volume of rotated solid Find the volume of the solid obtained by rotating about the $y$-axis the region bounded
by the curves $y = e^{-2x^2}$, $y = 0,$ $x = 0,$ and $x = 1.$
I'm a little confused on what the bounds of the integral would be.
Since it is revolved around the $y$ axis would I use bounds that go on the $y$ axis? If I did I'm thinking that my bounds would be $0$ and $1$ on the $y$ axis?
Or am I wrong and do I use the $x$ axis as my bounds for the integral?
 A: We assume it is $y=e^{2x^2}$.  The easiest method is cylindrical shells. That gives
$$\int_0^1 2\pi xe^{2x^2}\, dx,$$
and the integral yields to the substitution $u=2x^2$.
Remark: To figure out the bounds, first make a sketch.
Suppose that we will use cylindrical shells. We are rotating about a vertical line, the $y$-axis. Then the "method" consists of taking a thin vertical strip, from $x$ to $x+dx$, and rotating that about the $y$-axis, and then "adding up" (integrating) over all $x$. And we were told $x$ goes from $0$ to $1$. That takes care of the bounds issue.
If we use the method of slicing (cross-sections) then we would need to take cross-sections perpendicular to the $y$-axis. In that case we end up integrating with respect to $y$. For this problem, that method is messier. For one thing, the geometry is different from $y=0$ to $y=1$ than it is later. 
Added: Since the answer above was posted, the function has changed to $y=e^{-2x^2}$.  Apart from the obvious change in the integrand, nothing much changes. Now it is best to use the substitution $u=-2x^2$. 
