I had a thought about volume measurement for shapes which are uniform along the $z$ axis. We represent the volume of a square prism as $s^2*h$. This is permissible becuase $s^2$ represents the cross-sectional area of the tube. However, this expression has an equivalent in integral form:
$\int_0^h{s^2}dh$, since this simply results in $s^2h$.
Now, we take the plane which forms the square cross section, and rotate it along one of the two parallel axes, in this case either $x$ or $y$. We give this rotation an angle, %\theta%. Thus, the cross-sectional area becomes a rectangle, instead of a square; the cross section is now cut by a rotated plane.
If we start with one point at the top of the tube, and continue integrating until the bottom of the tube, we will have cut off a section at the top, and have an additional section of equal area at the bottom. This means the total volume given by the rotated cross section should be equivalent to the original cross section.
The new area becomes $s^2\cos\theta$, since one edge of the cross section has grown in length. Using the inner angle of rotation, the this new lengthened edge has length $s\cos\theta$ (for $\theta\lt\pi/2$). Therefore, the new integral becomes $\int_0^h{s^2\cos\theta}dh$.
However, this doesn't make sense; though both formulas theoretically should equal the total volume of the rectangular tube, there appears to be a paradox, in that:
$\int_0^h{s^2}dh=\int_0^h{s^2\cos\theta}dh$, which, of course, is only true when $\cos\theta=0$.
What's going on here? Why, when the cross section has been rotated along one axis, does the total volume of the tube not work out?
Best Answer
First, I think you must have meant that the new area is $s^2/\cos\theta,$ not $s^2\cos\theta.$ The latter would be less than $s^2,$ not greater.
The bigger problem is that $s^2/\cos\theta\cdot dh$ is not the correct way of computing the volume of an infinitesimal slice. The oblique slice has volume $s^2\, dh,$ same as the perpendicular slice. The area of the base must be computed by projecting onto a plane perpendicular to the vertical axis of the prism.
One way to visualize why is to imagine dividing the $x\text{-}y$ plane into small squares of side $s/n$. Your square prism is then a bundle of $n^2$ smaller square prisms. A perpendicular slice of height $\Delta h$ consists of $n^2$ square prisms of dimensions $(s/n)\times(s/n)\times\Delta h.$ But so does an oblique slice, with the small modification that the top and bottom of each small prism isn't perpendicular to the vertical axis. As $n$ gets large, this modification is negligible. Furthermore, you can slice off the top of each small prism, and add it to the bottom to make it rectangular again.
The following images of the two-dimensional analog my make it clearer why the perpendicular slice and the oblique slice have the same volume.