Using cross sections to find the volume bounded by the above function about the $x-$axis, I get the following:
$$\pi\int_0^1x - x^2\,\mathrm{d}x = \pi\left[{x^2\over2} - {x^3\over3}\right]_0^1 = {\pi\over6}$$
However, using the shell method I get the following:
First rearrange to put into terms of $y$: $x = y^2$ and $x = y$. Moreover, these functions intersect at $0$ and $1$. So, we have:
$$2\pi\int_0^1y - y^2\,\mathrm{d}y = 2\pi\left[{y^2\over2} - {y^3\over3}\right]_0^1 = {\pi\over3}$$
The answers are different, but they're supposed to be the same. I am pretty sure my shell method is wrong, but I don't see why.
Thanks.
In shell integration, you're have to integrate every single shell, which is found through $\text{height} \cdot \text{radius}$. You've forgotten the radius of each shell.
In this case, each shell's height is given by $y-y^2$ because $x=y$ is to right of $x=y^2$. The radius is given by $y$, because the radius of the shell corresponds to the current $y$-value, so you have to multiply those together to get the integral:
$$2\pi\int_0^1 y(y-y^2) \, dy$$