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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![**Using Double Integrals to Find Volume:**
This exercise involves calculating the exact volume under the surface defined by \( z = x^2 + y^2 \) over a rectangular region in the xy-plane.
**Problem Statement:**
Compute the volume under the surface \( z = x^2 + y^2 \) above the rectangle \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \) using a double integral.
**Explanation:**
To solve this problem, set up the double integral of the function \( x^2 + y^2 \) over the given region. The limits of the integrals follow the constraints on \( x \) and \( y \), creating a rectangular region in the xy-plane:
\[
\int_{0}^{1} \int_{0}^{1} (x^2 + y^2) \, dy \, dx
\]
1. **Integrate with respect to \( y \):**
- Keep \( x \) constant while integrating \( x^2 + y^2 \) with respect to \( y \) from 0 to 1.
2. **Integrate with respect to \( x \):**
- After integrating over \( y \), integrate the resulting expression with respect to \( x \) from 0 to 1.
The solution to this integral will yield the volume under the surface \( z = x^2 + y^2 \) above the specified rectangular region.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F22ace5cb-53da-4a61-aefb-a5b307082c8e%2Fc6c1c688-e04a-4565-9392-df5b7d0b39cb%2Fg9gkfay_processed.png&w=3840&q=75)
Transcribed Image Text:**Using Double Integrals to Find Volume:**
This exercise involves calculating the exact volume under the surface defined by \( z = x^2 + y^2 \) over a rectangular region in the xy-plane.
**Problem Statement:**
Compute the volume under the surface \( z = x^2 + y^2 \) above the rectangle \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \) using a double integral.
**Explanation:**
To solve this problem, set up the double integral of the function \( x^2 + y^2 \) over the given region. The limits of the integrals follow the constraints on \( x \) and \( y \), creating a rectangular region in the xy-plane:
\[
\int_{0}^{1} \int_{0}^{1} (x^2 + y^2) \, dy \, dx
\]
1. **Integrate with respect to \( y \):**
- Keep \( x \) constant while integrating \( x^2 + y^2 \) with respect to \( y \) from 0 to 1.
2. **Integrate with respect to \( x \):**
- After integrating over \( y \), integrate the resulting expression with respect to \( x \) from 0 to 1.
The solution to this integral will yield the volume under the surface \( z = x^2 + y^2 \) above the specified rectangular region.
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