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Chapter 15: Problem 37

Integrate \(\quad f(x, y)=\) \(\left[\ln \left(x^{2}+y^{2}\right)\right] / \sqrt{x^{2}+y^{2}}\) over the region \(1 \leq x^{2}+y^{2} \leq e\)

Short Answer

Expert verified
The integral evaluates to \(2\pi(2-\sqrt{e})\).

Step by step solution

01

Identify the Region for Integration

The given exercise asks us to integrate the function \( f(x, y) = \frac{\ln(x^2 + y^2)}{\sqrt{x^2 + y^2}} \) over the region where \(1 \leq x^2 + y^2 \leq e\). This describes an annular region in the polar coordinates, with inner radius \( r = 1 \) and outer radius \( r = \sqrt{e} \).
02

Transform to Polar Coordinates

Using polar coordinates, where \( x = r \cos\theta \) and \( y = r \sin\theta \), we have \( x^2 + y^2 = r^2 \). Thus, \( f(x, y) \) becomes \( \frac{\ln(r^2)}{r} \) or equivalently \( \frac{2\ln(r)}{r} \) in terms of \( r \). Also, the differential area element \( dA = dx\,dy \) converts to \( r\,dr\,d\theta \).
03

Set Up the Integral in Polar Coordinates

The integral becomes: \[\int_0^{2\pi} \int_1^{\sqrt{e}} \frac{2\ln(r)}{r} \cdot r\,dr\,d\theta = 2\int_0^{2\pi} \int_1^{\sqrt{e}} \ln(r)\,dr\,d\theta\] We simplify by canceling the \( r \) in the denominator with the \( r \) in \( r\,dr \).
04

Integrate with Respect to \( r \)

First, focus on the inner integral: \( \int_1^{\sqrt{e}} \ln(r)\,dr \). This requires integration by parts, where we let \( u = \ln(r) \), and \( dv = dr \). Then, \( du = \frac{1}{r}dr \), and \( v = r \). The integral becomes:\[ \int \ln(r)\,dr = r\ln(r) - \int r\cdot \frac{1}{r}\,dr = r\ln(r) - r + C \]Evaluating this from 1 to \( \sqrt{e} \), we find:\[ \left. r\ln(r) - r \right|_1^{\sqrt{e}} = \left(\sqrt{e}\ln(\sqrt{e}) - \sqrt{e}\right) - (1\ln(1) - 1) \]Simplifying, \( \ln(\sqrt{e}) = \frac{1}{2} \ln(e) = \frac{1}{2} \), so:\[ \sqrt{e} \times \frac{1}{2} - \sqrt{e} + 1 \approx \frac{\sqrt{e}}{2} - \sqrt{e} + 1 \]
05

Integrate with Respect to \( \theta \)

Since the integral is independent of \( \theta \), the integral over \( \theta \) is simply \( \int_0^{2\pi} d\theta = 2\pi \).
06

Combine Results

Multiply the results of the two integrals:\[ 2 \times \left(\frac{\sqrt{e}}{2} - \sqrt{e} + 1\right) \times 2\pi \approx 2\pi(\sqrt{e} - 2\sqrt{e} + 2) \]Simplifying within the brackets gives:\[ 2\pi(2 - \sqrt{e}) \]

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Key Concepts

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

Polar Coordinates
When dealing with functions of two variables, polar coordinates can offer a simplified way to handle certain types of regions. In polar coordinates, each point in the plane is represented by a radius and an angle. The relationship between Cartesian coordinates \(x, y\) and polar coordinates \(r, \theta\) is given by:
  • \(x = r \cos\theta\)
  • \(y = r \sin\theta\)
  • \(x^2 + y^2 = r^2\)
Using these transformations can make it easier to integrate functions over circular or annular regions, as Cartesian equations involving \(x\) and \(y\) often simplify to expressions involving \(r\) alone.
Integration by Parts
Integration by parts is a technique often used to integrate products of functions, particularly when a natural choice for differentiation and integration of one of those products exists. The formula for integration by parts is derived from the product rule for derivatives and is given by:
  • \(\int u\, dv = uv - \int v\, du\)
For the function \(\ln(r)\), we let \(u = \ln(r)\) and \(dv = dr\). Differentiating and integrating gives us \(du = \frac{1}{r} dr\) and \(v = r\). Applying the integration by parts formula simplifies integrals that appear complex at first glance.
Double Integrals
Double integrals allow us to integrate functions over two-dimensional regions. These are useful in calculating things like area and volume. When converting to polar coordinates for double integration:
  • The differential area element \(dA = dx \, dy\) changes to \(r \cdot dr \cdot d\theta\).
  • The limits of integration for \(r\) and \(\theta\) must account for the specific region being integrated.
In the exercise, the function \(f(x, y)\) was integrated over an annular region using polar coordinates, making the setup more straightforward.
Annular Region
An annular region is a two-dimensional area that resembles a ring, defined by two bounded circles: an outer and an inner circle. In polar coordinates, this region is described by an inner radius and an outer radius:
  • The inner boundary occurs where \(r = \ ext{inner radius}\).
  • The outer boundary occurs where \(r = \ ext{outer radius}\).
For the problem, the annular region was defined by \(1 \leq x^2 + y^2 \leq e\), indicating an inner radius of \(r = 1\) and an outer radius of \(r = \sqrt{e}\), allowing for a clear path to set up the integral in polar coordinates.

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Most popular questions from this chapter

Let \(P_{0}\) be a point inside a circle of radius \(a\) and let \(h\) denote the distance from \(P_{0}\) to the center of the circle. Let \(d\) denote the distance from an arbitrary point \(P\) to \(P_{0} .\) Find the average value of \(d^{2}\) over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and \(P_{0}\) on the \(x\) -axis.)

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$\int_{0}^{\sqrt{3}} \int_{0}^{\tan ^{-1} y} \sqrt{x y} d x d y$$

Find the area enclosed by one leaf of the rose \(r=12 \cos 3 \theta\)

Sketch the region of integration and write an equivalent double integral with the order of integration reversed. $$\int_{0}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} 6 x d y d x$$

Sketch the region of integration, reverse the order of integration, and evaluate the integral. \(\iint_{R} x y d A\) where \(R\) is the region bounded by the lines \(y=x, y=2 x,\) and \(x+y=2\)
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