Using the method of cylindrical shells, set up but do not evaluate an integral for the volume of the solid generated when the region R in the first quadrant bounded by the graphs of y = V4 – x², y = 0, and x = 0 is revolved about (a) the linex = 2 and (b) the line y = -2. (a) V = 2x(2 – x)V4 – x² åx V = #(2 – x)V4 – x² åx V = : 2«(2 – x)(4 – x²) ix V = :| 2x(2+ x)V4 – x år (b) v = [ 2a(2 + y»\/4 = y° ãy V = 2 + y)▼
Using the method of cylindrical shells, set up but do not evaluate an integral for the volume of the solid generated when the region R in the first quadrant bounded by the graphs of y = V4 – x², y = 0, and x = 0 is revolved about (a) the linex = 2 and (b) the line y = -2. (a) V = 2x(2 – x)V4 – x² åx V = #(2 – x)V4 – x² åx V = : 2«(2 – x)(4 – x²) ix V = :| 2x(2+ x)V4 – x år (b) v = [ 2a(2 + y»\/4 = y° ãy V = 2 + y)▼
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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![Using the method of cylindrical shells, set up but do not evaluate an integral for the volume of the solid generated when the region R in
the first quadrant bounded by the graphs of y = V4 - x, y = 0, and x = 0 is revolved about (a) the line x = 2 and (b) the line
y = -2.
(a)
2m(2 – x)V4 – x² áx
V =
V =
1(2 – x)V4 – x² dx
27(2 – x)(4 – x) áx
V =
V =
2x(2 + x)V4 – ? åx
(b)
V =
2л(2 + у) /4 — у? dу
я(2 + у)у
|4 – y² ay
V =
V =
27(2 – y)(4 – y) ảy
4
2я(2 + у) у4 — у dy
V =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F761c46be-c490-4f68-900f-dd39c070c263%2F51976cf5-6234-4bad-afb4-b77fedcb608d%2Fxhgv7k_processed.png&w=3840&q=75)
Transcribed Image Text:Using the method of cylindrical shells, set up but do not evaluate an integral for the volume of the solid generated when the region R in
the first quadrant bounded by the graphs of y = V4 - x, y = 0, and x = 0 is revolved about (a) the line x = 2 and (b) the line
y = -2.
(a)
2m(2 – x)V4 – x² áx
V =
V =
1(2 – x)V4 – x² dx
27(2 – x)(4 – x) áx
V =
V =
2x(2 + x)V4 – ? åx
(b)
V =
2л(2 + у) /4 — у? dу
я(2 + у)у
|4 – y² ay
V =
V =
27(2 – y)(4 – y) ảy
4
2я(2 + у) у4 — у dy
V =
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