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Let R be the rectangle in the xy-plane with vertices (+1,±1). Find the volume of the solid lying above R and below the graph of z = 2 – x² – y².

Let R be the rectangle in the xy-plane with vertices (+1,±1). Find the volume of the solid lying above R and below the graph of z = 2 – x² – y².

Calculus: Early Transcendentals
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 Statement:** Let \( R \) be the rectangle in the \( xy \)-plane with vertices \((\pm 1, \pm 1)\). Find the volume of the solid lying above \( R \) and below the graph of \( z = 2 - x^2 - y^2 \). **Explanation:** - **Vertices of Rectangle \( R \):** The rectangle \( R \) is in the \( xy \)-plane with corners at coordinates \((1, 1), (1, -1), (-1, 1), (-1, -1)\). - **Equation of Surface:** The surface is defined by the equation \( z = 2 - x^2 - y^2 \), which represents a paraboloid opening downwards. - **Objective:** Find the volume of the solid that is bounded by the surface above and the rectangle \( R \) below. This typically involves setting up a double integral over the region \( R \). In mathematical terms, the volume \( V \) can be found using the double integral: \[ V = \int_{-1}^{1} \int_{-1}^{1} (2 - x^2 - y^2) \, dy \, dx \] This integral sums up the infinitesimal volumes below the surface over the area of the rectangle \( R \).
Transcribed Image Text:**Problem Statement:** Let \( R \) be the rectangle in the \( xy \)-plane with vertices \((\pm 1, \pm 1)\). Find the volume of the solid lying above \( R \) and below the graph of \( z = 2 - x^2 - y^2 \). **Explanation:** - **Vertices of Rectangle \( R \):** The rectangle \( R \) is in the \( xy \)-plane with corners at coordinates \((1, 1), (1, -1), (-1, 1), (-1, -1)\). - **Equation of Surface:** The surface is defined by the equation \( z = 2 - x^2 - y^2 \), which represents a paraboloid opening downwards. - **Objective:** Find the volume of the solid that is bounded by the surface above and the rectangle \( R \) below. This typically involves setting up a double integral over the region \( R \). In mathematical terms, the volume \( V \) can be found using the double integral: \[ V = \int_{-1}^{1} \int_{-1}^{1} (2 - x^2 - y^2) \, dy \, dx \] This integral sums up the infinitesimal volumes below the surface over the area of the rectangle \( R \).
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