Find volume of region bound by $y=x, y=x^2$ around x-axis Here is the problem in my textbook: 

Find the volume of the solid obtained by rotating the region bounded by
  the curves $y=x, y=x^2$ about x-axis.

Here is my solution : 
Because equation $x = x^2$ has two roots : $0$ and $1$. we have:
$$ V= \int_0^1{2\pi x(x^2-x)}dx = \frac{\pi}{6}$$
But the solution in my textbook is $\frac{2\pi}{15}$. I think the hole in my solution is :  I haven't use the region rotate around x-axis yet. But, I don't know how to use this statement in solution when couting volume.
Thanks :)
 A: Consider the diagram

For your problem, this is equivalent to rotating $R_2$ about the line $y=0$.
Since neither of the curves are touching the axis we are rotating about, we must use washers.
For washers, the volume is given by $$ V = \pi\int^b_a (R(x))^2-(r(x))^2 \ dx$$
Where $R(x)$ is the curve farthest away from our axis of rotation - the top function $y = x$ and $r(x)$ is the curve closest to the axis of rotation - the bottom function $y=x^2.$
As you found the intersecting points, we can set up our integral as 
$$V = \pi\int^1_0 (x)^2-(x^2)^2 \ dx$$
$$V = \pi\int^1_0 x^2-x^4 \ dx$$
After integrating and evaluating, you should get $\frac{2\pi}{15}$.
A: Your formula should be $$V=\pi\int_0^1x^2-(x^2)^2\,dx,$$ which will give you the correct answer. You were trying to use the cylinder method (which would be appropriate if rotating about the $y$-axis), when you should have been using the washer method (since we are integrating along the axis of rotation, rather than perpendicular to it).
A: In this problem, it is probably easiest to use the method of slicing. However,
if you are going to use the "cylindrical shells" method, while rotating about the $x$-axis, you have to integrate with respect to $y$.
Take a thin horizontal strip going from height $y$ to height $y+dy$. The strip is more or less at height $y$. It has length $\sqrt{y}-1$ (draw a picture). The volume of the shell obtained by rotating this strip is about $2\pi y(\sqrt{y}-y)\,dy$. Thus
our volume is
$$\int_{y=0}^1 2\pi y(\sqrt{y}-y)\,dy.$$
So we want
$$2\pi\int_{y=0}^1 (y^{3/2}-y^2)\,dy.$$
Integrate. We get $2\pi\left(\frac{2}{5}-\frac{1}{3}\right)$. 
A: The volume of the solid bounded by $y= f(x)$ rotating about the $x$-axis is $V=\pi\int_a^b(f(x))^2dx$. That is, you want to calculate $\pi (\int_0^1x^2dx -\int_0^1x^4dx)$
