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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![### Volume Calculation of a Region Bounded by a Surface and Rectangle
**Problem Statement:**
Find the volume of the region bounded above by the surface \( z = 2 \cos x \cos y \) and below by the rectangle \( R: 0 \leq x \leq \frac{\pi}{4}, \; 0 \leq y \leq \frac{\pi}{2} \).
---
\[
\vdots
\]
\[
V = \boxed{\phantom{00000}}
\]
*(Simplify your answer. Type an exact answer, using radicals as needed. Type your answer in factored form. Use integers or fractions for any numbers in the expression.)*
---
### Explanation of Problem Components
- **Surface Equation:** The surface is defined by the function \( z = 2 \cos x \cos y \). This represents a 3D shape where the height of the surface above any point \((x, y)\) in the \(xy\)-plane is given by this equation.
- **Rectangle Constraints:**
- The variable \( x \) ranges from \( 0 \) to \( \frac{\pi}{4} \).
- The variable \( y \) ranges from \( 0 \) to \( \frac{\pi}{2} \).
### Approach to Solution
To find the volume of the region, integrate the function \( 2 \cos x \cos y \) over the defined rectangular region \( R \):
\[ V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} 2 \cos x \cos y \, dx \, dy \]
The integration should be performed first with respect to \( x \) and then with respect to \( y \).
1. **Perform Inner Integral** \(\int_{0}^{\frac{\pi}{4}} 2 \cos x \, dx\).
2. **Perform Outer Integral** \(\int_{0}^{\frac{\pi}{2}} \left[ \text{Result of Inner Integral} \right] \, dy\).
By evaluating these integrals, the exact volume \( V \) can be calculated.
*(Note: The detailed steps to solve the integral are omitted for brevity but should be included in a complete educational explanation.)*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc1840ebe-b0df-4fe8-9210-d8e3dcfa32cc%2F5863faf8-f35a-4a46-8476-ad030a18b0f6%2F37tlize_processed.png&w=3840&q=75)
Transcribed Image Text:### Volume Calculation of a Region Bounded by a Surface and Rectangle
**Problem Statement:**
Find the volume of the region bounded above by the surface \( z = 2 \cos x \cos y \) and below by the rectangle \( R: 0 \leq x \leq \frac{\pi}{4}, \; 0 \leq y \leq \frac{\pi}{2} \).
---
\[
\vdots
\]
\[
V = \boxed{\phantom{00000}}
\]
*(Simplify your answer. Type an exact answer, using radicals as needed. Type your answer in factored form. Use integers or fractions for any numbers in the expression.)*
---
### Explanation of Problem Components
- **Surface Equation:** The surface is defined by the function \( z = 2 \cos x \cos y \). This represents a 3D shape where the height of the surface above any point \((x, y)\) in the \(xy\)-plane is given by this equation.
- **Rectangle Constraints:**
- The variable \( x \) ranges from \( 0 \) to \( \frac{\pi}{4} \).
- The variable \( y \) ranges from \( 0 \) to \( \frac{\pi}{2} \).
### Approach to Solution
To find the volume of the region, integrate the function \( 2 \cos x \cos y \) over the defined rectangular region \( R \):
\[ V = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} 2 \cos x \cos y \, dx \, dy \]
The integration should be performed first with respect to \( x \) and then with respect to \( y \).
1. **Perform Inner Integral** \(\int_{0}^{\frac{\pi}{4}} 2 \cos x \, dx\).
2. **Perform Outer Integral** \(\int_{0}^{\frac{\pi}{2}} \left[ \text{Result of Inner Integral} \right] \, dy\).
By evaluating these integrals, the exact volume \( V \) can be calculated.
*(Note: The detailed steps to solve the integral are omitted for brevity but should be included in a complete educational explanation.)*
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