`y = x^3, y = 0, x = 1` Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves...

The shell has the radius `1 - y` , the cricumference is `2pi*(1 - y)` and the height is `1 - x` , hence, the volume can be evaluated, using the method of cylindrical shells, such that:

`V = 2pi*int_(0)^1 (1 - y)*(1-x) dy`

You need to evaluate x from equation `y = x^3 => x = root(3)y`

`V = 2pi*int_(0)^1 (1 - y)*(1-root(3)y) dy`

`V = 2pi*int_(0)^1 (1 - root(3)y - y + y*root(3)y)dy`

`V = 2pi*(int_(0)^1 dy - int_(0)^1 root(3)y dy - int_(0)^1 ydy + int_(0)^1 y*root(3)y dy)`

Using the formula `int x^n dx = (x^(n+1))/(n+1)` yields:

`V = 2pi*(y - (3/4)*y^(4/3) - y^2/2 + (3/7)*y^(7/3))|_0^1`

`V = 2pi*(1 - (3/4)*1^(4/3) - 1^2/2 + (3/7)*1^(7/3) - 0)`

`V = 2pi*(1 - 3/4 - 1/2 + 3/7)`

`V = 2pi*(28- 21 - 14 + 12)/28`

`V = (5pi)/14`

Hence, evaluating the volume, using the method of cylindrical shells, yields `V = (5pi)/14.`