The two curves intersect when [tex]y=0[/tex] and [tex]y=1[/tex]. Over the range [tex]0\le y\le1[/tex], the curve [tex]x=12y^2[/tex] lies above [tex]x=12y^3[/tex]. Hence the area of the shaded region is
[tex]\displaystyle \int_0^1 (12y^2 - 12y^3) \, dy = 4y^3 - 3y^4\bigg|_0^1 = (4-3)-(0-0) = \boxed{1}[/tex]
