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Find the length of the given curve. 3y2/3 x = 3y4/3 , -64 ≤ y ≤ 512 32 The length is equal to the integral L = = √1 [ h(y) dy where C= -64 d = 512 and h(y)= | To evaluate this integral we need to split it up at y = where the integral is improper. After evaluating the integral, we find that L =

Find the length of the given curve. 3y2/3 x = 3y4/3 , -64 ≤ y ≤ 512 32 The length is equal to the integral L = = √1 [ h(y) dy where C= -64 d = 512 and h(y)= | To evaluate this integral we need to split it up at y = where the integral is improper. After evaluating the integral, we find that L =

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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Find the length of the given curve. 3y2/3 x = 3y4/3 , -64 ≤ y ≤ 512 32 The length is equal to the integral L = = √1 [ h(y) dy where C= -64 d = 512 and h(y)= | To evaluate this integral we need to split it up at y = where the integral is improper. After evaluating the integral, we find that L =
Transcribed Image Text:Find the length of the given curve. 3y2/3 x = 3y4/3 , -64 ≤ y ≤ 512 32 The length is equal to the integral L = = √1 [ h(y) dy where C= -64 d = 512 and h(y)= | To evaluate this integral we need to split it up at y = where the integral is improper. After evaluating the integral, we find that L =
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