Find the volume of the solid of revolution formed by revolving the planar region, $\displaystyle f(x)=x^3$ and $\displaystyle f(x)=x$, about the y-axis. You may choose either the shell or the disk method.

Here is what I did:

v = pi*[integral from 0 to 1 of (y^2 - (cube root(y))^2)dy]

= pi*[integral from 0 to 1 of (y^2 - y^(2/3))dy]

= pi*(1/3y^3 - 3/5y^5/3)

= pi*(1/3 - 3/5)

= pi*(5/15 - 9/15)

= -4pi/15

I'm pretty sure I can't have a negative volume, can I? What did I do wrong?

Also, how do you use the math tags to write a definite integral?