Find the volume of the solid obtained by rotating the region bounded by y= x^(1/3), and y= x about the line y=1

Answer 1

here is a picture of what we are doing

http://www.wolframalpha.com/input/?i=plot+y%3D+x%5E%281%2F3%29+%2C+y+%3D+x%2C+y+%3D+1

clearly y = x and y = x^(1/3) intersect at (0,0) and (1,1)
so

outer radius = R = x-1 -->R^2 = x^2-2x+1
inner radius = r = x^(1/3) - 1) --> r^2 = x^(2/3)-2x^(1/3) + 1)

V = π∫(x^2 - 2x + 1 - x^(2/3) + 2x^(1/3) - 1) dx from 0 to 1
= π∫(x^2 - 2x - x^(2/3) + 2x^(1/3) dx from 0 to 1
= π [ x^3/3 - x^2 - (3/5)x^(5/3) + (3/2)x^(4/3) ] from 0 to 1
= π( 1/3 - 1 - 3/5 + 3/2 - 0 ...)
= 7π/30

check my steps, I should have written them out on paper first.

Answer 2

we can check using shells.

V = ∫[0,1] 2πrh dy
where r = 1-y and h = y-y^3
V = 2π∫[0,1] (1-y)(y-y^3) dy
= 2π∫[0,1] y^4-y^3-y^2+y dy
= 2π(1/5 y^5 - 1/4 y^4 - 1/3 y^3 + 1/2 y^2) [0,1]
= 2π(1/5 - 1/4 - 1/3 + 1/2)
= 2π(7/60)
= 7π/30

So, looks like Reiny's done it again!