Asked by Anonymous
use the technique of cylindrical shells to find the volume created then the plane region bound by y=x^2 and y=x is revolved around y=2
Answers
Answered by
Steve
Draw yourself a picture. each shell has
radius 2-y
height sqrt(y)-y
thickness dy
so, Integral(2pi r h dy)[0,1]
2piInt(2-y)(sqrt(y)-y)dy[0,1]
2piInt(y^2 - y^3/2 - 2y + 2y^1/2)dy[0,1]
= 2pi(1/3 y^3 - 2/5 y^5/2 - y^2 + 4/3 y^3/2)[0,1]
= 2pi(1/3 - 2/5 - 1 + 4/3)
= 2pi(4/15)
= 8pi/15
radius 2-y
height sqrt(y)-y
thickness dy
so, Integral(2pi r h dy)[0,1]
2piInt(2-y)(sqrt(y)-y)dy[0,1]
2piInt(y^2 - y^3/2 - 2y + 2y^1/2)dy[0,1]
= 2pi(1/3 y^3 - 2/5 y^5/2 - y^2 + 4/3 y^3/2)[0,1]
= 2pi(1/3 - 2/5 - 1 + 4/3)
= 2pi(4/15)
= 8pi/15
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