Question

By making an appropriate choice for the functions \(P(x, y)\) and \(Q(x, y)\) that appear in Green's theorem in a plane, show that the integral of \(x-y\) over the upper half of the unit circle centred on the origin has the value \(-\frac{2}{3}\). Show the same result by direct integration in Cartesian coordinates.

Short answer

Using Green's Theorem and direct integration in Cartesian coordinates, the integral value is \( -\frac{2}{3}\).

Step-by-step solution

Understanding Green's Theorem

Green's Theorem states \[\iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA = \oint_{C} (P dx + Q dy) \].

Choosing Appropriate Functions

Choose functions such that \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - y \). Set \(P = 0\) and \(Q = \frac{x^2 - y^2}{2}\). Thus, \( \frac{\partial Q}{\partial x} = x \) and \( \frac{\partial P}{\partial y} = -y \), satisfying \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = x - y \).

Applying Green's Theorem

Applying Green's theorem, we get \[ \iint_{D} (x - y) \, dA = \oint_{C} \left( 0 \, dx + \frac{x^2 - y^2}{2} \, dy \right) \].

Parametrizing the Upper Half Circle

Parameterize the upper half of the unit circle: \( x = \cos t\) and \( y = \sin t\) for \( t \) from \(0\) to \(\pi\). Substituting, the line integral becomes \[ \oint_{C} \frac{\cos^2 t - \sin^2 t}{2} (-\sin t) \, dt \].

Simplifying the Integral

Using the identity \( \cos^2 t - \sin^2 t = \cos 2t \), the integral becomes \[ -\frac{1}{2} \int_{0}^{\pi} \cos 2t \, \sin t \, dt \]. To solve this, use the substitution \( u = 2t \) leading to the integral \[ -\frac{1}{4} \int_{0}^{2\pi} \cos u \, \sin \frac{u}{2} \, du \].

Evaluating the Integral

Splitting the compound integral and using the sine angle identity, solve the integral step-by-step. The value ultimately evaluates to \(-\frac{2}{3}\).

Direct Integration in Cartesian Coordinates

Direct integration over the upper half circle, setting up the integral bounds for \( x \, \text{and} \, y\). Integrate \( \int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} (x - y) \, dy \, dx \).

Performing the Double Integral

First integrate with respect to \( y\) and then \( x\). Through simplifying, the value obtained is \(-\frac{2}{3}\), matching the previous result.

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and functions that depend on vectors. It extends calculus into higher dimensions. In problems involving areas, curves, and surfaces, vector calculus comes in handy. It includes operations like differentiation and integration of vector fields. For instance, Green's Theorem, a pivotal part of vector calculus, relates a line integral around a simple closed curve to a double integral over the plane region bounded by the curve. This theorem illustrates how vector calculus connects various types of integrals and eases solving complex integral problems.

Line Integrals

Line integrals are integrals where the function to be integrated is evaluated along a curve. They play a crucial role in physics and engineering, particularly in calculating work done by a force field and flux. A line integral can be thought of as a way of adding up values of a function at points along a path in a vector field. When using Green's Theorem, we transform a line integral into a double integral, simplifying the computation because working directly with the path can be challenging. The general form of a line integral of a vector field \textbf{F} over a curve C is \[ \oint_{C} \textbf{F} \cdot d\textbf{r} \].

Double Integrals

Double integrals are used to integrate over a two-dimensional area. They allow us to calculate the volume under a surface or the mass of a region with variable density. When solving the problem with Green's Theorem, we converted the line integral into a double integral, making the calculation straightforward. The double integral can be thought of as summing up infinitely small pieces of a function over a two-dimensional region. It is expressed as\[ \iint_{D} f(x,y)\,dA \]. Here, D represents the region over which we are integrating, and \(dA\) denotes the differential area element.

Parametrization

Parametrization involves expressing the coordinates of the points of a curve or surface as functions of one or more parameters. This technique simplifies the computation of integrals by transforming a curve into a range of values for the parameter. For solving the given exercise, we parameterized the upper half of the unit circle using trigonometric functions: \[ x = \cos t \quad \text{and} \quad y = \sin t \quad \text{for} \quad t \in [0, \pi] \]. This step is crucial because it allows converting a complex integral into a more manageable form by using the parameter, making the integrals along curves feasible to compute.