Question

In Exercises \(3-10\), a function \(f(x, y)\) is given and a region \(R\) of the \(x-y\) plane is described. Set up and evaluate \(\iint_{R} f(x, y) d A\) using polar coordinates. \(f(x, y)=(x-y) /(x+y) ; R\) is the region enclosed by the lines \(y=x, y=0\) and the circle \(x^{2}+y^{2}=1\) in the first quadrant.

Short answer

The integral evaluates to \(\frac{1}{4} \ln(2)\).

Step-by-step solution

Convert the Region to Polar Coordinates

First, let's identify the region \(R\) defined by the lines \(y=x\), \(y=0\), and the circle \(x^2+y^2=1\) in the first quadrant. In polar coordinates, we have \(x=r\cos\theta\) and \(y=r\sin\theta\). The line \(y=x\) becomes \(r\sin\theta = r\cos\theta\) or \(\theta = \frac{\pi}{4}\). The line \(y=0\) corresponds to \(\theta=0\). The circle \(x^2+y^2=1\) is \(r=1\). Thus, the region \(R\) in polar coordinates is bounded by \(0 \leq \theta \leq \frac{\pi}{4}\) and \(0 \leq r \leq 1\).

Express the Function in Polar Coordinates

Now, convert the function \(f(x, y) = \frac{x-y}{x+y}\) into polar coordinates. Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\): \[ f(r, \theta) = \frac{r\cos\theta - r\sin\theta}{r\cos\theta + r\sin\theta} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}. \]

Set Up the Double Integral

We can now set up the double integral in polar coordinates. The region \(R\) is expressed with bounds for \(r\) and \(\theta\), and the integral in polar coordinates is: \[ \iint_{R} f(x, y) \, dA = \int_{0}^{\frac{\pi}{4}} \int_{0}^{1} \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} \cdot r \, dr \, d\theta. \]

Evaluate the Inner Integral

First, evaluate the inner integral with respect to \(r\): \[ \int_{0}^{1} r \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} \, dr = \left[ \frac{1}{2} r^2 \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} \right]_0^1 = \frac{1}{2} \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}. \]

Evaluate the Outer Integral

Now, evaluate the outer integral with respect to \(\theta\): \[ \int_{0}^{\frac{\pi}{4}} \frac{1}{2} \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} \, d\theta. \]To evaluate this integral, observe that the integrand can be simplified using the substitution \(u = \cos\theta + \sin\theta\), which yields \(du = (-\sin\theta + \cos\theta) \, d\theta\). Thus the integral becomes: \[ \frac{1}{2} \int \frac{\cos\theta - \sin\theta}{u} \, du = \frac{1}{2} \ln|u| + C. \]Evaluating from \(\theta = 0\) to \(\theta = \frac{\pi}{4}\), we have \(u = \sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)\), yielding: \[ \left[ \frac{1}{2} \ln|u| \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \ln(\sqrt{2}) - \frac{1}{2} \ln(1) = \frac{1}{2} \ln(\sqrt{2}) = \frac{1}{4} \ln(2). \]

Key concepts

Polar Coordinates Transformation

Polar coordinates transform a point defined in the Cartesian (x, y) plane using two variables: the radial distance from the origin \( r \) and the angle \( \theta \) measured from the positive x-axis. This is a powerful method in integration, particularly when dealing with circular or radial symmetry.

In polar coordinates, any point \( (x, y) \) is represented as \( (r\cos\theta, r\sin\theta) \). For this problem, where the region is described by lines and a circle, the transformation helps in setting up the integral limits. The line \( y = x \) is simplified to \( \theta = \frac{\pi}{4} \), and \( y = 0 \) corresponds to \( \theta = 0 \). The circle \( x^2 + y^2 = 1 \) translates to \( r = 1 \). This makes defining the region in polar coordinates easier, covering angles \( 0 \leq \theta \leq \frac{\pi}{4} \) while \( 0 \leq r \leq 1 \).

Transitioning to polar coordinates allows us to leverage symmetries and simplifies the computational process of double integrals, converting complex regions into more manageable intervals.

Region in First Quadrant

The region being considered in this exercise lies in the first quadrant. The first quadrant of the Cartesian coordinate system is defined where both \( x \) and \( y \) values are positive. This inherently makes integration problems simpler as there are no negative bounds to worry about.

Specifically, the problem defines R as enclosed by:

• The line \( y = x \)
• The line \( y = 0 \)
• The circle \( x^2 + y^2 = 1 \)

In polar coordinates, for the first quadrant, \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{2} \). However, with the constraints of our region, \( \theta \) is further limited from \( 0 \) to \( \frac{\pi}{4} \), as \( y = x \) and \( y = 0 \) form the boundaries for this particular wedge-shaped region. Exploring this region bounded by these lines and curve helps in visualizing and accurately setting up our integration limits.

Double Integral Evaluation

To solve a double integral using polar coordinates, it is crucial to set it up with the new bounds and expressions based on the transformation. For the integral \( \iint_{R} f(x, y) \, dA \), the function \( f(x, y) = \frac{x-y}{x+y} \) needs to be expressed in terms of \( r \) and \( \theta \). Recall that:

\[ f(r, \theta) = \frac{r\cos\theta - r\sin\theta}{r\cos\theta + r\sin\theta} = \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}. \]
This reformatting is necessary, as it transforms the function into a more straightforward form based on trigonometric functions, making the integral easier to evaluate.

The area element \( dA \) in polar coordinates is \( r \cdot dr \cdot d\theta \), which accounts for the radial expansion of the element over angle increments. The integral becomes:
\[ \int_{0}^{\frac{\pi}{4}} \int_{0}^{1} \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} \cdot r \, dr \, d\theta. \]
This sets the stage to solve the integral by evaluating the bounds defined for \( r \) and \( \theta \). Carefully performing each step respects the transformed polar relationships.

Integral Substitution Method

Within the evaluation process of the double integral, the substitution method plays a crucial role. This powerful technique simplifies the process of integrating functions that are otherwise difficult to integrate directly.

In the outer integral \( \int_{0}^{\frac{\pi}{4}} \frac{1}{2} \frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} \, d\theta \), substitution is used for simplifying the expression for integration. By substituting \( u = \cos\theta + \sin\theta \), we find \( du = (-\sin\theta + \cos\theta) \, d\theta \), aligning perfectly with our integrand. This substitution reduces the problem to integrating a simple logarithmic function:

\[ \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C. \]

The process of substitution not only streamlines the evaluation but also fractionates complex trigonometric relationships into more manageable algebraic expressions. Completing the evaluation gives the result \( \frac{1}{4} \ln(2) \), highlighting how substitution aids in reaching a clean solution.