Why does rotating a cross sectional area result in a different volume? 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?
 A: It still took a little trouble to understand what you wanted.  I think you are asking about slicing the square prism obliquely to the xy-plane.  The cross-sections would now indeed be rectangles.  However, if you now "slide" that upward through a "height" h , you do not produce a square prism, but what is called a "parallelopiped", the horizontal ends of which are rectangles, one pair of opposite vertical faces are rectangles, but the other pair of opposite faces are now parallelograms.  This is still a prism, but no longer a rectangular prism.  So it is reasonable that the volume is different.
Am I interpreting your description correctly? 
All right, in response to your comment:
If the slicing is done at an angle $\theta$ to the horizontal, the area of the rectangles will be $s \cdot \frac{s}{\cos\theta}$.  Run these oblique rectangles along the z-axis along a "height" $h$.  What is the volume now?
We can think of the "slices" as a stack of playing cards with area $\frac{s^2}{\cos\theta}$ which have been pushed so that the stack is now leaning (with the line through their centers having length $h$).  A little geometry shows that the axis of the stack makes an angle $(90^{\circ} - \theta)$ to the horizontal, so the height of the leaning stack is now $h \sin(90^{\circ} - \theta)  =  h \cos\theta$. Cavalieri's Principle then tells us that the volume of this stack is  $\frac{s^2}{\cos\theta} \cdot  h \cos\theta  =  s^2 h.$  If I'm understanding what you are describing, there is no paradox.
EDIT (had to go to a talk and just got back):  I was a little troubled by this argument because it seemed a little glib, but it does still work out.  Suppose you leave the original square prism as it is and take the oblique slices (at angle $\theta$ to the horizontal) through it.  You will have a parallelopided and two identical end wedges left over.  
The end wedges have square bases with areas $s^2$ and height $\frac{s}{\tan\theta}$. If you put them together, you get a square prism with volume $\frac{s^3}{\tan\theta}$.
The parallelopiped has a "height" along the z-direction of $h - \frac{s}{\tan\theta}$. The Cavalieri's Principle argument I gave still applies, but the volume of this solid is now 
$$(\frac{s^2}{\cos\theta}) \cdot (h - \frac{s}{\tan\theta}) \cdot (\cos\theta)$$
$$= s^2h - \frac{s^3}{\tan\theta} .$$
(Note: this solid can also be pictured as having a parallelogram "base" with sides $\frac{s}{\cos\theta}$ and $h - \frac{s}{\tan\theta}$, making an angle $\theta$ between them, and having a "height" $s$ -- as a result of laying the parallelopiped on its side.)
So the volume of this parallelopiped and the two end wedges does give us $s^2h$.
A: 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.

A: Start with a cuboid of dimensions $0 \le x \le 1$, $0 \le y \le 1$ and $0 \le z \le 2$. Next we will take planer cross-sections which are all perpendicular to the $z$-axis. As you say, for all $z$ from $0$ to $2$ you get a square as the cross-section and the integral becomes $1 \times 1 \times 2 = 2$.
However, when you rotate the cross-sectional plane, you don't always get the same shape as the cross-section. The cross-sectional shape varies as a function of $z$. For example, take the planes $x+z=k$. The plane $x+z=0$ plane meets the cuboid only along the $y$-axis with $0 \le y \le 1$. The for all $0 < k < 1$, the planes meet the cuboid in ever growing rectangles. For $1 \le k \le 2$, the planes meet the cuboid in fixed-sized rectanges. For $2 < k \le 3$ the cross sections shrink down, all the way to a line segment when $k=3$.
