Problem 34 - Thomas Calculus

Question

Converting to a polar integral Integrate \(f(x, y)=\) \(\left[\ln \left(x^{2}+y^{2}\right)\right] /\left(x^{2}+y^{2}\right)\) over the region \(1 \leq x^{2}+y^{2} \leq e^{2}\)

Step-by-Step Solution

Answer

The integral evaluates to \(2\pi\).

Step 1: Identify the region of integration

The problem states that we need to integrate over the region where \( 1 \leq x^2 + y^2 \leq e^2 \). This describes an annular region (ring) with inner radius 1 and outer radius \( e \).

Step 2: Set up polar coordinates

In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). Thus, \( x^2 + y^2 = r^2 \). The given region transforms to \( 1 \leq r^2 \leq e^2 \), or equivalently, \( 1 \leq r \leq e \).

Step 3: Convert the integrand to polar coordinates

Substitute \( x^2 + y^2 \) with \( r^2 \) in the integrand: \[ f(x, y) = \frac{\ln(r^2)}{r^2} = \frac{2\ln(r)}{r^2}.\]

Step 4: Determine the differential area element in polar coordinates

The differential area element in polar coordinates is \( dA = r \, dr \, d\theta \).

Step 5: Set up and evaluate the double integral

The integral becomes: \[ \int_0^{2\pi} \int_1^e \frac{2\ln(r)}{r^2} \cdot r \, dr \, d\theta.\]Simplify to:\[ 2 \int_0^{2\pi} \int_1^e \frac{\ln(r)}{r} \, dr \, d\theta.\]Integrate with respect to \( r \):\[ 2 \int_0^{2\pi} \left[ \frac{1}{2} (\ln(r))^2 \right]_1^e \, d\theta.\]Evaluate this to get:\[ 2 \int_0^{2\pi} \left( \frac{1}{2}(\ln(e))^2 - \frac{1}{2}(\ln(1))^2 \right) \, d\theta = 2 \int_0^{2\pi} \left( \frac{1}{2} \right) \, d\theta = 2 \left( \frac{1}{2} \times 2\pi \right) = 2\pi.\]

Step 6: Simplify the result

The integral evaluates to \( 2\pi \). This is the final area integrated over the given region.

Key Concepts

Polar Coordinates

Polar coordinates offer a way of describing points in a plane using a radius and angle instead of traditional Cartesian coordinates. This approach can simplify many problems involving circles, spirals, and other radial patterns. In our problem, we used polar coordinates to describe an annular region with an inner radius of 1 and an outer radius of \(e\).

The conversion from Cartesian to polar is straightforward:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)
• The distance from the origin is \(r = \sqrt{x^2 + y^2}\)

In polar coordinates, the area swept by the radius \(r\) as it rotates through \(\theta\) can be directly integrated using the formula \(dA = r \, dr \, d\theta\). This adaptation is particularly effective for our integration problem, as it allows us to transform the region defined by \(x^2 + y^2\) directly into bounds for \(r\).

Double Integrals

Double integrals are an extension of single-variable integration. They allow us to compute the volume under a surface described by a function \(f(x, y)\) over a two-dimensional region. In Cartesian coordinates, a double integral requires evaluating the integral over a rectangle or more complex shape by summing up infinitely small elements. For our problem, this region is an annular shape between two circles with different radii.

When performing double integration, one can either use iterated integrals (integrating one variable at a time) or switch to polar coordinates, which is more intuitive for circular regions. In our case, converting to polar significantly simplified the process. We only needed to calculate the integral:

• First with respect to \(r\)
• Then with respect to \(\theta\).

This method provides a structured way to evaluate areas that are challenging to describe with simple Cartesian bounds.

Integration Techniques

For integration, there are several techniques depending on the component functions and bounds involved. In problems like ours, converting the integrand and bounds from Cartesian to polar form can simplify complex algebra involved.

Here's the general process:

• Substitute \(x^2 + y^2\) with \(r^2\)
• Change \(dx \, dy\) to \(r \, dr \, d\theta\)

For our exercise, once in polar form, we tackled the integration by simplifying the expression into a form recognizable to standard integrals. Most notably, simplifying \(\frac{2\ln(r)}{r^2}\) for integration by recognizing it as a derivative and using techniques like substitution or parts as needed, though in this example direct integration was achieved after setup.

Mathematical Analysis

Mathematical analysis involves verifying, interpreting, and comprehending the implications of integrals and differential equations. By correctly setting up and evaluating our integral in polar coordinates, we were able to calculate the area of the annular region accurately.

This exercise highlights how breaking down a problem into smaller, more manageable parts can lead to successful outcomes. Understanding the role of limits, function behavior, and area elements in both coordinate systems ensures reliable results. The result, \(2\pi\), corresponds to the total area in question, illustrating the power of mathematical analysis in confirming theoretical and geometric concepts.

The area evaluation exemplifies how applicable mathematics can solve complex problems and how thorough analysis leads to simpler solutions by choosing the most fitting approach.