Problem 32 - Advanced Engineering Mathematics

Question

Evaluate the given iterated integral by changing to polar coordinates. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-y^{2}}} \frac{1}{1+\sqrt{x^{2}+y^{2}}} d x d y $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{2} - \frac{\pi}{2} \ln 2 \).

Step 1: Analyze the Region of Integration

The region of integration is defined by the bounds of the integral. Here, the limits for \( x \) are from \( 0 \) to \( \sqrt{1-y^2} \), and for \( y \), from \( 0 \) to \( 1 \). This describes the top half of a circle with a radius of 1 centered at the origin. In polar coordinates, this region can be defined by \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq \pi/2 \).

Step 2: Change to Polar Coordinates

In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \), and the area element \( dx \, dy \) becomes \( r \, dr \, d\theta \). The function \( \frac{1}{1 + \sqrt{x^2 + y^2}} \) becomes \( \frac{1}{1 + r} \). So, the integral can be rewritten as \( \int_{0}^{\pi/2} \int_{0}^{1} \frac{1}{1 + r} r \, dr \, d\theta \).

Step 3: Integrate with Respect to r

Consider the integral \( \int_{0}^{1} \frac{r}{1+r} \, dr \). Use substitution: let \( u = 1+r \), then \( du = dr \) and when \( r = 0, u = 1 \) and when \( r = 1, u = 2 \). The integral becomes \( \int_{1}^{2} \frac{u-1}{u} \, du = \int_{1}^{2} \left( 1 - \frac{1}{u} \right) \, du \). This can be split into \( \int_{1}^{2} 1 \, du - \int_{1}^{2} \frac{1}{u} \, du \).

Step 4: Solve Integrals after Substitution

For \( \int_{1}^{2} 1 \, du \), evaluate to get \( u \big|_{1}^{2} = 2-1 = 1 \). For \( \int_{1}^{2} \frac{1}{u} \, du \), compute as \( \ln |u| \big|_{1}^{2} = \ln 2 - \ln 1 = \ln 2 \). Thus, the full integral evaluates to \( 1 - \ln 2 \).

Step 5: Integrate with Respect to θ

Now integrate the result over \( \theta \): \( \int_{0}^{\pi/2} (1 - \ln 2) \, d\theta \). Since the integrand is a constant, this evaluates to \( (1 - \ln 2) \theta \big|_{0}^{\pi/2} = (1 - \ln 2)(\pi/2 - 0) = \frac{\pi}{2} - \frac{\pi}{2} \ln 2 \).

Step 6: Write the Final Answer

The evaluated iterated integral in polar coordinates is \( \frac{\pi}{2} - \frac{\pi}{2} \ln 2 \).

Key Concepts

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by a distance and an angle. Instead of using x and y, polar coordinates use the radius (r) and the angle (\(\theta\)). This system is extremely useful in scenarios involving circular regions or where symmetry around a point is present.
In our exercise, we noticed that the region of integration forms half of a circle centered at the origin with radius 1. In situations like this, converting from Cartesian to polar coordinates can simplify the problem significantly.
This conversion is done using the relationships:

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

These conversions also change the integration element from \(dx \, dy\) to \(r \, dr \, d\theta\). This change accounts for the circular nature of the region.

Change of Variables

Changing variables in an integral can make a problem simpler by transforming complex boundaries into straightforward ones. In this exercise, changing from Cartesian to polar coordinates made the bounds of integration clearer.
Initially, the given Cartesian bounds required translating the integral over a semi-circular region. By converting to polar coordinates, the bounds become simpler:

• For \(r\), the limits range from 0 to 1 (representing the size of the circle).
• For \(\theta\), they range from 0 to \(\pi/2\), indicating the top half of our circle.

This transformation not only simplifies the integral's boundaries but often makes the integral itself easier to evaluate, especially when dealing with symmetries or circular regions.

Integration Techniques

Integration involves evaluating integrals to find areas, volumes, and other properties. Various techniques can be applied to simplify and solve integrals.
For this exercise, the integral \(\int_{0}^{1} \frac{r}{1+r} \, dr\) was tackled using substitution. Here, we set \(u = 1 + r\) transforming the integral's complex fraction into a more manageable form.
The substitution method is a key technique, helping transform less readily integrable functions into simpler forms, allowing easy evaluation that manual calculation or common methods can't easily achieve. It's particularly helpful when the integral includes expressions with squares, square roots, or linear denominators.

Mathematical Analysis

Mathematical analysis explores the infinite depth of calculus, seeking not just solutions but deep understanding.
In iterated integrals, we evaluate the integral by integrating with respect to one variable before the other, simplifying the process in layered computational steps.
This task involves first integrating with respect to \(r\), finding \(\int_{1}^{2} \left(1 - \frac{1}{u}\right) \, du\), and then integrating over \(\theta\).
The logic of mathematical analysis reveals itself in this systematic, ordered solving, converting complex interrelated problems into digestible steps. Each integral is thus evaluated in its own right using constants or simple functions to facilitate further integration steps.