[ \oint_C \mathbf{F} \cdot \mathbf{dr} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \] where \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \). For outward flux, it is similar: \[ \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) \, dA \].
02

Identify Components of \(\mathbf{F}\)

For the vector field \( \mathbf{F} = (x+y)\mathbf{i} - (x^2+y^2)\mathbf{j} \), we identify \( M = x + y \) and \( N = -(x^2 + y^2) \).
03

Apply Green's Theorem for Circulation

Using Green's Theorem for circulation: \[ \oint_C \mathbf{F} \cdot \mathbf{dr} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \ dA \] Calculate the derivatives: \( \frac{\partial N}{\partial x} = -2x \) and \( \frac{\partial M}{\partial y} = 1 \). Thus, \( \int \int_R (-2x - 1) \, dA \).
04

Describe Region \(R\)

Region \(R\) is the triangle bounded by \(y=0\), \(x=1\), and \(y=x\). The limits of integration for \(y\) are from \(0\) to \(x\), and for \(x\) are from \(0\) to \(1\).
05

Calculate Double Integral for Circulation

Evaluate the double integral \( \int_0^1 \int_0^x (-2x - 1) \, dy \, dx \):1. Integrate with respect to \(y\): \( \int_0^x (-2x - 1) \, dy = (-2x - 1)y \bigg|_0^x = (-2x - 1)x = -2x^2 - x \).2. Integrate with respect to \(x\): \( \int_0^1 (-2x^2 - x) \, dx \).Calculate: \( \left[ -\frac{2}{3}x^3 - \frac{1}{2}x^2 \right]_0^1 = -\frac{2}{3} - \frac{1}{2} = -\frac{7}{6} \).
06

Apply Green's Theorem for Flux

For the outward flux: \[ \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) \, dA \] Calculate the derivatives: \( \frac{\partial M}{\partial x} = 1 \) and \( \frac{\partial N}{\partial y} = -2y \). Thus, the integral becomes: \( \int \int_R (1 - 2y) \, dA \).
07

Calculate Double Integral for Flux

Evaluate the double integral \( \int_0^1 \int_0^x (1 - 2y) \, dy \, dx \):1. Integrate with respect to \(y\): \( \int_0^x (1 - 2y) \, dy = \left[ y - y^2 \right]_0^x = x - x^2 \).2. Integrate with respect to \(x\): \( \int_0^1 (x - x^2) \, dx \).Calculate: \( \left[ \frac{1}{2}x^2 - \frac{1}{3}x^3 \right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \).

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Key Concepts

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

Line Integral
Line integrals play a crucial role in understanding Green's Theorem as well as many other mathematical and physical applications. A line integral helps to calculate the integral of a vector field along a curve. Essentially, it adds up vector values at small pieces along the curve.
In this specific exercise, you find the line integral around a simple closed curve \( C \), which is the boundary of a region \( R \). Using a line integral, you can determine the work done by a force field along a specific path or calculate circulation of fluid around the path.
For vector fields, the line integral's formula looks like this:
  • \( \oint_C \mathbf{F} \cdot \mathbf{dr} \)
Here, \( \mathbf{F} \) is a vector field in the plane, and \( \mathbf{dr} \) is the infinitesimal displacement vector along the curve. This specific exercise focuses on using Euler's form of line integral in conjunction with Green's Theorem to find circulation and flux around triangle perimeter \( C \).
Double Integral
Double integrals are fundamental for determining areas, volumes, and other sums over a two-dimensional region. In Green's Theorem, the double integral over region \( R \) transformed the line integral around \( C \) to a simpler form.
In the exercise, we deal with the double integral representing the sum of tiny parts of area \( dA \) over the triangular region. These are used to calculate circulation and outward flux. Evaluation of such integrals uses limits determined by the region's boundaries, here given as a triangle bounded by three lines.
The formula for the double integral, in context, given by Green's Theorem is:
  • \( \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \) to find circulation.
  • \( \iint_R \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) \, dA \) for flux.
They involve computing partial derivatives \( \frac{\partial N}{\partial x} \), \( \frac{\partial M}{\partial y} \), \( \frac{\partial M}{\partial x} \), and \( \frac{\partial N}{\partial y} \) to find their values and then integrate over defined limits.
Vector Field
A vector field is a mathematical structure where each point in a space is assigned a vector. In simpler terms, it resembles a map where every point has a direction and magnitude, think of wind speeds on a weather map.
In the exercise, the given vector field is \( \mathbf{F} = (x+y) \mathbf{i} - (x^2+y^2) \mathbf{j} \). Each vector is defined by horizontal \( \mathbf{i} \) and vertical \( \mathbf{j} \) components based on functions, \( M \) and \( N \).
Here's how the vector field is expressed:
  • \( M = x + y \)
  • \( N = -(x^2 + y^2) \)
This tells us that at any point \( (x,y) \), the vector's direction and length are governed by expressions for \( M \) and \( N \). Using these components in Green's Theorem, students can unravel the complete interaction of the field with the boundary curve \( C \) to compute values like flux or circulation.

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Most popular questions from this chapter

Conservation of mass \(\quad\) Let \(\mathbf{v}(t, x, y, z)\) be a continuously differ- entiable 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 \(v\) 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.)

In Exercises \(47-52,\) use a CAS to perform the following steps for finding the work done by force \(\mathbf{F}\) over the given path: a. Find \(d \mathbf{r}\) for the path \(\mathbf{r}(t)=g(t) \mathbf{i}+h(t) \mathbf{j}+k(t) \mathbf{k}\) b. Evaluate the force \(\mathbf{F}\) along the path. c. Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) $$ \begin{array}{l}{\mathbf{F}=x y^{6} \mathbf{i}+3 x\left(x y^{5}+2\right) \mathbf{j} ; \quad \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(\sin t) \mathbf{j}} \\\ {0 \leq t \leq 2 \pi}\end{array} $$

Zero curl, yet field not conservative Show that the curl of $$ \mathbf{F}=\frac{-y}{x^{2}+y^{2}} \mathbf{i}+\frac{x}{x^{2}+y^{2}} \mathbf{j}+z \mathbf{k} $$ is zero but that $$ \begin{array}{l}{\oint_{C} \mathbf{F} \cdot d \mathbf{r}} \\ {c}\end{array} $$ is not zero if \(C\) is the circle \(x^{2}+y^{2}=1\) in the \(x y\) -plane. (Theorem 6 does not apply here because the domain of \(\mathbf{F}\) is not simply connected. The field \(\mathbf{F}\) is not defined along the \(z\) -axis so there is no way to contract \(C\) to a point without leaving the domain of \(\mathbf{F} . )\)

Let \(\mathbf{n}\) be the outer unit normal (normal away from the origin) of the parabolic shell $$ \text { S: } 4 x^{2}+y+z^{2}=4, \quad y \geq 0 $$ and let $$ \mathbf{F}=\left(-z+\frac{1}{2+x}\right) \mathbf{i}+\left(\tan ^{-1} y\right) \mathbf{j}+\left(x+\frac{1}{4+z}\right) \mathbf{k} $$ Find the value of $$ \iint_{S} \nabla \times \mathbf{F} \cdot \mathbf{n} d \sigma $$

Integrate \(g(x, y, z)=x \sqrt{y^{2}+4}\) over the surface cut from the parabolic cylinder \(y^{2}+4 z=16\) by the planes \(x=0, x=1\) and \(z=0\)

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