Volume formed by rotating square by 360 deg

Question

If I know the area of the square and I rotate it around one of the edges by $360$ deg, it forms a solid cylinder. Why then is the volume of this cylinder not equal to "area of the Square" multiplied by $360$? Or is it "area of the Square" multiplied by circumference multiplied by diameter?

Answer 1

Hints: Volume of the cylinder is the area of the (circular) base times the height.

As you rotate your square, what determines the area of the base; and what determines the height?

Answer 2

Assume that the square has side lengths $x$. Then, the cylinder has height $x$ and radius $x$, so its volume is $\pi x^3$.

There are many problems with the formula in the question.

First, 360 is an angle measurement, not a distance, so the product of the area of the square and 360 will not have the right units.

Also, area of the square times the circumference times the diameter (which I assume is the circumference and diameter of the cylinder) also has the wrong units: it is a product of four length-units.

One way to compute the correct value following your thinking is to notice that when you rotate the square, some of the points move more than others. In particular, point on the edge of revolution do not move and points on the opposite edge move a distance of $2\pi x$. If you average these two distances, you find that an "average" point moves $\pi x$ during the rotation. Multiplying this by the area $x^2$ will give you $\pi x^3$ as desired.

Answer 3

Imagine a cylinder with height $h$ and radius $r$. By definition, the formula for the volume of a cylinder is $$V= \pi r^2 h $$ Notice that the formula has nothing to do with "multiplying by $360^\circ$". The formula should make intuitive sense. The $\pi r^2$ that appears in $V$ is the cross-sectional area of the cylinder, so multiplying that quantity by $h$ will "stretch" that cross-section into the $3$-d volume we recognize as a cylinder.

Now imagine a square $S$ with base and height of length $s$. If you rotate $S$ by $360^\circ$ about one of its edges, then you will indeed generate a cylinder. Convince yourself that $h=s$ and $r=s$. Hence the volume of this cylinder must be $$V_S = \pi(s^2)s = \pi s^3$$

Answer 4

If I understand the intuition behind the question, the idea is that the square swept out a certain volume through space, so if you multiply the area of the square times the distance the square "traveled" then you will have the volume that was swept out.

This certainly works if the square moves along a straight line perpendicular to its plane, sweeping out a square prism.

What happens in this case, however, is that one edge of the square of side $s$ travels a distance $2\pi s$, while the opposite edge does not travel at all. So while it would be accurate to say that a small region of the square very near the fast-moving edge sweeps out a volume that is (almost) $2\pi s$ times the area of the region, most parts of the square sweep out far smaller volumes relative to their areas.

One could formulate this by setting up an integral using the shell method, each shell being generated by a thin strip of the square parallel to the axis of rotation.

One could say (very informally) that the "average" distance traveled is the distance traveled by a point at the center of the square, $2\pi\left(\frac12 s\right) = \pi s$, and it turns out that the true volume of the cylinder, $\pi s^3$, is equal to $s^2 \times \pi s$. But you should regard this as a happy coincidence rather than as a means of computing the volume, because in general "average distance traveled" is a slippery concept. (In fact I chose the "average distance" in this case specifically so that it would match the formula for volume, in other words, the average is a consequence of the volume of the cylinder rather than the other way around.)