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Chapter 5: Problem 422

Evaluate the following integrals. \(\iint_{D} \sin \left(x^{2}+y^{2}\right) d A\) where \(D\) is a disk of radius 2 centered at the origin

Short Answer

Expert verified
The integral evaluates to \( \pi (1 - \cos(4)) \).

Step by step solution

01

Set Up the Double Integral in Polar Coordinates

We start by converting the given double integral into polar coordinates. Recall that in polar coordinates, the Cartesian coordinates \(x\) and \(y\) are related by \(x = r \cos \theta\) and \(y = r \sin \theta\). The area element \(dA\) in polar coordinates is \(r \, dr \, d\theta\). Since \(D\) is a disk of radius 2 centered at the origin, \(r\) ranges from 0 to 2, and \(\theta\) ranges from 0 to \(2\pi\). The integral can be expressed as: \[ \iint_{D} \sin(x^2 + y^2) \, dA = \int_{0}^{2\pi} \int_{0}^{2} \sin(r^2) \cdot r \, dr \, d\theta. \]
02

Integrate with Respect to \(r\)

First, integrate the inner integral with respect to \(r\). Remember that \(\int \sin(u) \, du = -\cos(u)\), so set \(u = r^2\) which means \(du = 2r \, dr\) or \( r \, dr = \frac{1}{2}du\). Substitute and integrate:\[ \int_0^2 \sin(r^2) r \, dr = \frac{1}{2} \int_0^{4} \sin(u) \, du = \frac{1}{2} [-\cos(u)]_0^{4} = \frac{1}{2} [-\cos(4) + \cos(0)]. \]Since \(\cos(0) = 1\), the expression becomes:\[ \frac{1}{2} (1 - \cos(4)). \]
03

Integrate with Respect to \(\theta\)

Next, integrate the result of the \(r\)-integration with respect to \(\theta\). The integral becomes:\[ \int_0^{2\pi} \frac{1}{2} (1 - \cos(4)) \, d\theta = \frac{1}{2} (1 - \cos(4)) \int_0^{2\pi} 1 \, d\theta. \]Since \(\int_0^{2\pi} 1 \, d\theta = 2\pi\), the final integral simplifies to:\[ \pi (1 - \cos(4)). \]
04

Final Evaluation of the Integral

The expression from the previous step, \(\pi (1 - \cos(4))\), represents the evaluated double integral over the disk \(D\). Thus, the final result for the integral is\[ \pi (1 - \cos(4)). \]

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Key Concepts

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

Polar Coordinates
Polar coordinates offer a unique way to express points on a plane using angles and distances. Unlike Cartesian coordinates, which use x and y to determine a location, polar coordinates use radius \( r \) and angle \( \theta \). This system is particularly effective for circular or rotational systems, as it simplifies calculations. In polar terms, any point usually has coordinates \((r, \theta)\), where \( r \) is the radial distance from the origin, and \( \theta \) is the angle from the positive x-axis.
  • Conversion to Cartesian: \( x = r \cos \theta \) and \( y = r \sin \theta \).
  • Area Element: In polar coordinates, the area element \( dA \) becomes \( r \, dr \, d\theta \).
These transformations make polar coordinates invaluable for solving problems with radial symmetry, as seen in this integral over a disk.
Integration
Integration is the process of finding the accumulated area under curves, and in this case, it's about integrating a function over a particular region. Double integrals extend this concept to functions of two variables, integrating over a surface.For the given problem, we perform integration in polar coordinates. The setup involves two integrals:
  • Inner Integral: Integrate with respect to \( r \), accounting for the radial distance.
  • Outer Integral: Integrate with respect to \( \theta \), covering the rotational angle.
The computed integral explores the function \( \sin(x^2 + y^2) \), transformed into polar coordinates as \( \sin(r^2) \). The integration process deals with such functions over well-defined regions and gives insights into total accumulated values over those regions.
Disk Region
A disk region is a circular area defined in the parameter space. It is determined by a radius and a center point, typically at the origin for simplicity.
  • In Cartesian coordinates, a disk is expressed by \( x^2 + y^2 \leq R^2 \).
  • In Polar coordinates, it's much simpler: \( 0 \leq r \leq R \) and \( 0 \leq \theta \leq 2\pi \).
In this exercise, the disk is described with radius 2 centered at the origin. This setup translates to \( r \) ranging from 0 to 2 and \( \theta \) from 0 to \( 2\pi \), providing a neat framework to apply polar integration. Calculations on disk regions often leverage such symmetrical properties, making polar coordinates an optimal choice.
Trigonometric Integration
Trigonometric integration involves integrating functions with trigonometric expressions. These integrals often require substitution techniques for simplification.In the given exercise, the integration involves \( \sin(r^2) \). To simplify, we use substitution by setting \( u = r^2 \), leading to \( du = 2r \, dr \) or \( r \cdot dr = \frac{1}{2} du \). This substitution transforms the integral into a more manageable form:
  • Evaluate \( \int \sin(u) \, du = -\cos(u) \).
  • Substitute back: This results in evaluating from \( 0 \) to 4, forming: \( \frac{1}{2}[-\cos(4) + \cos(0)] \).
Understanding the nuances of trigonometric integration is crucial, as similar techniques often apply to different parts of calculus and differential equations.

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Most popular questions from this chapter

A solid \(Q\) has a volume given by \(\iint_{D} \int_{a}^{b} d A d z\), where \(D\) is the projection of the solid onto the \(x y\) -plane and \(a

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The midpoint rule for the triple integral III \(f(x, y, z) d V\) over the rectangular solid box \(B\) is a generalization of the midpoint rule for double integrals. The region \(B\) is divided into subboxes of equal sizes and the integral is approximated by the triple Riemann sum \(\sum_{i=1}^{l} \sum_{j=1}^{m} \sum_{k=1}^{n} f\left(\bar{x}_{i}, \bar{y}_{j}, \bar{z}_{k}\right) \Delta V,\) where \(\left(\bar{x}_{i}, \bar{y}_{j}, \bar{z}_{k}\right)\) is the center of the box \(B_{i j k}\) and \(\Delta V\) is the volume of each subbox. Apply the midpoint rule to approximate \(\varliminf_{B} x^{2} d V\) over the solid \(B=\\{(x, y, z)|0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1|\) by using a partition of eight cubes of equal size. Round your answer to three decimal places.

In the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the first two groups of Exercises. a. Find the moments of inertia \(I_{x}, I_{y},\) and \(I_{0}\) about the \(x\) -axis, \(y\) -axis, and origin, respectively. b. Find the radii of gyration with respect to the \(x\) -axis, \(\quad y\) -axis, and origin, respectively. $$ R \text { is the unit disk; } \rho(x, y)=3 x^{4}+6 x^{2} y^{2}+3 y^{4} \text { . } $$

In the following exercises, the transformations \(T: S \rightarrow R\) are one- to-one. Find their related inverse transformations \(T^{-1}: R \rightarrow S\). $$ \begin{aligned} &\text x=u+v+w, y=3 v, z=2 w,\\\ &\text { where }\\\ &S=R=\mathrm{R}^{3} . \end{aligned} $$
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