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 2:** Use polar coordinates to evaluate the double integral
\[
\int_{0}^{2} \int_{0}^{\sqrt{4-y^2}} \sin(x^2 + y^2) \, dx \, dy.
\]
**Answer:**
\[
\frac{\pi}{4} [1 - \cos(4)].
\]
**Explanation:**
The problem involves evaluating a double integral using polar coordinates. The function inside the integral, \(\sin(x^2 + y^2)\), suggests a natural transformation to polar coordinates since \(x^2 + y^2 = r^2\) in polar terms.
1. **Conversion to Polar Coordinates:**
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(dx \, dy = r \, dr \, d\theta\)
2. **Limits of Integration:**
- The given limits for \(x\) and \(y\) define a region in the first quadrant of the unit circle with radius 2.
- Thus, \(r\) ranges from 0 to 2, and \(\theta\) ranges from 0 to \(\pi/2\).
3. **Transformed Integral:**
- Replace \(x^2 + y^2\) with \(r^2\), and the integral becomes:
\[
\int_{0}^{\pi/2} \int_{0}^{2} \sin(r^2) \, r \, dr \, d\theta.
\]
4. **Calculation:**
- Integrate with respect to \(r\) first and then with respect to \(\theta\).
The final calculation yields:
\[
\frac{\pi}{4} [1 - \cos(4)],
\]
as the result of the original double integral problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F43d7ff2f-486e-4301-9ac6-c1e1dcf694a8%2F53628cad-b9ca-4a11-99b0-2a2a571bdcd7%2Fl4iqci_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 2:** Use polar coordinates to evaluate the double integral
\[
\int_{0}^{2} \int_{0}^{\sqrt{4-y^2}} \sin(x^2 + y^2) \, dx \, dy.
\]
**Answer:**
\[
\frac{\pi}{4} [1 - \cos(4)].
\]
**Explanation:**
The problem involves evaluating a double integral using polar coordinates. The function inside the integral, \(\sin(x^2 + y^2)\), suggests a natural transformation to polar coordinates since \(x^2 + y^2 = r^2\) in polar terms.
1. **Conversion to Polar Coordinates:**
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(dx \, dy = r \, dr \, d\theta\)
2. **Limits of Integration:**
- The given limits for \(x\) and \(y\) define a region in the first quadrant of the unit circle with radius 2.
- Thus, \(r\) ranges from 0 to 2, and \(\theta\) ranges from 0 to \(\pi/2\).
3. **Transformed Integral:**
- Replace \(x^2 + y^2\) with \(r^2\), and the integral becomes:
\[
\int_{0}^{\pi/2} \int_{0}^{2} \sin(r^2) \, r \, dr \, d\theta.
\]
4. **Calculation:**
- Integrate with respect to \(r\) first and then with respect to \(\theta\).
The final calculation yields:
\[
\frac{\pi}{4} [1 - \cos(4)],
\]
as the result of the original double integral problem.
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