The region bounded by y = x and y = √x lies in the first quadrant with x in the interval [0,1].
Take a thin vertical cross section of the region with width dx. Rotating the cross section about the line y = 1(a horizontal line) generates a washer with outer radius 1-x and inner radius 1-√x.
Note: With the disk/washer method, we always take cross sections perpendicular to the axis of rotation.
Volume of typical washer = π(1-x)2dx - π(1-√x)2dx
Volume of solid = π∫from 0 to 1[x2+2√x-3x]dx
= π[(1/3)x3+(4/3)x3/2-(3/2)x2]from 0 to 1
= π[(1/3)+(4/3)-(3/2)] = π/6
Ab: I guess that we are both right (or both wrong).