V = ∫1/22 2πrhdx
= 2π ∫1/22 (2 - x)(x - (-x + 1)) dx
= 2π ∫1/22 (2 - x)(2x - 1)) dx
= 2π ∫1/22 (-2x2 + 5x - 2) dx
= 2π [ -2/3x3 + 5/2x2 - 2x ]1/22
= 2π [2/3 - (-11/24)]
= 9π/4
Leah H.
asked • 04/21/24use the shell method to find the volume of region enclosed by y=x, y=-x+1, revolved about the line x=2
Sketch and find the volume, using the shell method, of the region enclosed by y=x, y=-x+1, revolved about the line x=2.
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2 Answers By Expert Tutors
HIRDESHWAR S. answered • 04/21/24
Tutor
New to Wyzant
M.Sc. & Ph.D. in Mathematics. Experienc 40 years
V= Int.[2(Pi)(x+1-x)(2-x)dx (from x= 0 to x=2) = 2Pi [2x-(1/2)x^2}(X=0 t0 x=2)]= 4Pi
Doug C.
I think the line x = 2 was not a boundary, but just the axis of revolution. Probably need to assume y = 0 was the other boundary line for the region revolved around x = 2. Using washer method instead of shell I got 3pi/4 as volume.
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04/22/24
Josh F.
tutor
I agree with Hirdeshwar's interpretation: the region to be revolved is triangular, bounded by y = x on top, y = - x + 1 below, and by the axis of revolution, x = 2, on the right.
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04/22/24
Doug C.
In that case I agree with Josh's answer 9pi/4, using both disk and shell method here: desmos.com/calculator/kseunrssjn
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04/23/24
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HIRDESHWAR S.
Region Bounded by y= x and y = -x+1 and line x=2. The lines intersect at x = -x +1 , x=1/2. V= Int. [ 2pi{x- (-x+1)}(2-x)dx from x=0.5 to x= 2 = int. [2Pi {(2x-1)(2-x)]dx = Int[2pi (-2x^2+5x-2)]dx = = 2pi[-(2/3)x^3 + (5/2)x^2 -2x](x=0.5 to x=2) = = (Pi/3)[(-32+20-24) - (-1/2 + 15/4 -6)] = Pi/3 = 37Pi/1204/21/24