Are you talking of the area of a surface of revolution? If so, then;
\(\displaystyle S \ = \ 2\pi\int_{a}^{b}r(x)\sqrt(1+[f ' (x)]^{2})dx\)
\(\displaystyle In \ your\ \ case \ f(x) \ = \ x^{3}, \ ergo \ \ S \ = \ 2\pi\int_{0}^{1}x\sqrt(1+9x^{4})dx, \ revolved \ around \ y \ axis.\)
Note: It is immaterial which axis it is revolved around (horizontal or vertical)
If revolved around y axis r(x) usually is x, if revolved around x axis, r(x) is usually f(x), r(x) being the distance between the graph of f and the axis of revolution.
