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Chapter 16: Problem 5

Let \(\boldsymbol{f}=\langle y,-x\rangle\) and let \(D\) be given by \(x^{2}+y^{2} \leq 1 .\) Compute \(\int_{\partial D} \boldsymbol{f} \cdot d \boldsymbol{r}\) and \(\int_{\partial D} \boldsymbol{f} \cdot \boldsymbol{N} d s .\)

Short Answer

Expert verified
\(\int_{\partial D} \boldsymbol{f} \cdot d \boldsymbol{r} = -2\pi\) and \(\int_{\partial D} \boldsymbol{f} \cdot \boldsymbol{N} \, ds = 0\).

Step by step solution

01

Identify the given vector field and region

The vector field given is \( \boldsymbol{f} = \langle y, -x \rangle \). The region \( D \) is a circle defined by the inequality \( x^2 + y^2 \leq 1 \), which is a unit circle centered at the origin.
02

Apply Green's Theorem

Green's Theorem states that for a region \( D \), \( \int_{\partial D} \boldsymbol{f} \cdot d\boldsymbol{r} = \int \int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \), where \( \boldsymbol{f} = \langle P, Q \rangle \). Here, \( P = y \) and \( Q = -x \).
03

Calculate the partial derivatives

Compute the partial derivatives: \( \frac{\partial Q}{\partial x} = \frac{\partial (-x)}{\partial x} = -1 \), and \( \frac{\partial P}{\partial y} = \frac{\partial y}{\partial y} = 1 \).
04

Evaluate the double integral over the region D

Substitute the partial derivatives to find \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -1 - 1 = -2 \). The area integral becomes \( \int \int_{D} -2 \, dA = -2 \times \text{area of } D = -2 \times \pi \). Since the area of the unit circle \( D \) is \( \pi \), the integral yields \( -2\pi \).
05

Compute \(\int_{\partial D} \boldsymbol{f} \cdot \boldsymbol{N} \, ds\)

Notice that for a closed curve on the plane, the vector field \( \boldsymbol{f} = \langle y, -x \rangle \) is tangent to the circle, meaning \( \boldsymbol{f} \perp \boldsymbol{N} \), the outward normal vector. Therefore, the integral \( \int_{\partial D} \boldsymbol{f} \cdot \boldsymbol{N} \, ds = 0 \), because the dot product of perpendicular vectors is zero.

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Key Concepts

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

Vector Field
A vector field is a mathematical construct where each point in a space is assigned a vector. Think of it like attaching an arrow, representing a vector, to every point in a region. In this exercise, the vector field is given as \( \boldsymbol{f} = \langle y, -x \rangle \). This means the vector at each point \((x, y)\) is \(y\) in the x-direction and \(-x\) in the y-direction.
In simpler terms:
  • At any point, \(y\) dictates how far and which way to step in the x-direction.
  • Similarly, \(-x\) dictates the y-direction movement.
By understanding these vectors, you can visualize how a field applies a force or transformation throughout the space. Consequently, vector fields are fundamental in modeling a wide array of physical phenomena.
Unit Circle
The unit circle is a special circle in mathematics, representing all points that are exactly one unit away from a central point, usually at the origin \((0, 0)\). This is represented by the equation \(x^2 + y^2 \leq 1\). Here, within the inequality, all points inside (and on) the circle fit the rule where their coordinates provide the square sum less than or equal to one.
  • The 'unit' refers to the radius having a length of one.
  • This circle is centered at the origin, simplifying many calculations.
The unit circle is integral to this example because it's the specific region \(D\) that our vector field \(\boldsymbol{f}\) interacts with. Understanding this basic structure is crucial for visualizing how the vector field acts around its boundary.
Partial Derivatives
Partial derivatives represent the rate of change of a multivariable function, focusing on one variable at a time while holding others constant. They're crucial tools in calculus, allowing us to explore how changing one dimension (or direction) affects a function.
Within this exercise:
  • The vector field is \( \boldsymbol{f} = \langle P, Q \rangle = \langle y, -x \rangle \), where \(P = y\) and \(Q = -x\).
  • We compute the partial derivatives: \(\frac{\partial Q}{\partial x} = -1\) and \(\frac{\partial P}{\partial y} = 1\).
This means, focusing on the x-direction for function \(Q = -x\), you find it changes by \(-1\) for each unit change in x. Similarly, function \(P = y\) changes by \(1\) for each unit change in y. These derivatives are pivotal in applying Green's Theorem, allowing the area integral to be evaluated.
Outward Normal Vector
An outward normal vector is a vector that is perpendicular to a curve or surface, pointing away from it. In the context of a circle, it's a line coming straight out from the surface at any given point, like spikes of a hedgehog.
In this problem, the outward normal vector is also crucial for calculating certain integrals. When dealing with the integral \( \int_{\partial D} \boldsymbol{f} \cdot \boldsymbol{N} \, ds \), the vector field \( \boldsymbol{f} = \langle y, -x \rangle \) is tangent to the circle.
  • This tangency means the vectors are perpendicular to their normal vectors.
  • As perpendicular vectors have a dot product of zero, the integral over the boundary is zero.
Understanding these principles help us conclude why the dot product of these perpendicular vectors results in zero, a key realization in solving such exercises.

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