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
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?