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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![**Topic: Evaluating a Double Integral**
**Content:**
**Integral Expression:**
Evaluate the integral:
\[
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{\sin \theta}} r^2 \cos(\theta) \, dr \, d\theta
\]
**Explanation:**
This is a double integral in polar coordinates. The inner integral is with respect to \( r \), the radial coordinate, ranging from \( 0 \) to \( \frac{1}{\sin \theta} \). The outer integral is with respect to \( \theta \), the angular coordinate, ranging from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \).
The integrand function is \( r^2 \cos(\theta) \), which combines both the radial distance and the cosine of the angle \( \theta \).
---
This integral might be used to calculate the volume under a surface defined in polar coordinates. Solving it would involve first integrating with respect to \( r \), and then integrating the result with respect to \( \theta \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12c62ea9-2423-4a35-a6cd-74646c6bbd41%2F38d58004-3441-47e3-8123-478637146a31%2Fqyc1bxh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Evaluating a Double Integral**
**Content:**
**Integral Expression:**
Evaluate the integral:
\[
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{\sin \theta}} r^2 \cos(\theta) \, dr \, d\theta
\]
**Explanation:**
This is a double integral in polar coordinates. The inner integral is with respect to \( r \), the radial coordinate, ranging from \( 0 \) to \( \frac{1}{\sin \theta} \). The outer integral is with respect to \( \theta \), the angular coordinate, ranging from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \).
The integrand function is \( r^2 \cos(\theta) \), which combines both the radial distance and the cosine of the angle \( \theta \).
---
This integral might be used to calculate the volume under a surface defined in polar coordinates. Solving it would involve first integrating with respect to \( r \), and then integrating the result with respect to \( \theta \).
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