I'm trying to solve this calculus problem:
A tower is constructed with a square base and square horizontal cross-sections. Viewed from any direction perpendicular to a side, the tower has base $y = 0$ and profile lines $y = (x – 1)^2$ and $y = (x+1)^2$. (See shaded region in picture.) Find the volume of the solid.
Here's what I tried:
Consider only to the right of the $y$-axis ($x > 0$): my idea is to use the method of shells, taking the sum of volumes of small hollow cylindrical chunks obtained by rotating green vertical strips of different radii, about the $y$-axis. Each such cylindrical shell would have radius $x$, height $(x-1)^2$ and thickness $dx$. The radii would range from $0$ to $1$ (since at $x = 1$, $(x-1)^2 = 0$), and so the total volume would be given by:
$$V = \int_{0}^{1} 2 \pi x(x-1)^2 dx = \frac{\pi}{6}$$
But this turns out to be incorrect. Where am I going wrong in my reasoning? What is the right way to think about this problem?
Best Answer
At height $y$, the horizontal cross-section of the tower is a square of side $2(1-\sqrt y)$, and area $4(1-2\sqrt y+y)$. So just integrate this area over $0\le y\le 1$.