he problem is shown on the screen and the answer is on paper. For the integral I thought the y^2/3 was supposed to be y^1/3. Why is it y^2/3?

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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The problem is shown on the screen and the answer is on paper. For the integral I thought the y^2/3 was supposed to be y^1/3. Why is it y^2/3?

2m
6.3.27 V = T
(1 – y²/3) dy =
n
,5/3
y -
5.
01
Transcribed Image Text:2m 6.3.27 V = T (1 – y²/3) dy = n ,5/3 y - 5. 01
27. y=x, y = 0, and x = 1; about the y-axis
y A
y= x³
Transcribed Image Text:27. y=x, y = 0, and x = 1; about the y-axis y A y= x³
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To let know why in the integral, it is y23 and not y13

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