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Chapter 6: Problem 192

Use Green's theorem to evaluate line integral \(\int_{C} \sin y d x+x \cos y d y\) where \(C\) is ellipse \(x^{2}+x y+y^{2}=1\) oriented in the counterclockwise direction.

Short Answer

Expert verified
The line integral evaluates to zero.

Step by step solution

01

Green's Theorem Identification

Green's Theorem states that for a curve \(C\) enclosing a region \(R\), \( \oint_{C} (Pdx + Qdy) = \int \int_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \). In the given problem, \(P = \sin y\) and \(Q = x\cos y\).
02

Calculate Partial Derivatives

Compute \( \frac{\partial Q}{\partial x}\) and \( \frac{\partial P}{\partial y}\):- \( \frac{\partial Q}{\partial x} = \cos y \)- \( \frac{\partial P}{\partial y} = \cos y \).Substituting these, we find \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \cos y - \cos y = 0 \).
03

Evaluate Double Integral

The double integral becomes \( \int \int_{R} 0 \, dA \). Since the integrand is zero, the value of the integral over the region \(R\) (bounded by the ellipse \(x^2 + xy + y^2 = 1\)) is zero.
04

Conclusion

Since the double integral evaluated to zero, it implies that the original line integral \(\int_{C} \sin y \, dx + x \cos y \, dy\) is also zero.

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Key Concepts

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

Line Integral
A line integral is a type of integral where a function is evaluated along a curve. Generally, this involves integrating a function over a path in the plane or in space. In our context, we have a function that's dependent on both the position and an external function as we move along a curve.
  • For line integrals involving vector fields, as in our problem, we integrate a combination of functions over a path.
  • The situation often looks like this: an integral of the form \( \int_C P \, dx + Q \, dy \) for a given curve \( C \).
  • This is seen in physical applications, like calculating the work done by a force along a path.
In the context of Green's Theorem, line integrals around a closed curve can be related to a double integral over the region encompassed by the curve. This powerful theorem simplifies calculations by transforming complex line integrals into typically simpler double integrals.
Partial Derivatives
Partial derivatives are pivotal in multivariable calculus, where functions depend on more than one variable. Simply put, a partial derivative measures how a function changes as one variable changes, keeping the others constant.
  • For the function \(Q = x \cos y\), \( \frac{\partial Q}{\partial x} \) tells us how \(Q\) changes as \(x\) changes. Here, \( \frac{\partial Q}{\partial x} = \cos y\), emphasizing that this derivative does not depend on \(x\), but only on \(y\), showing the relationship between the variables.
  • Similarly, for \(P = \sin y\), \( \frac{\partial P}{\partial y} \) reveals how \(P\) changes with respect to \(y\). We find \(\frac{\partial P}{\partial y} = \cos y\), which again is independent of \(x\).
In scenarios like our problem, these derivatives help us evaluate the behaviors of functions over regions enclosed by curves. By utilizing partial derivatives, Green's Theorem facilitates a transition from line integrals to potentially simpler forms.
Ellipse
An ellipse is a geometric shape that resembles a stretched circle. It's defined by a specific algebraic equation that describes the set of all points whose distances to two fixed points (the foci) sum to a constant.
  • The given equation \(x^2 + xy + y^2 = 1\) defines an ellipse in this problem, which influences the path \(C\) along which our line integral is evaluated.
  • Counterclockwise orientation means we traverse the boundary of this ellipse in such a way that the region enclosed is always on our left. This orientation is significant due to Green's Theorem's convention.
  • In both physics and engineering, ellipses can model various real-world phenomena, from planetary orbits to signal focal points in acoustics and optics.
Understanding the role of the ellipse in the context of Green's Theorem allows us to visualize better how the region \(R\) between the curve \(C\) affects the evaluation of the double integrals, and how ultimately, certain properties, like symmetry, lead to simplified results.

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Most popular questions from this chapter

Find \(\int_{C} y^{2} d x+\left(x y-x^{2}\right) d y\) along \(C: y=3 x\) from (0,0) to (1,3).

Use the divergence theorem to calculate the flux of \(\mathbf{F}(x, y, z)=x^{3} \mathbf{i}+y^{3} \mathbf{j}+z^{3} \mathbf{k} \quad\) through sphere \(x^{2}+y^{2}+z^{2}=1\)

IT] Use a CAS to find the flux of vector field \(\mathbf{F}(x, y, z)=\left(e^{y}+x\right) \mathbf{i}+(3 \cos (x z)-y) \mathbf{j}+z \mathbf{k} \quad\) through surface \(S\), where \(S\) is given by \(z^{2}=4 x^{2}+4 y^{2}\) from \(0 \leq z \leq 4,\) oriented so the unit nomal vector points downward.

For the following exercises, use a CAS and the divergence theorem to compute the net outward flux for the vector fields across the boundary of the given regions \(D\). \(|\mathrm{T}| \mathbf{F}=\langle z-x, x-y, 2 y-z\rangle ; D\) is the region between spheres of radius 2 and 4 centered at the origin.

For the following exercises, use Stokes' theorem to find the circulation of the following vector fields around any smooth, simple closed curve C. $$ \mathbf{F}=\nabla\left(x \sin y e^{z}\right) $$
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