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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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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