Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Title: Evaluating Integrals Using Polar Coordinates**
**Problem Statement:**
Evaluate the given integral by changing to polar coordinates:
\[
\int \int_{D} \cos\left(\sqrt{x^2 + y^2}\right) \, dA
\]
where \( D \) is the disk with center at the origin and radius 9.
**Solution Approach:**
To solve the problem, we will transform the given integral from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\). In polar coordinates, the expression \(x^2 + y^2\) becomes \(r^2\), and \(dA = r \, dr \, d\theta\).
The limits for \(r\) will be from 0 to 9, and the limits for \(\theta\) will be from 0 to \(2\pi\) because we are considering the entire disk.
Thus, the integral in polar coordinates becomes:
\[
\int_{0}^{2\pi} \int_{0}^{9} \cos(r) \, r \, dr \, d\theta
\]
This is now a straightforward double integral that can be evaluated using standard techniques.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb314b2da-8126-41f7-9949-20c1694a622a%2F9cb741ef-3b6c-43eb-9d4f-e7a854ae7001%2F74nxm6tj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Evaluating Integrals Using Polar Coordinates**
**Problem Statement:**
Evaluate the given integral by changing to polar coordinates:
\[
\int \int_{D} \cos\left(\sqrt{x^2 + y^2}\right) \, dA
\]
where \( D \) is the disk with center at the origin and radius 9.
**Solution Approach:**
To solve the problem, we will transform the given integral from Cartesian coordinates \((x, y)\) to polar coordinates \((r, \theta)\). In polar coordinates, the expression \(x^2 + y^2\) becomes \(r^2\), and \(dA = r \, dr \, d\theta\).
The limits for \(r\) will be from 0 to 9, and the limits for \(\theta\) will be from 0 to \(2\pi\) because we are considering the entire disk.
Thus, the integral in polar coordinates becomes:
\[
\int_{0}^{2\pi} \int_{0}^{9} \cos(r) \, r \, dr \, d\theta
\]
This is now a straightforward double integral that can be evaluated using standard techniques.
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