# Volumes of Rotation-- space and its compliment don't have same volume in same boundaries

These two problems come from the MIT 18.01 problem sets. I believe they describe two complimentary spaces in the same boundary, so I would have thought they had the same volume. They don't. I feel insane.

Consider the volume of the region described by $0 \le y \le x^2,$ $x \le 1$ rotated around the y axis. I had no problem finding the answer: it's $\frac\pi 2$.

Already I feel crazy, because the volume of the cylinder described by those boundaries is $\pi$-- meaning that the space described above is half the volume. But, just for the area between 0 and 1, the non-rotated space described is $\frac13$ the area of the corresponding square. I would imagine the volume would have the same ratio, but it doesn't. That's kind of a side issue. Here's the kicker:

Consider the volume of the region described by $\sqrt x \le y \le 1,$ $x \ge 0$ rotated around the y axis. Again, the answer isn't hard to get-- it's $\frac\pi 5$. Of course, $y = x^2$ is just the inverse of $y=\sqrt x$ between 0 and 1. In that domain and range, the area below $y = x^2$ has the same volume as the area above $y=\sqrt x$. Shouldn't they have the same volume?

But they don't. One of these statements is false, but I don't know which one it is:

1. The two volumes described above are compliments where $-1 \le x \le 1$ and $0 \le y \le 1$.
2. Complimentary volumes in the same space should be equivalent.

Help? And if someone could clear up my confusion described above about $\frac 1 3$ of the area can create $\frac 1 2$ of the volume when rotated, I'd really appreciate it. Like I said, I feel crazy. Thanks!