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Chapter 16: Problem 8

Use Green's Theorem to evaluate the line integral along the given positively oriented curve. \(\int_{c} y^{4} d x+2 x y^{3} d y, \quad C\) is the ellipse \(x^{2}+2 y^{2}=2\)

Short Answer

Expert verified
The line integral evaluates to 0 using Green's Theorem.

Step by step solution

01

Identify the Given Functions

The vector field components represented in the line integral are given by \( P(x, y) = y^4 \) and \( Q(x, y) = 2xy^3 \).
02

Set Up Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the region \( R \) it encloses: \( \int_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).
03

Calculate Partial Derivatives

Calculate \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2xy^3) = 2y^3 \) and \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y^4) = 4y^3 \).
04

Evaluate the Expression Inside the Double Integral

Substitute the partial derivatives into the expression: \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2y^3 - 4y^3 = -2y^3 \).
05

Set Up the Double Integral

The region \( R \) is the ellipse defined by \( x^2 + 2y^2 = 2 \). We will convert this to polar coordinates or use a suitable substitution. Letting \( x = \sqrt{2}\cos\theta \) and \( y = \sin\theta \), we describe the ellipse fully.
06

Convert to Polar Coordinates

Although we convert for ease, integrals can often use standard forms: \( x = \sqrt{2}r\cos\theta, \; y = r\sin\theta \), where \( 0 \leq \theta < 2\pi \) and apply the Jacobian for integration.
07

Perform the Double Integral

The transformed integral simplifies to an integral over the circle \( x^2 + y^2 = 1 \). The Jacobian determinant conversion is necessary here, translating to simpler polar integration \( r \). Integration of \(-2y^3\) will show circular symmetry providing zero contribution over full bounds.

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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 allows us to evaluate a function along a curve. It's particularly useful for calculating quantities like work done by a force field on a path. In this context, Green's Theorem helps transform a line integral into a more manageable double integral, involving vector fields.
  • A line integral on a curve considers how a vector field affects a path.
  • It can be thought of as summing up small contributions along the curve.
  • The curve, in this exercise, is oriented positively, meaning counterclockwise.
  • Mathematically, for a vector field F = (P, Q), the line integral is expressed as: \[ \int_{C} P \, dx + Q \, dy \]
Understanding the relationship between the line integral and the enclosed area via Green's Theorem offers a path from seemingly complex integrals solving over curves to simpler double integrals over planar regions.
Ellipse
In this exercise, the curve of interest is an ellipse. Understanding its equation and properties is essential for applying Green鈥檚 Theorem accurately. The ellipse given is:
  • Its equation is \(x^2 + 2y^2 = 2\), which slightly differs from the standard form of an ellipse \(ax^2 + by^2 = c\).
  • In this specific equation, the coefficients of \(x^2\) and \(2y^2\) indicate the semi-axes of the ellipse.
  • To convert this into a more mathematically favorable form, polar coordinate substitution is introduced.
This conversion into polar coordinates simplifies the computation, especially for double integration, as the relationship involves easier symmetries and circular substitution aspects rather than handling elliptical descent directly.
Vector Field
Vector fields are essential in understanding line integrals and their applications. In this scenario, the vector field F is given by its components: \(P(x, y) = y^4\) and \(Q(x, y) = 2xy^3\).
  • A vector field assigns a vector to every point in a plane or space.
  • For Green's Theorem, these components split into horizontal (\(P\)) and vertical (\(Q\)) directions.
  • Green鈥檚 Theorem interrelates these components to transform complex path integrals into double integrals, using the vector nature of fields.
  • It's necessary to compute the partial derivatives of these components, a crucial step in Green鈥檚 Theorem application.
Partial derivatives help determine the change rates along different field lines and are vital for the transition from line to double integral as seen in this exercise.
Double Integral
Double integrals provide a way to evaluate functions over a two-dimensional region. They are used in Green's Theorem to replace the line integral with one over an area:
  • Green's Theorem applies to convert the line integral into a double integral over the enclosed region, reducing complexity in evaluation.
  • The setup involves finding the expressions \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\), analytical steps required for double integration.
  • In our exercise, these expressions simplify to an integrable form in more convenient coordinates or findings due to symmetry.
  • The final form of integral involves standard trigonometric transformations, switching to polar for simplification.
Thus, the double integral allows us to sum little pieces of an area, replacing tedious curve traversal by utilizing area-under-the-curve concepts more aligned to Calculus.

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Most popular questions from this chapter

Use a computer to graph the parametric surface. Get a printout and indicate on it which grid curves have \(u\) constant and which have \(v\) constant. $$ \begin{array}{l}{x=\cos u, \quad y=\sin u \sin v, \quad z=\cos v} \\ {0 \leqslant u \leqslant 2 \pi, \quad 0 \leqslant v \leqslant 2 \pi}\end{array} $$

Find the area of the part of the sphere \(x^{2}+y^{2}+z^{2}=a^{2}\) that lies inside the cylinder \(x^{2}+y^{2}=a x\)

Determine whether the points \(P\) and \(Q\) lie on the given surface. $$ \begin{array}{l}{\mathbf{r}(u, v)=\langle u+v, u-2 v, 3+u-v\rangle} \\ { P(4,-5,1), Q(0,4,6)}\end{array} $$

Maxwell's equations relating the electric field \(\mathbf{E}\) and magnetic field \(\mathbf{H}\) as they vary with time in a region containing no charge and no current can be stated as follows: $$ \operatorname{div} \mathbf{E}=0 \quad \quad \text { div } \mathbf{H}=0 $$ $$ \operatorname{curl} \mathbf{E}=-\frac{1}{c} \frac{\partial \mathbf{H}}{\partial t} \quad \operatorname{curl} \mathbf{H}=\frac{1}{c} \frac{\partial \mathbf{E}}{\partial t} $$ where \(c\) is the speed of light. Use these equations to prove the following: $$ \begin{array}{l}{\text { (a) } \nabla \times(\nabla \times \mathbf{E})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}} \\\ {\text { (b) } \nabla \times(\nabla \times \mathbf{H})=-\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}} \\ {\text { (c) } \nabla^{2} \mathbf{E}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}} \quad[\text {Hint: Use Exercise } 29 .]} \\ {\text { (d) } \nabla^{2} \mathbf{H}=\frac{1}{c^{2}} \frac{\partial^{2} \mathbf{H}}{\partial t^{2}}}\end{array} $$

Find the mass and center of mass of a wire in the shape of the helix \(x=t, y=\cos t, z=\sin t, 0 \leqslant t \leqslant 2 \pi\), if the density at any point is equal to the square of the distance from the origin.
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