Find volume of a revolved solid by integrating wedges. So, lets say that I wanted to find the volume of the solid formed by rotating the area between
$f(x)=\sqrt{1-x^2}, 0<x<1$ and the $x$ axis around the $y$ axis. (This example is simply a hemisphere).  
Now normally, I would use geometry, or the "disk method", so the area would simply be $\pi\int_0^1(1-y^2)dy=\frac{2\pi}{3}$.  
I was thinking about this and I was wondering if it would be possible to find the answer by integrating wedges of this volume from $0$ to $2\pi$. This seems to be an approach that more closely resembles the premise of the problem. At first I thought that this might be as easy as $\frac{1}{2}\int_0^{2\pi}[\int_0^1f(x)dx]^2d\theta$, essentially integrating a polar circle with radius of the area that is revolved around the y axis. However, when I tried this, I did not get my expected answer. I calculated the volume to be $\frac{\pi^3}{16}$, however, I should have found the volume to be $\frac{2\pi}{3}$.
Can anyone help me understand why my approach was not successful, and also explain a successful method of evaluating the volume in this way?

You can certainly integrate using wedges instead of disks, but what you've written down doesn't correspond to wedges. One thing you can check to see that this can't be right is dimensions – your expression has dimensions of length to the fourth power whereas a volume has length to the third. That's not entirely suprising when you say "with radius of the area" – it doesn't make sense to use an area as a radius (at least not usually :-).
In a wedge with angular extent $\mathrm d\theta$, an area $\mathrm dS$ of the rotated quarter-circle contributes $x\mathrm dS\mathrm d\theta$ to the volume of the wedge, so the volume is
$$
\begin{align}
\int_0^{2\pi}\left[\int x\mathrm dS\right]\mathrm d\theta
&=\int_0^{2\pi}\left[\int_0^1xf(x)\mathrm dx\right]\mathrm d\theta
\\
&=\int_0^{2\pi}\left[\int_0^1x\sqrt{1-x^2}\mathrm dx\right]\mathrm d\theta
\\
&=\int_0^{2\pi}\left[-\frac13\sqrt{1-x^2}^3\mathrm dx\right]_0^1\mathrm d\theta
\\
&=\frac{2\pi}3
\end{align}
$$
as expected.
