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Chapter 6: Problem 91

For the following exercises, find the flux. Let \(\mathbf{F}=\left(x^{2}+y^{3}\right) \mathbf{i}+(2 x y) \mathbf{j}\) . Calculate flux \(\mathbf{F}\) orientated counterclockwise across curve \(C : x^{2}+y^{2}=9\)

Short Answer

Expert verified
The flux is \(-81\pi\).

Step by step solution

01

Understand the Problem

We need to find the flux of the vector field \( \mathbf{F} = (x^2 + y^3) \mathbf{i} + (2xy) \mathbf{j} \) across a circular curve \( C : x^2 + y^2 = 9 \) oriented counterclockwise.
02

Use Green's Theorem

Green's Theorem relates the flux of a vector field across a closed curve to a double integral over the region it encloses. It states: \[ \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] where \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \). Here, \( P = x^2 + y^3 \) and \( Q = 2xy \).
03

Compute the Partial Derivatives

To apply Green's Theorem, find the partial derivatives: \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2xy) = 2y \) and \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^3) = 3y^2 \).
04

Set Up the Double Integral

Insert these into the expression from Green's Theorem: \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2y - 3y^2 \). The region \( R \) is the circle \( x^2 + y^2 \leq 9 \).
05

Convert to Polar Coordinates

Since the region is a circle, converting to polar coordinates simplifies the integral. In polar, \( x = r\cos\theta \), \( y = r\sin\theta \), and \( dA = r \, dr \, d\theta \). The circle \( R \) becomes \( r \leq 3 \).
06

Evaluate the Double Integral

The integral becomes: \[ \int_0^{2\pi} \int_0^3 (2r\sin\theta - 3r^2\sin^2\theta) r \, dr \, d\theta \]. Integrate with respect to \( r \) first: \[ \int_0^3 (2r^2\sin\theta - 3r^3\sin^2\theta) \, dr = \left[ \frac{2r^3}{3}\sin\theta - \frac{3r^4}{4}\sin^2\theta \right]_0^3 \].
07

Simplify and Finalize the Calculation

Evaluate the bounds: \[ \left(18\sin\theta - (81\sin^2\theta)\right) = (18\sin\theta - 81\sin^2\theta) \]. Integrate over \( \theta \): \[ \int_0^{2\pi} (18\sin\theta - 81\sin^2\theta) \, d\theta \]. After solving, observe that \( \int_0^{2\pi} \sin\theta \, d\theta = 0 \) and \( \int_0^{2\pi} \sin^2\theta \, d\theta = \pi \). Thus, the final result is \(-81\pi\).

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Key Concepts

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

Green's Theorem
Green's Theorem connects the work along a curve to a double integral over the region the curve encloses. It's a cornerstone of vector calculus. Think of it as a bridge between a line integral around a closed curve and a double integral over the plane region inside the curve.

Green's Theorem states:
  • Use it for vector fields \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \).
  • The theorem equates \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds \) to \( \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).
This means that the line integral of \( \mathbf{F} \) across a closed curve \( C \) is equal to the sum of the partial derivatives over the enclosed region \( R \). In practice, it makes solving complex curves easier by transforming them into area problems.
Polar Coordinates
When we work with problems involving circular regions, polar coordinates can simplify our calculations. They are a way of expressing points in the plane using angles and radii, which can be very handy for circular symmetry.

In polar coordinates:
  • \( x = r\cos\theta \)
  • \( y = r\sin\theta \)
  • The area element is \( dA = r \, dr \, d\theta \)
With these substitutions, circular regions often become easier to work with than using traditional \( x, y \) coordinates. The key is to transform the limits of integration and the functional form, which can lead to more straightforward integration in many cases.
Double Integral
A double integral allows us to integrate functions over a two-dimensional region, capturing the sum of function values throughout a specified region across both dimensions.

In polar coordinates:
  • The region is defined by limits, often \( r \) for radius and \( \theta \) for angle.
  • This becomes a powerful tool in evaluating quantities like area or flux because it captures contributions from infinitesimally small elements over the whole region.
We decompose the integral into two single integrals: one over the radius \( r \) and the other over the angle \( \theta \). Solving them separately can simplify computations, especially for symmetric domains like circles.
Partial Derivatives
Partial derivatives measure how a multivariable function changes when one variable changes while others are held constant. They are crucial in vector field analysis, especially for applications of Green's Theorem.

Here's how they are used:
  • For a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), measures the rate of change in the \( x \) direction.
  • Similarly, \( \frac{\partial f}{\partial y} \) is the rate of change in the \( y \) direction.
In exercises like the flux problem, these derivatives help us compute the necessary integral by indicating the behavior of the field within the region. Understanding these concepts thoroughly unlocks deeper insights into multivariate calculus challenges.

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