Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded byy = 8 andy16-16 about thex-axis. (A) V=R = π *((16-1)-64) dx = 14,336 √/2π 15 ℗ - RF ((16-) - 64) dx = 14,36 /2π 15 E V=n TS ((16- 2)²-64) dx = 28,672 √/2π 16 15 © V = - π (16--64)dx = 7,168 √2π 15 - ((16-1)-64) dx = 28,672 √/2π TLV V=R
Set up and evaluate the integral that gives the volume of the solid formed by revolving the region bounded byy = 8 andy16-16 about thex-axis. (A) V=R = π *((16-1)-64) dx = 14,336 √/2π 15 ℗ - RF ((16-) - 64) dx = 14,36 /2π 15 E V=n TS ((16- 2)²-64) dx = 28,672 √/2π 16 15 © V = - π (16--64)dx = 7,168 √2π 15 - ((16-1)-64) dx = 28,672 √/2π TLV V=R
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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Answer these 2 calculus problem please

Transcribed Image Text:Set up and evaluate the integral that gives the volume of the solid formed by revolving the
region bounded byy = 8 andy = 16-
about thex-axis.
^ V = π ((16-³-64) dx = 14,336 √/2π
[ 1"
15
B
D
E
T
16
= πt
[W/((16- )* - 64) dx = ¹
14,336
15
√2π
((16--64) dx = 28,672 √2
15
V=R
= π ((16- )-64)dx = 7.158 √/2π
7/16
TV ((16-1)-64) dx = 28,672 √21
V=T
15

Transcribed Image Text:Find the volume of the solid generated by revolving the region bounded by the graphs of the
equations 8x², y = 0, andx = 2 about the liney = 32,
A
B
D
E
1,792
T
15
14,336
15
14,336
15
1,792
15
-T
15,488
15
-T
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