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Chapter 15: Problem 20

Change the Cartesian integral into an equivalent polar integral. Then evaluate the polar integral. $$\int_{-1}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} \ln \left(x^{2}+y^{2}+1\right) d x d y$$

Short Answer

Expert verified
The evaluated polar integral is \( 2\pi \ln 2 - \pi \).

Step by step solution

01

Understand the Cartesian Integral

The given Cartesian integral is \( \int_{-1}^{1} \int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}} \ln (x^{2}+y^{2}+1) \, dx \, dy \). This integral is over a circular region defined by \( x^2 + y^2 \leq 1 \), which will be easier to evaluate using polar coordinates.
02

Convert Cartesian to Polar Coordinates

Recall the transformations: \( x = r \cos \theta \) and \( y = r \sin \theta \), and \( dx \, dy = r \, dr \, d\theta \). The region \( x^2 + y^2 \leq 1 \) transforms to \( 0 \leq r \leq 1 \) and \( 0 \leq \theta \leq 2\pi \). The integral becomes \( \int_{0}^{2\pi} \int_{0}^{1} \ln(r^2+1) \cdot r \, dr \, d\theta \).
03

Integrate with respect to \( r \)

First, focus on \( \int_{0}^{1} \ln(r^2+1) \cdot r \, dr \). Use the substitution \( u = r^2+1 \), then \( du = 2r \, dr \) or \( r \, dr = \frac{1}{2} du \). Update limits: when \( r = 0 \), \( u = 1 \); when \( r = 1 \), \( u = 2 \). Thus, \[ \int_{0}^{1} \ln(r^2+1) \cdot r \, dr = \frac{1}{2} \int_{1}^{2} \ln(u) \, du. \]
04

Evaluate the Substituted Integral

Integrate \( \int \ln(u) \, du \) using integration by parts: let \( v = \ln(u) \), \( dv = \frac{1}{u} \, du \), \( w = u \), and \( dw = du \). Then, \[ \int \ln(u) \, du = u \ln(u) - \int u \cdot \frac{1}{u} \, du = u \ln(u) - u.\]Evaluate from 1 to 2:\[ (2 \ln 2 - 2) - (1 \ln 1 - 1) = 2 \ln 2 - 1. \]
05

Integrate with respect to \( \theta \)

Now return to the full expression for integration:\[ \frac{1}{2} \int_{0}^{2\pi} (2 \ln 2 - 1) \, d\theta. \]This simplifies to:\[ (\ln 2 - \frac{1}{2}) \int_{0}^{2\pi} \, d\theta = (\ln 2 - \frac{1}{2}) \times 2\pi = 2\pi \ln 2 - \pi. \]
06

Final Evaluation

The evaluated polar integral is found to be \( 2\pi \ln 2 - \pi \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cartesian Coordinates
Cartesian coordinates are a foundational concept in mathematics used to define positions or points in a 2-dimensional plane using pairs of numerical values. Each point in the Cartesian coordinate system is uniquely identified by an ordered pair, \( (x, y) \), where \( x \) represents the horizontal position, and \( y \) represents the vertical position.
The Cartesian coordinate system is arranged such that:
  • The horizontal axis is called the x-axis.
  • The vertical axis is called the y-axis.
  • The point of intersection, where both axes meet, is the origin, denoted as \( (0, 0) \).
  • Regions can be described by equations in terms of \( x \) and \( y \), like the circle described by \( x^2 + y^2 \leq 1 \).
For complex integration problems, Cartesian coordinates can sometimes be inefficient or complicated, making alternate systems like polar coordinates more suitable.
Integration by Parts
Integration by parts is a mathematical technique used to integrate products of functions. It is derived from the product rule of differentiation and helps to simplify the integration process when faced with the integral of a product of two functions, \( \int u \, dv \).
The formula for integration by parts is:\[\int u \, dv = uv - \int v \, du\]Where:
  • \( u \) is a function that is chosen for differentiation.
  • \( dv \) is a function to be integrated.
  • \( du \) is the derivative of \( u \).
  • \(( v \) is the integral of \( dv \).
In our example, this rule helped simplify the complex log function integrated with respect to \( r \) after converting \( u = \ln(u) \) and rearranging terms to evaluate between specified limits.
Coordinate Transformation
Coordinate transformation, such as switching from Cartesian to polar coordinates, simplifies solving integrals over circular or symmetric regions. In a coordinate transformation:
  • Cartesian coordinates \( (x, y) \) transform into polar coordinates \( (r, \theta) \).
  • \( x \) is replaced by \( r \cos(\theta) \), and \( y \) is replaced by \( r \sin(\theta) \).
  • The area element transforms as \( dx \, dy = r \, dr \, d\theta \).
  • Limits of integration often change to account for the region鈥檚 shape, like converting a circle to limits from \( 0 \) to \( 1 \) for \( r \) and \( 0 \) to \( 2\pi \) for \( \theta \).
This method streamlines integration and is particularly effective when the described region naturally fits in a different coordinate system, simplifying expressions and easing the computational burden.
Definite Integral Evaluation
Definite integral evaluation is a process of finding the accumulation of quantities across a given interval. Unlike indefinite integrals, definite integrals have limits or bounds, which provide a specific numerical value representing the area under a curve across an interval.
Steps involved include:
  • Identifying the function to integrate and the integration bounds.
  • Applying integration techniques, such as substitution or by parts, to solve the integral.
  • Evaluating the antiderivative at the upper and lower limits of integration.
  • Subtracting the results to find the net area.
In the spherical problem above, the definite integral helped evaluate the computed area after transforming the region, simplifying integration using polar transformation and integration by parts, finally applying limits to find \( 2\pi \ln 2 - \pi \)).

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