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Evaluate (x² + y²) d.S, where S' is the surface z = 4 - æ, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2. S (036² + y²) ds = 16pisqrt2

Evaluate (x² + y²) d.S, where S' is the surface z = 4 - æ, 0 ≤ x ≤ 2, 0 ≤ y ≤ 2. S (036² + y²) ds = 16pisqrt2

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** Evaluate the surface integral: \[ \iint_S (x^2 + y^2) \, dS \] where \( S \) is the surface defined by the equation \( z = 4 - x \) with the constraints \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 2 \). **Solution** The value of the integral is: \[ \iint_S (x^2 + y^2) \, dS = 16\pi \sqrt{2} \]
Transcribed Image Text:**Problem Statement** Evaluate the surface integral: \[ \iint_S (x^2 + y^2) \, dS \] where \( S \) is the surface defined by the equation \( z = 4 - x \) with the constraints \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 2 \). **Solution** The value of the integral is: \[ \iint_S (x^2 + y^2) \, dS = 16\pi \sqrt{2} \]
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