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:

Calculus: Early Transcendentals
8th Edition
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Author:James Stewart
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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 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
r(x)=x²+1
(1,2)
(-x+3)² - (x² + 1)²) dx
= r(=x
-2
= #(8
6x - x²- x²) dx
Interval of
integration
(a)
(-2.5)
ROX)
-x+3
(1, 2)
+1
=π 8x
3x²------- 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 ?
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 r(x)=x²+1 (1,2) (-x+3)² - (x² + 1)²) dx = r(=x -2 = #(8 6x - x²- x²) dx Interval of integration (a) (-2.5) ROX) -x+3 (1, 2) +1 =π 8x 3x²------- 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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