Question

Miscellaneous integrals Sketch the region of integration and evaluate the following integrals, using the method of your choice. \(\iint_{R} \frac{x-y}{x^{2}+y^{2}+1} d A ; R\) is the region bounded by the unit circle centered at the origin.

Short answer

Question: Sketch the region of integration and evaluate the double integral \(\iint_{R} \frac{x-y}{x^{2}+y^{2}+1} d A\), where R is the unit circle centered at the origin. Answer: The region of integration is a unit circle centered at the origin, and after converting to polar coordinates and evaluating the integral, we find that the integral is equal to 0.

Step-by-step solution

Sketching the region of integration

The given region, R, is defined as the unit circle centered at the origin. This is a simple circle with radius 1: 1. Draw the x-y coordinate plane. 2.List the equations for the region. - x²+y² ≤ 1(agebraic equation for a unit circle) 3. Draw the circle using the center (0,0) and radius 1.

Converting to polar coordinates

Since the region is a circle, it is more convenient to use polar coordinates to evaluate the integral. We will convert the given function and the region R into polar coordinates: \(x = r\cos\theta\) , \(y = r\sin\theta\) , \(dA = rdrd\theta\) The function becomes: \(\frac{x-y}{x^{2}+y^{2}+1} = \frac{r\cos\theta-r\sin\theta}{r^{2}\cos^2\theta+r^{2}\sin^2\theta+1}\) The region R in polar coordinates: \(r \in [0, 1]\) (radius is between 0 and 1) \(\theta \in [0, 2\pi]\) (the angle covers a full rotation)

Evaluating the integral

Now we will evaluate the integral using the converted function and region: \(\iint_{R} \frac{x-y}{x^{2}+y^{2}+1} d A = \int_{0}^{2\pi} \int_{0}^{1} \frac{r\cos\theta-r\sin\theta}{r^{2}+1} rdrd\theta\) First, we integrate with respect to r: \(\int_{0}^{1} \frac{r\cos\theta-r\sin\theta}{r^{2}+1} rdr = \int_{0}^{1} \frac{(r\cos\theta-r\sin\theta)r}{r^{2}+1}dr\) Since \(\cos\theta\) and \(\sin\theta\) are constants with respect to r, we can separate the integrals: \(= \cos\theta \int_{0}^{1} \frac{r^2}{r^{2}+1}dr - \sin\theta \int_{0}^{1} \frac{r^2}{r^{2}+1}dr\) Note that both integrals are the same except for the constant factors. Calculate one of them: \(\int_{0}^{1} \frac{r^2}{r^{2}+1}dr = \frac{1}{2}\int_{0}^{1} \frac{(r^{2}+1)-1}{r^{2}+1}dr = \frac{1}{2}\int_{0}^{1} \left(1 - \frac{1}{r^{2}+1}\right)dr\) Evaluate the integral: \(\int_{0}^{1} \left(1 - \frac{1}{r^{2}+1}\right)dr = r\Big|_0^1 - \tan^{-1}(r)\Big|_0^1 =1-\frac{\pi}{4}\) Now, apply it to our previous expression: \(= \cos\theta \left( 1-\frac{\pi}{4}\right) - \sin\theta \left( 1-\frac{\pi}{4} \right)\) Finally, we integrate with respect to \(\theta\): \(\int_{0}^{2\pi} \left(\cos\theta \left( 1-\frac{\pi}{4}\right) - \sin\theta \left( 1-\frac{\pi}{4} \right)\right)d\theta\) Since the integrals are with respect to \(\theta\), we can separate them: \(= \left( 1-\frac{\pi}{4}\right) \left(\int_{0}^{2\pi}\cos\theta d\theta - \int_{0}^{2\pi}\sin\theta d\theta\right) = \left( 1-\frac{\pi}{4}\right) (0) = 0\) So, the value of the integral is 0.

Key concepts

Polar Coordinates

When dealing with circular or symmetrical regions, polar coordinates often provide a more straightforward approach to integration. In polar coordinates, each point in the plane is represented by a distance from the origin \(r\) and an angle \(\theta\).

The relationship between polar and Cartesian coordinates is given by:

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)

In our exercise, the region \(R\) is a unit circle centered at the origin, which makes polar coordinates particularly useful. The circle can be described as \(r \in [0, 1]\) and \(\theta \in [0, 2\pi]\), where \(r\) is the radius, and \(\theta\) is the angle around the circle.

This conversion eases the computational effort as instead of integrating over a rectangular region you'd encounter in Cartesian coordinates, we integrate over a circular path that simplifies the limits. This is especially helpful when the function is already naturally decomposed into radial and angular components.

Region of Integration

The region of integration in our exercise is crucial as it dictates the bounds over which we perform the integral. Understanding the shape and extent of this region is key to correctly setting up and evaluating the integral.

The region \(R\) given in the exercise is described by the inequality \(x^2 + y^2 \leq 1\), which outlines a unit circle centered at the origin. We need to sketch this region to visualize the domain over which the double integral will apply:

• The unit circle is centered at the origin, \((0, 0)\), and has a radius of 1. This means it includes all points \((x, y)\) such that the distance from the center is less than or equal to 1.
• In polar coordinates, this region simplifies to \(r\) ranging from 0 to 1 and \(\theta\) from 0 to 2\(\pi\), capturing the full 360 degrees around the circle.

By accurately defining this region, you ensure that the integral accounts for all values within what is visually described as a circular disk.

Trigonometric Functions

Trigonometric functions like \(\cos\theta\) and \(\sin\theta\) are pivotal in converting between Cartesian and polar coordinates.

These functions represent the projections of the radius vector onto the x and y axes, respectively. They allow us to express the position of a point within the circular region in terms of angles and radii, which significantly simplifies many integration problems.

• Utilizing \(\cos\theta\) and \(\sin\theta\), our integral function \(\frac{x-y}{x^2+y^2+1}\) transforms to \(\frac{r\cos\theta - r\sin\theta}{r^2 + 1}\). This transformation showcases the advantage of working with trigonometric expressions.
• In our integral calculation, trigonometric identities help break down the expression into simpler components where each involves constant multipliers of \(r\) terms. This separation is crucial as it allows integrations over \(r\) to be straightforwardly executed.

Additionally, trigonometric functions hold properties such as periodicity and symmetry, which can be exploited to further simplify computations, especially when evaluating full circles or periodic functions from 0 to \(2\pi\).