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Line integrals Use Green's Theorem to evaluate the following line integrals. Assume all curves are oriented counterclockwise. A sketch is helpful. \(\oint_{c} f d y-g d x,\) where \(\langle f, g\rangle=\langle 0, x y\rangle\) and \(C\) is the triangle with vertices \((0,0),(2,0),\) and (0,4)

Short Answer

Expert verified
To summarize, we were given a vector field \(\vec{F} = \langle 0, xy \rangle\) and asked to evaluate the line integral over the curve \(C\), which is a triangle with vertices at \((0,0), (2,0),\) and \((0,4)\). By parametrizing the curve and applying Green's Theorem \(\oint_C (f \, dy - g \, dx) = \iint_D (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) dA\), we found that the line integral value is \(\frac{16}{3}\).

Step by step solution

01

Parametrizing the Curve C

The curve \(C\) is a triangle with vertices at \((0,0), (2,0),\) and \((0,4)\). This triangle can be divided into three line segments (going counterclockwise): 1. \(C_1: (0,0) \to (2,0)\). Parametrize this line as \(\vec{r}_1(t) = \langle 2t, 0 \rangle\) with \(0 \leq t \leq 1\). 2. \(C_2: (2,0) \to (0,4)\). Parametrize this line as \(\vec{r}_2(t) = \langle 2 - 2t, 4t \rangle\) with \(0 \leq t \leq 1\). 3. \(C_3: (0,4) \to (0,0)\). Parametrize this line as \(\vec{r}_3(t) = \langle 0, 4 - 4t \rangle\) with \(0 \leq t \leq 1\). Now that we have parametric equations for all three line segments, we can proceed further.
02

Finding the Partial Derivatives

We need to find the partial derivatives of \(f\) and \(g\) for Green's theorem. Our vector field is given by \(\vec{F}=\langle f,g \rangle = \langle 0, xy \rangle\). So the partial derivatives are: \(\frac{\partial g}{\partial x} = \frac{\partial (xy)}{\partial x} = y\) \(\frac{\partial f}{\partial y} = \frac{\partial (0)}{\partial y} = 0\) Now we can apply Green's Theorem.
03

Applying Green's Theorem

According to Green's theorem: \(\oint_C (f \, dy - g \, dx) = \iint_D (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) dA\) In our case, this simplifies to: \(\oint_C (-g \, dx ) = \iint_D y \, dA\) Now, we need to evaluate the integral.
04

Evaluating the Integral

As the domain \(D\) is a triangle, we can set up the iterated integral limits as follows: \(\int_0^2 \int_0^{2x} y \, dy \, dx\) Evaluating the inner integral: \(\int_0^2 ( \frac{1}{2} y^2 \Big|_0^{2x} ) \, dx\) Evaluating the outer integral: \( \int_0^2 (2x^2) \, dx = \frac{2}{3} x^3 \Big|_0^2 = \frac{16}{3}\) So, the line integral is: \(\oint_C (-g \, dx) = \frac{16}{3}\)

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Key Concepts

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

Line Integral Parametrization
When dealing with line integrals, an essential step is to describe the path of integration in a mathematical way. This process is known as parametrization. It allows us to express a curve in terms of a single variable, typically denoted as t. Parametrization is not just a method to simplify calculations but also a powerful tool to visualize complex paths.

The curve C in our example represents a triangle, and it has been broken down into three line segments for which parametric equations are established. Each of these segments is described using a vector function \( \vec{r}(t) \) that captures both the direction and the shape of the path.
  • For the horizontal segment \( C_1 \), the parametrization is purely in the x-direction, reflecting that there is no vertical movement.
  • Segment \( C_2 \) illustrates a decrease in the x-coordinate while the y-coordinate increases, a clear depiction of a line slanted down and to the right.
  • Finally, for the vertical segment \( C_3 \), the movement is solely in the y-direction.
Each parametrization encompasses a range for t from 0 to 1, simplifying the calculations since it corresponds to the proportion of the path covered. This common range makes it easier to piece together the contributions of each segment while maintaining the curve's orientation.
Vector Field Partial Derivatives
To apply Green's Theorem, we need to explore the concept of partial derivatives of vector fields. In the context of our problem, we have a vector field \( \vec{F} = \langle f, g \rangle = \langle 0, xy \rangle \). A vector field associates a vector to every point in a space, describing, for example, the magnitude and direction of a force at different points in a field.

In our case, we determine how the function g, which is part of the vector field, changes as x, and separately as y, changes. This is the partial derivative. The partial derivative of g with respect to x (\( \frac{\partial g}{\partial x} \)) is y, signifying that as x changes, the value of g changes at a rate proportional to y. Conversely, the partial derivative of f with respect to y is zero, indicating that f remains constant regardless of how y changes.

