Find the volume formed by rotating the region enclosed by $x = 36y$ and $y^3 = x$ with $y \ge 0$ about the y-axis .
My try : We can conclude from the conditions that $x=0 $ , $y = 0$ , $x = 6^3$ and $y = 6$ . So the answer is a cylinder with $r = 6^3$ , $h = 6$ and the volume is $6^7 \times \pi$ .
Is my answer right ? If not , what's the solution ?
HINT
Find the intersection point $x=a>0$ then by the shell method the set up is
$$V=\int_0^a 2\pi x\left(\sqrt[3] x-\frac x {36}\right) dx$$