/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Line integrals Use Green's Theor... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 34

Line integrals Use Green's Theorem to evaluate the following line integrals. Assume all curves are oriented counterclockwise. A sketch is helpful. \(\int_{c} \frac{1}{1+y^{2}} d x+y d y,\) where \(C\) is the boundary of the triangle with vertices \((0,0),(1,0),\) and (1,1)

Short Answer

Expert verified
Question: Evaluate the line integral: $$\oint_C \left(\frac{1}{1+y^2} dx + y dy\right)$$ where C is the boundary of the triangle with vertices (0,0), (1,0), and (1,1). Answer: The line integral is equal to: $$-\frac{\pi}{4} + 1$$

Step by step solution

01

For P(x, y) = \(\frac{1}{1+y^{2}}\), \(\frac{\partial P}{\partial y} = - \frac{2y}{(1+y^2)^2}\). For Q(x, y) = y, \(\frac{\partial Q}{\partial x} = 0\). #Step 2: Set up the double integral using Green's theorem# According to Green's Theorem, the line integral can be expressed as the double integral of the region. Since the curve C is the boundary of the triangle with vertices (0,0), (1,0), and (1,1), we can describe the region D as $$0\leq x\leq1, 0\leq y\leq x$$. $$\oint_C \left(\frac{1}{1+y^2} dx + y dy\right) = \iint_D\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dA$$ Now we can plug in the values of the partial derivatives and set up the double integral with respect to x and y. $$= \int_0^1\int_0^x (0 - (- \frac{2y}{(1+y^2)^2})) dy dx$$ #Step 3: Evaluate the double integral# Now, let's solve the integral by evaluating the inner integral first and then the outer integral.

For the inner integral, we have $$\int_0^x \frac{2y}{(1+y^2)^2} dy .$$ Let $$u = 1+y^2,$$ then $$du = 2ydy$$. The integral becomes $$\int \frac{1}{u^2} du$$ With the limits of integration changing as: When y = 0, u = 1 (lower limit) When y = x, u = 1+x^2 (upper limit) So the new integral becomes: $$\int_1^{1+x^2} \frac{1}{u^2} du$$ Now we can evaluate this integral: $$= -\frac{1}{u} \Big|_1^{1+x^2} = -\frac{1}{1+x^2} + 1$$ Now, we need to evaluate the outer integral: $$\int_0^1 (-\frac{1}{1+x^2} + 1) dx$$ Separate the integrals to make it easier to evaluate: $$ = \int_0^1 -\frac{1}{1+x^2} dx + \int_0^1 dx$$ Evaluate the integrals: $$ = \left[-\arctan(x)\right]_0^1 + [x]_0^1 = (-\arctan(1) + \arctan(0)) + (1-0) = -\frac{\pi}{4} + 1$$ So, the given line integral is: $$\oint_C \left(\frac{1}{1+y^2} dx + y dy\right) = -\frac{\pi}{4} + 1$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Double Integral
A double integral is a mathematical operation used to calculate the volume under a surface over a two-dimensional region. This concept is often used in physics and engineering to determine quantities like mass, volume, and center of mass. In the context of Green's Theorem, which connects line integrals with double integrals, it transforms a complicated line integral along a curve into a more manageable double integral over the region enclosed by the curve.

In the solution steps, we convert the line integral \(\int_{c} \frac{1}{1+y^{2}} dx + y dy\) into a double integral over the region D. The region D is defined by the bounds of integration, which for the given triangle are \(0 \leq x \leq 1\), \(0 \leq y \leq x\). This step simplifies the problem by allowing us to evaluate the area under the curve of the partial derivative differences within region D, rather than manually tracing the curve C.
Partial Derivatives
Partial derivatives are used to measure how a function changes as only one of its input variables changes while the others remain fixed. They are foundational in multivariate calculus and are critical when applying Green's Theorem to line integrals.

