Calculus: Early Transcendentals, Chapter 6, 6.3, Section 6.3, Problem 19

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.