It is helpful to first draw a picture: 
Notice that $x = x^2$ implies that $x^2 - x = 0$ so that $x = 0$ or $x = 1$. So we are rotating the area around the line $y = 2$. We think of the volume we are seeking as a difference of two volumes. First, we rotate the outermost function ($\color{blue}{y = x^2}$) around the line $y = 2$. Then, we rotate the innermost function ($\color{green}{y = x}$) around the line $y = 2$. Taking the difference will give the volume we are seeking:
$ V = \pi \int_0^1 (2 - x^2)^2 dx - \pi \int_0^1 (2 - x)^2 dx = \pi\int_0^1 ((2 - x^2)^2 - (2 - x)^2)dx $
Again, to give you a better idea of what we are doing, we start by partitioning the area between the curves with rectangles. Next, imagine we pick up each rectangle and rotate it around the line $y = 2$. When we do this with each rectangle, we get a solid that looks like a washer (hence the name). Finally, the volume of a disk of width $\Delta x$ and height $y = f(x)$ is $\pi f(x)^2 \Delta x$. Adding these up give:
$ V \approx \sum_{k=1}^n \pi f(x_k)^2 \Delta x \to \pi \int_a^b f(x)^2 dx \qquad (\text{as }n \to \infty) $