Easy to see that the intersection of y = x and y = √x is (0,0) and (1,1)
In that interval y = √x is above y = x, so we have
Vol = π ∫ (x - x^2) dx from 0 to 1
= π [ x^2/2 - x^3/3 | from 0 to 1
= π( 1/2 - 1/3 - 0)
= π/6 units^3
Find the volume V of the solid obtained by rotating the region bounded
by curves y=x and y= √𝒙 about the x-axis
3 answers
Well they cross at (1,1) so I suppose you mean between x = 0 and x = 1
so the outer radius is ro = x^.5
and the inner radius is ri = x
and we want the integral from x = 0 to x = 1 of
[ pi ro^2 - pi ri^2 ] dx
= pi [ x - x^2 ] dx
= pi [ (1/2) x^2 - (1/3) x^3] at x = 1 - at x = 0
= pi [1/2 - 1/3] = pi / 6
so the outer radius is ro = x^.5
and the inner radius is ri = x
and we want the integral from x = 0 to x = 1 of
[ pi ro^2 - pi ri^2 ] dx
= pi [ x - x^2 ] dx
= pi [ (1/2) x^2 - (1/3) x^3] at x = 1 - at x = 0
= pi [1/2 - 1/3] = pi / 6
using shells of thickness dy, we have
v = ∫[0,1] 2πrh dy
where r = y and h = y-y^2
v = ∫[0,1] 2πy(y-y^2) dy = π/6
v = ∫[0,1] 2πrh dy
where r = y and h = y-y^2
v = ∫[0,1] 2πy(y-y^2) dy = π/6
1 answer