Understanding these rates of change is critical because they inform us about the curl of the vector field, which plays a central role in applying Green's Theorem. The curl can be interpreted as a measure of the rotation of the field around a point, and it's precisely what we integrate over the region enclosed by the curve C.
Iterated Integrals
The core idea behind iterated integrals is to evaluate multiple integrals in a step-by-step fashion, one after the other. They are fundamental in calculating areas, volumes, and in our case, using Green's Theorem to convert a line integral into an easier-to-solve double integral over a region D.

In our example, the domain D is a triangle, and the iterated integrals involve integrating first with respect to y and then with respect to x. This order is not arbitrary; it follows from the limits of integration which are, in turn, dictated by the shape of D. It's important to note that the limits for y are variable – they depend on x (0 to 2x), while the limits for x are constants (0 to 2), indicating that x spans the entire width of the triangle base.
By evaluating the inner integral first, we condense the two-dimensional problem into a one-dimensional integral. The result from the inner integral is then used as the integrand for the outer integral. Completing both steps gives us the area under the curve—and in the particular application for Green's Theorem, the value of the line integral around the curve C.

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Most popular questions from this chapter

Average circulation Let \(S\) be a small circular disk of radius \(R\) centered at the point \(P\) with a unit normal vector \(\mathbf{n}\). Let \(C\) be the boundary of \(S\). a. Express the average circulation of the vector field \(\mathbf{F}\) on \(S\) as a surface integral of \(\nabla \times \mathbf{F}\) b. Argue that for small \(R\), the average circulation approaches \(\left.(\nabla \times \mathbf{F})\right|_{P} \cdot \mathbf{n}(\text { the component of } \nabla \times \mathbf{F} \text { in the direction of } \mathbf{n}\) evaluated at \(P\) ) with the approximation improving as \(R \rightarrow 0\)

Gravitational potential The gravitational force between two point masses \(M\) and \(m\) is $$\mathbf{F}=G M m \frac{\mathbf{r}}{|\mathbf{r}|^{3}}=G M m \frac{\langle x, y, z\rangle}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}$$ where \(G\) is the gravitational constant. a. Verify that this force field is conservative on any region excluding the origin. b. Find a potential function \(\varphi\) for this force field such that \(\mathbf{F}=-\nabla \varphi\). c. Suppose the object with mass \(m\) is moved from a point \(A\) to a point \(B\), where \(A\) is a distance \(r_{1}\) from \(M,\) and \(B\) is a distance \(r_{2}\) from \(M .\) Show that the work done in moving the object is \(G M m\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right)\). d. Does the work depend on the path between \(A\) and \(B ?\) Explain.

Heat flux The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k > 0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\) $$T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}} ; S \text { is the sphere } x^{2}+y^{2}+z^{2}=a^{2}$$

Gauss' Law for gravitation The gravitational force due to a point mass \(M\) at the origin is proportional to \(\mathbf{F}=G M \mathbf{r} /|\mathbf{r}|^{3},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(G\) is the gravitational constant. a. Show that the flux of the force field across a sphere of radius \(a\) centered at the origin is \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=4 \pi G M\) b. Let \(S\) be the boundary of the region between two spheres centered at the origin of radius \(a\) and \(b,\) respectively, with \(a

Conservation of energy Suppose an object with mass \(m\) moves in a region \(R\) in a conservative force field given by \(\mathbf{F}=-\nabla \varphi,\) where \(\varphi\) is a potential function in a region \(R .\) The motion of the object is governed by Newton's Second Law of Motion, \(\mathbf{F}=m \mathbf{a},\) where a is the acceleration. Suppose the object moves from point \(A\) to point \(B\) in \(R\) a. Show that the equation of motion is \(m \frac{d \mathbf{v}}{d t}=-\nabla \varphi\) b. Show that \(\frac{d \mathbf{v}}{d t} \cdot \mathbf{v}=\frac{1}{2} \frac{d}{d t}(\mathbf{v} \cdot \mathbf{v})\) c. Take the dot product of both sides of the equation in part (a) with \(\mathbf{v}(t)=\mathbf{r}^{\prime}(t)\) and integrate along a curve between \(A\) and B. Use part (b) and the fact that \(\mathbf{F}\) is conservative to show that the total energy (kinetic plus potential) \(\frac{1}{2} m|\mathbf{v}|^{2}+\varphi\) is the same at \(A\) and \(B\). Conclude that because \(A\) and \(B\) are arbitrary, energy is conserved in \(R\).

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