Problem 10 - Thomas Calculus

Question

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. \(\int_{-1}^{1} \int_{-\sqrt{1-y^{2}}}^{0} \frac{4 \sqrt{x^{2}+y^{2}}}{1+x^{2}+y^{2}} d x d y\)

Step-by-Step Solution

Answer

The integral evaluates to \(2\pi(1 - \ln(2))\).

Step 1: Understand the Problem

We need to convert the given Cartesian double integral into a polar integral. The integral is from \(-1\) to \(1\) for \(y\) and from \(-\sqrt{1-y^{2}}\) to \(0\) for \(x\). The integral involves the function \(\frac{4 \sqrt{x^{2}+y^{2}}}{1+x^{2}+y^{2}}\).

Step 2: Define Polar Coordinates

In polar coordinates, \(x\) and \(y\) are expressed as \(x = r \cos \theta\) and \(y = r \sin \theta\). The region corresponds to a semicircle in the left half-plane. Here the angle \(\theta\) varies from \(\pi/2\) to \(3\pi/2\), and the radius \(r\) varies from 0 to 1. The differential area element \(dx \, dy\) becomes \(r \, dr \, d\theta\).

Step 3: Transform the Integral

Substitute the polar coordinates into the integral. The integral becomes \(\int_{\pi/2}^{3\pi/2} \int_{0}^{1} \frac{4r}{1+r^{2}} \, r \, dr \, d\theta\), because \(\sqrt{x^2 + y^2} = r\).

Step 4: Simplifying the Integral Expression

The function \(\frac{4r^2}{1+r^2}\) can be simplified, and we proceed to evaluate the inner radial integral first, \(\int_{0}^{1} \frac{4r^2}{1+r^2} \, dr\).

Step 5: Evaluate the Inner Integral Over \(r\)

To solve \(\int_{0}^{1} \frac{4r^2}{1+r^2} \, dr\), let \(u = 1 + r^2\) so that \(du = 2r \, dr\). Then the limits of \(r\) change from 0 to 1, giving us limits of \(u\) from 1 to 2. The integral becomes \(\int_{1}^{2} 2 \frac{(u-1)}{u} \, du\).

Step 6: Simplify and Calculate the Substitution Integral

The integral \(2\int_{1}^{2} \left(1 - \frac{1}{u}\right) \, du\) can be split into two parts: \(2\left(u - \ln|u| \right)\) evaluated from 1 to 2. This simplifies to \(2[2 - \ln(2)] - 2[1 - \ln(1)]\), giving \(2(1 - \ln(2))\).

Step 7: Evaluate the Outer Integral Over \(\theta\)

Now integrate over \(\theta\) from \(\pi/2\) to \(3\pi/2\). We have \(\int_{\pi/2}^{3\pi/2} 2(1-\ln(2)) d\theta\). This evaluates to \([\theta]_{\pi/2}^{3\pi/2} \times 2(1-\ln(2))\), which yields \([\pi] \times 2(1-\ln(2))\).

Step 8: Calculate the Final Answer

The final evaluation of the integral is \(2\pi(1 - \ln(2))\).

Key Concepts

Cartesian to Polar Conversion

Cartesian and polar coordinates are two different systems for expressing points in a plane. In a Cartesian system, points are determined using the coordinates \(x\) and \(y\). However, for many integrals, especially in circular regions, using polar coordinates can simplify the calculations.
To convert from Cartesian to polar coordinates, we use the formulas \(x = r \cos \theta\) and \(y = r \sin \theta\), where \(r\) is the radius or distance from the origin, and \(\theta\) is the angle from the positive \(x\)-axis.
Another crucial step in this conversion is adjusting the differential area element. In Cartesian coordinates, the area element is \(dx \, dy\), but in polar coordinates, it becomes \(r \, dr \, d\theta\).
So, when converting a double integral from Cartesian to polar form, we should replace \(dx \, dy\) with \(r \, dr \, d\theta\), and express the integral's limits and function in terms of \(r\) and \(\theta\).

Double Integrals

Double integrals allow us to calculate the volume under a surface spread over a given region. They are an extension of single-variable integrals to two dimensions and are represented as \(\int \int f(x, y) \, dx \, dy\).
In practice, double integrals involve finding the value of the function \(f(x, y)\) at numerous points within a specific region and summing them up. This requires understanding how to set the limits of integration to accurately capture the desired area.
When integrating in polar coordinates, the limits of \(r\) and \(\theta\) define the region of interest. The order of integration can be crucial—typically, we evaluate the inner integral first to simplify the expression before tackling the outer integral.
Double integrals can be particularly advantageous for solving problems involving symmetry, such as circular or semicircular regions, by simplifying the visualization and calculation process.

Semicircle Region

A semicircle is half of a circle, cut along its diameter. In coordinate terms, if the full circle is centered at the origin, the semicircle can be described succinctly using inequalities for \(x\) and \(y\).
For a semicircle in the left half-plane with radius 1, such as in our given problem, \(y\) ranges from \(-1\) to \(1\), and for each \(y\), \(x\) ranges from \(-\sqrt{1 - y^2}\) to \(0\).
When dealing with semicircle problems in polar coordinates, it helps to note that \(\theta\) will vary over half the circle, such as from \(\frac{\pi}{2}\) to \(\frac{3\pi}{2}\), while \(r\) ranges from \(0\) to the circle's radius, which is \(1\). This simplifies the integration bounds and often leads to more straightforward calculations.

Integral Evaluation

Evaluating integrals involves calculating the definite integral within specific limits. In the context of double integrals, this process involves two main steps: evaluating the inner integral and then the outer one.
For the problem at hand, we first deal with the radial part, due to the shift from Cartesian to polar coordinates. This means solving the inner integral \(\int_{0}^{1} \frac{4r^2}{1+r^2} \, dr\).

• Use substitution to simplify the function.
• Evaluate the expression within the specified limits.

Once the inner integral is calculated, the result becomes the integrand for the outer integral over the angular component \(\theta\).

• In this case, \(\int_{\pi/2}^{3\pi/2} 2(1-\ln(2)) \, d\theta\) finalizes the calculation.
• The challenge lies in careful bookkeeping of the integral bounds and functions.

Mathematical Substitution

Mathematical substitution is a powerful technique to simplify integrals by changing variables. When dealing with complex expressions, such as in our integral \(\frac{4r^2}{1+r^2}\), substitution makes solving manageable.
One effective strategy is to express one part of the formula in terms of a new variable. For example, letting \(u = 1 + r^2\) helps streamline the integral computation by transforming it into something easier to solve.
With substitution, don't forget:

• Calculating the differential (e.g., \(du = 2r \, dr\)).
• Changing the limits of integration to match the new variable's scale.
• Rewriting the function accurately in terms of the new variable.

These steps keep the integrity of the integral intact while simplifying calculations, which can be particularly helpful in tackling complex integrals encountered in polar coordinate transformations.