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cos (2? + y')dA where D = {(x, y) | 16 s a² + y² < 36} Find

cos (2? + y')dA where D = {(x, y) | 16 s a² + y² < 36} Find

Calculus: Early Transcendentals
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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**Problem Statement:** Find \(\iint_D \cos(x^2 + y^2) \, dA\) where \(D = \{(x, y) \mid 16 \leq x^2 + y^2 \leq 36\}\). **Explanation:** This problem involves calculating a double integral over the region \(D\), which is defined as the set of points \((x, y)\) satisfying the inequality \(16 \leq x^2 + y^2 \leq 36\). The region \(D\) describes an annular region (a ring-shaped area) in the xy-plane. The inequality \(16 \leq x^2 + y^2 \leq 36\) represents the area between two concentric circles: - The inner circle has a radius of \(\sqrt{16} = 4\). - The outer circle has a radius of \(\sqrt{36} = 6\). The goal is to integrate the function \(\cos(x^2 + y^2)\) over this annular region. **Conceptual Steps:** 1. **Describe the Region \(D\):** - It's a ring between circles with radii 4 and 6 centered at the origin. 2. **Polar Coordinates Conversion:** - A typical way to evaluate this integral is by transforming it into polar coordinates because the region is circular. - Let \(x = r \cos \theta\) and \(y = r \sin \theta\). Then \(x^2 + y^2 = r^2\). - The Jacobian of the transformation from Cartesian to polar coordinates is \(r\), so \(dA = r \, dr \, d\theta\). 3. **Set up the Integral in Polar Coordinates:** - Replace \(x^2 + y^2\) with \(r^2\). - The limits for \(r\) will be from 4 to 6, and for \(\theta\) from \(0\) to \(2\pi\). 4. **Evaluate:** - Compute the integral \(\int_0^{2\pi} \int_4^6 \cos(r^2) \cdot r \, dr \, d\theta\). The result of this integration will yield the value of the double integral
Transcribed Image Text:**Problem Statement:** Find \(\iint_D \cos(x^2 + y^2) \, dA\) where \(D = \{(x, y) \mid 16 \leq x^2 + y^2 \leq 36\}\). **Explanation:** This problem involves calculating a double integral over the region \(D\), which is defined as the set of points \((x, y)\) satisfying the inequality \(16 \leq x^2 + y^2 \leq 36\). The region \(D\) describes an annular region (a ring-shaped area) in the xy-plane. The inequality \(16 \leq x^2 + y^2 \leq 36\) represents the area between two concentric circles: - The inner circle has a radius of \(\sqrt{16} = 4\). - The outer circle has a radius of \(\sqrt{36} = 6\). The goal is to integrate the function \(\cos(x^2 + y^2)\) over this annular region. **Conceptual Steps:** 1. **Describe the Region \(D\):** - It's a ring between circles with radii 4 and 6 centered at the origin. 2. **Polar Coordinates Conversion:** - A typical way to evaluate this integral is by transforming it into polar coordinates because the region is circular. - Let \(x = r \cos \theta\) and \(y = r \sin \theta\). Then \(x^2 + y^2 = r^2\). - The Jacobian of the transformation from Cartesian to polar coordinates is \(r\), so \(dA = r \, dr \, d\theta\). 3. **Set up the Integral in Polar Coordinates:** - Replace \(x^2 + y^2\) with \(r^2\). - The limits for \(r\) will be from 4 to 6, and for \(\theta\) from \(0\) to \(2\pi\). 4. **Evaluate:** - Compute the integral \(\int_0^{2\pi} \int_4^6 \cos(r^2) \cdot r \, dr \, d\theta\). The result of this integration will yield the value of the double integral
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