In our example, we take the partial derivative of \(P(x, y)\) with respect to y, which yields \(\frac{\partial P}{\partial y} = - \frac{2y}{(1+y^2)^2}\). Similarly, the partial derivative of \(Q(x, y)\) with respect to x is zero, as Q(x, y) is simply \(y\). In the context of Green's Theorem, we're interested in the partial derivatives because the theorem states that under certain conditions, the line integral around a closed curve C is equal to the double integral of the difference between the partial derivatives over the region D enclosed by C.
Orientation of Curves
The orientation of a curve is crucial when applying Green's Theorem because the theorem assumes a positive orientation (counterclockwise) of the curve. The orientation determines the direction to traverse along the curve, which affects the sign of the line integral's value.

In the provided exercise, it is assumed that all curves are oriented counterclockwise, consistent with the standard orientation required by Green's Theorem. If the curve were oriented clockwise, it would change the sign of the line integral. This aspect is an integral part of applying Green's Theorem correctly; it ensures consistency and correctness when translating a line integral into a double integral over a particular region.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Zero flux fields. For what values of \(a\) and \(d\) does the vector field \(\mathbf{F}=\langle a x, d y\rangle\) have zero flux across the unit circle centered at the origin and oriented counterclockwise?

A beautiful flux integral Consider the potential function \(\varphi(x, y, z)=G(\rho),\) where \(G\) is any twice differentiable function and \(\rho=\sqrt{x^{2}+y^{2}+z^{2}} ;\) therefore, \(G\) depends only on the distance from the origin. a. Show that the gradient vector field associated with \(\varphi\) is \(\mathbf{F}=\nabla \varphi=G^{\prime}(\rho) \frac{\mathbf{r}}{\rho},\) where \(\mathbf{r}=\langle x, y, z\rangle\) and \(\rho=|\mathbf{r}|\)b. Let \(S\) be the sphere of radius \(a\) centered at the origin and let \(D\) be the region enclosed by \(S\). Show that the flux of \(\mathbf{F}\) across \(S\) is \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=4 \pi a^{2} G^{\prime}(a)\) c. Show that \(\nabla \cdot \mathbf{F}=\nabla \cdot \nabla \varphi=\frac{2 G^{\prime}(\rho)}{\rho}+G^{\prime \prime}(\rho)\) d. Use part (c) to show that the flux across \(S\) (as given in part (b)) is also obtained by the volume integral \(\iiint_{D} \nabla \cdot \mathbf{F} d V\) (Hint: Use spherical coordinates and integrate by parts.)

Let \(\mathbf{F}=\langle z, 0,-y\rangle\) a. Find the scalar component of curl \(\mathbf{F}\) in the direction of the unit vector \(\mathbf{n}=\langle 1,0,0\rangle\). b. Find the scalar component of curl \(\mathbf{F}\) in the direction of the unit vector \(\mathbf{n}=\left\langle\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right\rangle\). c. Find the unit vector \(\mathbf{n}\) that maximizes \(\operatorname{scal}_{\mathbf{n}}\langle-1,1,0\rangle\) and state the value of \(\operatorname{scal}_{\mathbf{n}}\langle-1,1,0\rangle\) in this direction.

Given a vector field \(\mathbf{F}=\langle f, 0\rangle\) and curve \(C\) with parameterization \(\mathbf{r}(t)=\langle x(t), y(t)\rangle,\) for \(a \leq t \leq b,\) we see that the line integral \(\int_{C} f d x+g d y\) simplifies to \(\int_{C} f d x\) a. Show that \(\int_{C} f d x=\int_{a}^{b} f(t) x^{\prime}(t) d t\) b. Use the vector field \(\mathbf{F}=\langle 0, g\rangle\) to show that \(\int_{C} g d y=\int_{a}^{b} g(t) y^{\prime}(t) d t\) c. Evaluate \(\int_{C} x y d x,\) where \(C\) is the line segment from (0,0) to (5,12) d. Evaluate \(\int_{C} x y d y,\) where \(C\) is a segment of the parabola \(x=y^{2}\) from (1,-1) to (1,1)

For what vectors \(\mathbf{n}\) is \((\operatorname{curl} \mathbf{F}) \cdot \mathbf{n}=0\) when \(\mathbf{F}=\langle y,-2 z,-x\rangle ?\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.