Solid of revolution (disk vs. shell method) intersecting parabolas This is a simple problem and I am trying to understand it from two approaches. The book I am working out of (Calculus with Analytic Geometry - Simmons) gives an example in chapter 7.4 of how the shell method can be used to calculate the region bounded by $y=x^2$ and $y=2-x^2$ by revolving around the y-axis.
Here is the approach:
The typical shell height is $y=(2-x^2)-x^2=2-x^2$ so then we have a differential element of volume equal to $dV=2\pi xy dx = 2\pi x(2-x^2)dx=4\pi(x-x^3)dx$.
The curves intersect at $\pm 1$ and thus
$$\begin{align*}V&=4\pi\int_0^1(x-x^3)dx\\
&=4\pi(\frac{1}{2}x^2-\frac{1}{4}x^4)\bigg|_0^1\\
&=\pi.
\end{align*}$$
Now this is all well and good. But I had a problem with the following claim following this result. From the book: "Notice that if we attempt to solve this problem by the disk method, then it is necessary to calculate two integrals- one referring to the volume below the points of intersection of the two curves, and the other to the volume above".
I understand that this is the case if we use horizontal strips via the disk method, but why don't vertical strips work via the disk method for this problem?
To me, it seems that the intersecting parabolas have a symmetric property (i.e. the upper curve is symmetrical to the lower curve). So why can't we imagine a line $y=1$ and then use the disk method on the area bound by the region $y=1$ and $y=2-x^2$ and then we can treat $2-x^2-1=1-x^2$ as the "radius" of our disk? And then $dV=\pi r^2 dx = \pi (1-x^2)^2 dx$.
When I try to integrate this solid of revolution however, this is the result I get:
$$\begin{align*}
\int_{-1}^1 \pi (1-x^2)^2 dx = \frac{16\pi}{15}
\end{align*}$$
result of above integration: (link to above result).
It seems my result is very close to $\pi$ but is incorrect. Can someone see where I am going wrong here? Perhaps I am looking at the symmetry/geometry of this problem incorrectly.
 A: Your problem seems to stem from a misunderstanding of the disk and/or shell method, so I'll give a quick explanation of the two and then explain what is incorrect with your disk method. 
The idea of the shell method is integrating across "shells" of cylinders at constant $x$ values, each with a radius $\Delta y$ (when rotating around the $y$ axis). The idea of the disk, or washer, method, is to integrate along circles at constant $y$ values, each with a radius of the $x$ value at that certain $y$ value. Here are some animations to help if you still don't quite understand what is going on: Shell Animations,  Disk Animations.
To properly solve this using the disk method, we must find the radius, or $x$ value, of the circle in terms of the $y$ value, then integrate across $y$. To integrate, we must first find the radius of the circle at each $y$ value, must find $x$ in terms of $y$. For the bottom half, this is $x=\sqrt{y}$ and $dV = \pi(\sqrt{y})^2dy$, and for the top half, this is $x=\sqrt{2-y}$ and $dV=\pi(\sqrt{2-y})^2dy$. As you noticed, there is symmetry here, so we can just integrate along the bottom half and then multiply by 2. Doing so, we get the same volume as the shell method.
$$2\int_0^1 \pi (y) dy = 2\pi\frac{y^2}{2}\bigg|_0^1=\pi$$
Now here is the problem with your solution. You took $\Delta y = (2-x^2) - 1 = 1-x^2$ as the "radius" and then attempted to integrate this across $x$. This is what we would do for the shell method, but for the disk method we are integrating the radius $x$ across $y$, as we are integrating across circles centered at the $y$ axis. What you have done here with $dV = \pi ((2-x^2)-1)^2 dx$ is find the volume of the region rotated around the line $y=1$ (Try to convince yourself of this).
Lastly, the statement in the book does turn out to be incorrect in this case as there is symmetry across $y=1$. But a slight change to the problem, say $y=2-2x^2$ as the upper curve, would cause us to solve two separate integrals, one for the top half and one for the bottom half, whereas the shell method still only requires one integral (since there is still symmetry across the $y$ axis). 
