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Evaluate the surface integral. Jy y ds = ²/3 (x³/² + y³/2), 0 ≤ x ≤ 5,0 ≤ y ≤ 3 S is the surface Z =

Evaluate the surface integral. Jy y ds = ²/3 (x³/² + y³/2), 0 ≤ x ≤ 5,0 ≤ y ≤ 3 S is the surface Z =

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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Question
Evaluate the surface integral. JY y ds = (x³/2² + y³/²), 0 ≤ x ≤ 5,0 ≤ y ≤ 3 s S is the surface z =
Transcribed Image Text:Evaluate the surface integral. JY y ds = (x³/2² + y³/²), 0 ≤ x ≤ 5,0 ≤ y ≤ 3 s S is the surface z =
Evaluate the surface integral. J S is the surface with parametric equations x = u², y = u sin(3w), z = u cos(3w), 0 ≤ u ≤ 1,0 ≤ w ≤ T yz ds
Transcribed Image Text:Evaluate the surface integral. J S is the surface with parametric equations x = u², y = u sin(3w), z = u cos(3w), 0 ≤ u ≤ 1,0 ≤ w ≤ T yz ds
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