The region bounded by the curve y = x² + 1 and the line y-x+3 is revolved about the x-axis to generate a solid. Find the volume of this solid. Solution: x² + 1 = x + 3 x²+x-2=0 (x+2)(x-1) = 0 x=-2, x = 1 Outer radius: R(x) = x+3 Inner radius: r(x) = x² + 1 = L" = ([R(x)]² - [r{(x)]³) dx V = (-2,5) R(x)--x+3 y--x+3 (1,2) (-x+3)² - (x² + 1)²) dx = 」7(=x = 7(8 6x - x²-x4) dx Interval of integration (a) i need clear ans and solve very very fast in 20 min and thank you | DYBALA ROX) ra (1, 2)) 8x 3x²------² - 117 = Washer cross section Outer radius: R(x)--x+3 Inner radius: r(x) = x²+1 (b) H.W. 1: Solve the previous example by rotating the region about the line y = -1 ?
The region bounded by the curve y = x² + 1 and the line y-x+3 is revolved about the x-axis to generate a solid. Find the volume of this solid. Solution: x² + 1 = x + 3 x²+x-2=0 (x+2)(x-1) = 0 x=-2, x = 1 Outer radius: R(x) = x+3 Inner radius: r(x) = x² + 1 = L" = ([R(x)]² - [r{(x)]³) dx V = (-2,5) R(x)--x+3 y--x+3 (1,2) (-x+3)² - (x² + 1)²) dx = 」7(=x = 7(8 6x - x²-x4) dx Interval of integration (a) i need clear ans and solve very very fast in 20 min and thank you | DYBALA ROX) ra (1, 2)) 8x 3x²------² - 117 = Washer cross section Outer radius: R(x)--x+3 Inner radius: r(x) = x²+1 (b) H.W. 1: Solve the previous example by rotating the region about the line y = -1 ?
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...
Related questions
Question
![The region bounded by the curve y = x² + 1 and the line y-x+3 is
revolved about the x-axis to generate a solid. Find the volume of this solid.
Solution:
x² + 1 = x + 3
x²+x-2=0
(x+2)(x-1) = 0
x=-2, x = 1
Outer radius: R(x) = x+3
Inner radius: r(x) = x² + 1
= L" = ([R(x)]² - [r{(x)]³) dx
V =
(-2,5)
R(x)--x+3
y--x+3
(1,2)
(-x+3)² - (x² + 1)²) dx
= 」7(=x
= 7(8 6x - x²-x4) dx
Interval of
integration
(a)
i need clear ans and solve very very fast
in 20 min and thank you | DYBALA
ROX)
ra
(1, 2))
8x
3x²------² - 117
=
Washer cross section
Outer radius: R(x)--x+3
Inner radius: r(x) = x²+1
(b)
H.W. 1:
Solve the previous example by rotating the region about the
line y = -1 ?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0baaf4e4-5654-4278-a18e-0b3d1e9bbff7%2F8cec95f7-3fce-42d7-ad4a-5082179fcfa1%2F0wigxwy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The region bounded by the curve y = x² + 1 and the line y-x+3 is
revolved about the x-axis to generate a solid. Find the volume of this solid.
Solution:
x² + 1 = x + 3
x²+x-2=0
(x+2)(x-1) = 0
x=-2, x = 1
Outer radius: R(x) = x+3
Inner radius: r(x) = x² + 1
= L" = ([R(x)]² - [r{(x)]³) dx
V =
(-2,5)
R(x)--x+3
y--x+3
(1,2)
(-x+3)² - (x² + 1)²) dx
= 」7(=x
= 7(8 6x - x²-x4) dx
Interval of
integration
(a)
i need clear ans and solve very very fast
in 20 min and thank you | DYBALA
ROX)
ra
(1, 2))
8x
3x²------² - 117
=
Washer cross section
Outer radius: R(x)--x+3
Inner radius: r(x) = x²+1
(b)
H.W. 1:
Solve the previous example by rotating the region about the
line y = -1 ?
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