find the volume of the solid formed by revolving the region bounded by the graphs of y=x^3 y=1 x=2 about the y-axis

Answer 1

when y = 1, x = 1
when y = 8, x = 2so we seem to be talking about the sort of triangle between (1,1) (2,1) and (2,8)

could be vertical cylinders of wall thickness dx and height (x^3-1) and radius x from x = 1 to x = 2
v = integral [2 pi x(x^3-1)]dx from x = 1 to x = 2

2 pi [ x^5/5 - x^2/3 ]

2 pi [ 32/5 - 4/3 ] - 2 pi [1/5-1/3]

2 pi [ 31/5 - 1 ]
check my arithmetic !

Answer 2

using shells of thickness dx, we have

v = ∫[1,2] 2πrh dx
where r=x and h=y-1
v = ∫[1,2] 2πx(x^3-1) dx = 47π/5

using discs (washers) of thickness dy,

v = ∫[1,8] π(R^2-r^2) dy
where R=2 and r=x
v = ∫[1,8] π(4-y^(2/3)) dy = 47π/5

Answer 3

Thank you!!!!