Find the volume formed by rotating the region enclosed by the following functions about the y-axis.
The curves intersects at $\displaystyle (0,0)$ and $\displaystyle (\sqrt{3},3\sqrt{3})$, so using cilindrical shel method, the volume is given by $\displaystyle V=2\pi\int_0^{3\sqrt{3}}x\left(\sqrt[3]{x}-\frac{x}{3}\right)dx$.
You may also see the problem like rotating the area between $\displaystyle y=3x $ and $\displaystyle y=x^3$ about the x-axis from $\displaystyle (0,0)$ to $\displaystyle (\sqrt{3},3\sqrt{3})$.
intersection values ...
$\displaystyle 0 = y^3 - 3y$
$\displaystyle 0 = y(y + \sqrt{3})(y - \sqrt{3})
$
two symmetrical regions in quads I and III
using the method of washers and taking advantage of symmetry ...
$\displaystyle V = \pi \int_0^{\sqrt{3}} (3y)^2 - (y^3)^2 \, dy$
I'm telling you, don't embarrass yourself. You know how to use the calculator for definite integrals right? Try using Abu-Khali's "dx" version and compare it to your own "dy" version. Then try using my own "dy" version and you will see that you are wrong, I promise you.
Edit: Sorry if I sounded like I was flaming you, that's certainly not my intention. It's just that I know I'm right, factually. Try have an open mind.
2nd Edit: yes, pony, you're right, that's the right answer.