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Chapter 9: Problem 13

Use Green's theorem to evaluate the given line integral. \(\oint_{C} \frac{1}{3} y^{3} d x+\left(x y+x y^{2}\right) d y\), where \(C\) is the boundary of the region in the first quadrant determined by the graphs of \(y \quad 0\), \(x \quad y^{2}, x \quad 1-y^{2}\)

Short Answer

Expert verified
The value of the line integral is 0.

Step by step solution

01

Understand Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \). Formally, it states \( \oint_{C} M\,dx + N\,dy = \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \). To apply it, identify \( M = \frac{1}{3}y^3 \) and \( N = xy + xy^2 \).
02

Compute Partial Derivatives

Calculate \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \). \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(xy + xy^2) = y + y^2 \). Calculate \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(\frac{1}{3}y^3) = y^2 \).
03

Set Up Double Integral

The double integral is \( \iint_{D} (y + y^2 - y^2) \, dA = \iint_{D} y \, dA \). The partial derivatives simplified to \( y \).
04

Determine the Region of Integration

The region \( D \) is bounded by \( y = 0 \), \( x = y^2 \), and \( x = 1-y^2 \). Therefore, integrate with respect to \( y \) from \( y=0 \) to \( y=1 \), and with respect to \( x \) from \( x = y^2 \) to \( x = 1-y^2 \).
05

Evaluate the Double Integral

The integral becomes: \( \int_{0}^{1} \int_{y^2}^{1-y^2} y \, dx \, dy \). First, integrate with respect to \( x \): \( \int_{y^2}^{1-y^2} y \, dx = y(x) \bigg|_{y^2}^{1-y^2} = y(1-y^2 - y^2) = y(1-2y^2) \).
06

Integrate with Respect to \( y \)

Evaluate \( \int_{0}^{1} y(1-2y^2) \, dy = \int_{0}^{1} (y - 2y^3) \, dy \). Solve for the antiderivative: \( \frac{1}{2}y^2 - \frac{1}{2}y^4 \) evaluated from \(0\) to \(1\).
07

Compute the Definite Integral

Plug in the limits to find \( \left( \frac{1}{2}(1)^2 - \frac{1}{2}(1)^4 \right) - \left( \frac{1}{2}(0)^2 - \frac{1}{2}(0)^4 \right) = \left( \frac{1}{2} - \frac{1}{2} \right) = 0 \).

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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 the function to be integrated is evaluated along a curve. In simpler terms, think of it as adding up tiny values along a path. Green's Theorem relates the line integral around a closed curve to a certain kind of double integral over the region it encloses. In the exercise, the line integral we began with is\[ \oint_{C} \frac{1}{3} y^3 \, dx + (xy + xy^2) \, dy. \]Line integrals are useful because they allow us to evaluate a function along a path within a field. Some key points about line integrals:
  • Path-dependent: They rely on the path taken, not just the endpoints.
  • Applications: Often used in physics for work done by a force field.
  • Requires parametric equations of the curve for computations.
Double Integral
A double integral is essentially an operation to compute a sum over a two-dimensional region. Such integrals generalize the notion of a one-dimensional definite integral.In our exercise with Green's Theorem, the line integral is converted into a double integral over a region \(D\): \[ \iint_{D} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA = \iint_{D} y \, dA. \]This transformation is useful because it changes a potentially difficult problem into a simpler one.Why double integrals matter:
  • Calculate area and volume: Especially in more complex shapes.
  • Transform into other coordinate systems: For ease of calculation.
  • Physics applications: Used in electromagnetism and fluid dynamics.
Partial Derivatives
Partial derivatives show how a function changes as its variables change. For functions with more than one variable, like our function with \(M = \frac{1}{3}y^3\) and \(N = xy + xy^2\), partial derivatives focus on one variable at a time while keeping others constant.In the process of applying Green's Theorem:
  • Find \(\frac{\partial N}{\partial x}\) for \(N = xy + xy^2\): This yields \(y + y^2\).
  • Find \(\frac{\partial M}{\partial y}\) for \(M = \frac{1}{3}y^3\): This simplifies to \(y^2\).
Finding these partial derivatives allow us to determine the integrand for the double integral, simplifying the equation to just \(y\). Importance of partial derivatives:
  • Core component in many calculus problems.
  • Used in optimization, physics, and in the study of dynamic systems.
  • Helps in understanding changes to multi-variable functions.
Region of Integration
The region of integration is the area over which we compute our double integral. In mathematical terms, it's often represented as a planar region \( D \) in coordinates.For our problem, the region \( D \) is defined by the curves:
  • \(y = 0\)
  • \(x = y^2\)
  • \(x = 1-y^2\)
These boundaries form a specific shape in the first quadrant, dictating where our calculations for the double integral will occur. Why is defining the region crucial?
  • Ensures correct limits of integration.
  • Affects the outcome of the integral values.
  • Visualizing it helps understand problem parameters.
Setting clear boundaries helps ensure you're evaluating the integral over the correct area.

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