Problem 11 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

\( \pi \ln 2 - \frac{\pi}{2} \)

Step 1: Understand the Region of Integration

The limits of the Cartesian integral are given for \( x \) and \( y \). The integral is evaluated from \( x = 0 \) to \( x = \sqrt{(\ln 2)^2 - y^2} \), and from \( y = 0 \) to \( y = \ln 2 \). This represents a quarter-circle in the first quadrant of radius \( \ln 2 \).

Step 2: Convert Cartesian Coordinates to Polar Coordinates

In polar coordinates, \( x = r \cos\theta \) and \( y = r \sin\theta \). The expression \( \sqrt{x^2 + y^2} \) becomes \( r \). The differential area element \( dx \, dy \) becomes \( r \, dr \, d\theta \) in polar coordinates.

Step 3: Determine Polar Coordinate Bounds

Given the circular region of radius \( \ln 2 \), \( r \) ranges from \( 0 \) to \( \ln 2 \). The angle \( \theta \) ranges from \( 0 \) to \( \frac{\pi}{2} \) as it covers the first quadrant.

Step 4: Set Up the Polar Integral

The polar integral becomes \( \int_{0}^{\frac{\pi}{2}} \int_{0}^{\ln 2} e^r \cdot r \, dr \, d\theta \). Here, \( r \, dr \, d\theta \) is the Jacobian for the transformation from Cartesian to polar coordinates.

Step 5: Evaluate the Inner Integral with respect to r

Evaluate \( \int_{0}^{\ln 2} e^r \cdot r \, dr \). Use integration by parts: let \( u = r \), \( dv = e^r \, dr \). Then \( du = dr \), \( v = e^r \). The integral becomes \( r e^r \bigg|_0^{\ln 2} - \int_0^{\ln 2} e^r \, dr \).

Step 6: Complete the Integration by Parts

Calculate \( r e^r \bigg|_0^{\ln 2} = (\ln 2) e^{\ln 2} - 0 = (\ln 2) \, 2 \). The second part is \( \int_0^{\ln 2} e^r \, dr = e^r \bigg|_0^{\ln 2} = 2 - 1 \). Thus, the result of the inner integral is \( 2 \ln 2 - 1 \).

Step 7: Integrate with Respect to θ

Integrate the result from Step 6 with respect to \( \theta \): \[ (2 \ln 2 - 1) \int_{0}^{\frac{\pi}{2}} d \theta \]. This simplifies to \( (2 \ln 2 - 1) \cdot \frac{\pi}{2} \).

Step 8: Simplify the Final Result

The integral evaluates to \( \frac{\pi}{2} (2 \ln 2 - 1) \). Simplifying, we get \( \pi \ln 2 - \frac{\pi}{2} \).

Key Concepts

Cartesian Coordinates

Cartesian coordinates are a customary way of representing points in a plane using two perpendicular axes: the x-axis (horizontal) and the y-axis (vertical). Every point on the plane is represented by an ordered pair \(x, y\). In the context of integration, these coordinates help in defining the limits and regions for integrating functions over a plane.

For the given exercise, the integral is presented in Cartesian coordinates with limits of integration for x and y. These limits help define a specific region of integration within the Cartesian plane. Here, the region under consideration resembles a quarter circle in the first quadrant, bounded by the positive x and y axes. Understanding these bounds is crucial to evaluating the integral correctly. The simplicity of using Cartesian coordinates allows for easy visualization of the integration region, making it feasible to translate into other coordinate systems as required.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation. It is particularly useful for integrating products of functions, where direct methods are inefficient. This technique helps in breaking down a difficult integral into simpler ones.

The formula for integration by parts is:

\[ \int u \, dv = uv - \int v \, du \]

In this exercise, integration by parts is employed to evaluate the inner integral over r. The choice of \(u = r\) and \(dv = e^r \, dr\) simplifies the integral significantly. Calculating \(v\) by integrating \(dv\), and then differentiating \(u\), results in the expression \(r e^r - \int e^r \, dr\), which can be simplified to obtain the values needed for further calculation.

This method is flexible and powerful, enabling the breakdown of integrals that involve the product of algebraic and exponential terms.

Transformation of Coordinates

The transformation of Cartesian coordinates to polar coordinates is essential when dealing with problems where symmetry or limits suggest circular or radial integration regions. Polar coordinates use two values: the radial distance \(r\) from the origin, and the angular coordinate \(\theta\).

In this exercise, transformation is vital because the integration region is circular, making polar coordinates more appropriate. The relationships used are:

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

Converting the Cartesian form to polar involves changing not only the function but also the differentials from \(dx\,dy\) to \(r\,dr\,d\theta\). This adjustment is necessary to accommodate the geometric considerations of a circular region and yields an integral that's simpler to evaluate.

These transformations offer a strategic advantage in evaluating integrals over circular or rotationally symmetric regions, allowing for more efficient computation.

Definite Integrals

Definite integrals allow for the calculation of the area under a curve within specified limits. These limits help in defining the start and end points of integration, crucial in evaluating the total accumulated value or area.

In the solution provided, the integration limits specify both Cartesian and polar integral bounds. Initially, the limits in Cartesian coordinates delineate the quarter-circle region to be integrated over — marked out by \(-\ln 2\) for \(y\) and \(\sqrt{(\ln 2)^2 - y^2}\) for \(x\).

Upon converting to polar coordinates, the bounds on \(r\) and \(\theta\) describe an equivalent quarter-circle with the limits \(0\) to \(\ln 2\) for \(r\) and \(0\) to \(\frac{\pi}{2}\) for \(\theta\). These transformations not only make the calculation easier but also maintain the integral's sinuous equivalence, leading to accurate results that reflect the geometry of the problem.