/*! 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 28 Apply Green's Theorem to evaluat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Apply Green's Theorem to evaluate the integrals in Exercises \(27-30 .\) $$\oint(3 y d x+2 x d y)$$ \(C:\) The boundary of \(0 \leq x \leq \pi, 0 \leq y \leq \sin x\)

Short Answer

Expert verified
The value of the integral is 2.

Step by step solution

01

Understand the Given Curve

The curve \( C \) encloses the region defined by the equations \( 0 \leq x \leq \pi \) and \( 0 \leq y \leq \sin x \). This region is the area under the sine curve from \( x = 0 \) to \( x = \pi \).
02

Apply Green's Theorem

Green's Theorem states that \( \oint_C (M \, dx + N \, dy) = \int\int_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \), where \( R \) is the region enclosed by \( C \). Here, \( M = 3y \) and \( N = 2x \).
03

Compute Partial Derivatives

Calculate the partial derivatives: \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2x) = 2 \) and \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3y) = 3 \).
04

Setup Double Integral

Substitute the partial derivatives into Green's Theorem: \( \int\int_R (2 - 3) \, dA = \int\int_R (-1) \, dA \). This simplifies the problem to finding the area of \( R \) under the sine curve between \( x = 0 \) and \( x = \pi \).
05

Evaluate the Double Integral

The double integral becomes \(-\int_0^\pi \int_0^{\sin x} 1 \, dy \, dx = -\int_0^\pi [y]_0^{\sin x} \, dx = -\int_0^\pi \sin x \, dx\). Evaluate this integral.
06

Compute Definite Integral

Calculate the definite integral: \(-\int_0^\pi \sin x \, dx = -[-\cos x]_0^\pi = -[(-1) - (1)] = 2\).
07

Conclusion

The integral around the curve \( C \) evaluates to \( 2 \).

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 91Ó°ÊÓ!

Key Concepts

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

Line Integrals
Line integrals are an essential concept in vector calculus that help in understanding how a vector field affects a curve. When performing a line integral, we integrate along a path or curve, which is denoted by the symbol \( \oint \) when the path forms a closed loop. Line integrals are used to calculate the work done by a force field as an object moves along a path or to find circulation and flux.

For example, in the given exercise, the line integral \( \oint (3y \, dx + 2x \, dy) \) represents integrating the expression over the closed curve \( C \), which is the boundary of the region \( 0 \leq x \leq \pi, 0 \leq y \leq \sin x \). This integral measures the work done by the vector field \( \mathbf{F} = (3y, 2x) \) along curve \( C \).
  • The function \( M = 3y \) multiplies the differential \( dx \), indicating changes in the \( x \)-direction.
  • The function \( N = 2x \) multiplies the differential \( dy \), indicating changes in the \( y \)-direction.

Such an integral can often be simplified using Green's Theorem, allowing us to convert it into a double integral, making it easier to evaluate.
Double Integrals
Double integrals allow for the calculation of volume under a surface over a region \( R \). These integrals extend the concept of calculating area under a curve to a two-dimensional region. In the context of Green's Theorem, double integrals help in determining the area or the total contribution of a field over a defined region.

In the given exercise, after applying Green's Theorem, the line integral is converted into a double integral \( \int\int_R (2 - 3) \, dA = \int\int_R (-1) \, dA \). Here, the partial derivatives computed from the vector field give us a simplified version for calculating the integral by integrating over region \( R \), which is under the sine curve between \( x = 0 \) and \( x = \pi \).
  • The negative sign in \( \int\int_R (-1) \, dA \) signifies subtraction, indicating the downward contribution or the negative orientation of the area.
  • The region \( R \) is set up using the limits of the sine function within the specified range.

Through evaluating this double integral, we calculate the area under the sine curve precisely.
Partial Derivatives
Partial derivatives are derivatives of multivariable functions with respect to one variable, holding other variables constant. They are crucial in understanding how a function changes at a point concerning each variable individually.

In Green's Theorem, partial derivatives are used to transform a line integral into a double integral. For the given vector field \( \mathbf{F} = (3y, 2x) \), we compute the partial derivatives with respect to \( x \) and \( y \):
  • \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(2x) = 2 \), showing how \( N = 2x \) changes in the \( x \)-direction.
  • \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3y) = 3 \), showing how \( M = 3y \) changes in the \( y \)-direction.

These derivatives are used to compute \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 2 - 3 = -1 \), simplifying the line integral into an easier-to-evaluate double integral. Partial derivatives thus provide insight into the internal structure of the vector field and guide us through the simplification process.

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

Use a parametrization to express the area of the surface as a double integral. Then evaluate the integral. (There are many correct ways to set up the integrals, so your integrals may not be the same as those in the back of the text. They should have the same values, however.) The lower portion cut from the sphere \(x^{2}+y^{2}+z^{2}=2\) by the cone \(z=\sqrt{x^{2}+y^{2}}\)

Gradient of a line integral Suppose that \(\mathbf{F}=\nabla f\) is a conservative vector field and $$ g(x, y, z)=\int_{(0,0,0)}^{(x, y, z)} \mathbf{F} \cdot d \mathbf{r} $$ Show that \(\nabla g=\mathbf{F}\)

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

Use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(\mathbf{F}\) around the simple closed curve C. Perform the following CAS steps. a. Plot \(C\) in the \(x y\) -plane. b. Determine the integrand (aN/ax) \(-(a M /\) ay ) for the tangential form of Green's Theorem. c. Determine the (double integral) limits of integration from your plot in part (a), and evaluate the curl integral for the circulation. $$\mathbf{F}=x e^{y} \mathbf{i}+\left(4 x^{2} \ln y\right) \mathbf{j}$$ C: The triangle with vertices \((0,0),(2,0),\) and (0,4)

Use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D\). Thick sphere \(\quad \mathbf{F}=(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) / \sqrt{x^{2}+y^{2}+z^{2}}\) \(D:\) The region \(1 \leq x^{2}+y^{2}+z^{2} \leq 4\)

See all solutions

Recommended explanations on Math Textbooks

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.