Why is arc length not in the formula for the volume of a solid of revolution?

Solution 1:

One way to gain the intuition behind this is to look at what happens in 2 dimensions. Here, rather than surface area and volume, we look at arc length and area under the curve. When we want to find the area under the curve, we estimate using rectangles. This is sufficient to get the area in a limit; one way to see why this is so is that both the error and the estimate are 2-dimensional, and so we aren't missing any extra information.

However, the analogous approach to approximating arc length is obviously bad: this would amount to approximating the curve by a sequence of constant steps (i.e. the top of the rectangles in a Riemann sum) and the length of this approximation is always just the length of the domain. Essentially, we are using a 1-dimensional approximation (i.e. only depending on $x$) for a 2-dimensional object (the curve), and so our approximation isn't taking into account the extra length coming from the other dimension. This is why the arc length is computed using a polygonal approximation by secants to the curve; this approximation incorporates both change in $x$ and change in $y$.

Why is this relevant to solids of revolution? Well, in essence, the volume and surface area formulae are obtained by simply rotating the corresponding 2-dimensional approximation rotated around an axis, and taking a limit. If it wouldn't work in 2 dimensions, it certainly won't work in 3 dimensions.

Solution 2:

Consider a cone. It is a rotated line segment $f(y)=\dfrac{r}{h}y$ (leaving it without a base). We know its volume (if it had a base) will be $\dfrac{\pi}{3}r^2h$, and its surface area (without the base) will be $\pi r\ell$.

To find the volume, we take the integral of the function of the areas of concentric circles. Essentially, this is like breaking up the cone into small quasi-cylinders, then taking the limit as they become infinitesimal.

Looking at a finite one of these, the volume is $\pi R^2\,\mathrm dy$. Notice that if there are $n$ of these slices, $n\,\mathrm dy$ will be $h$, the height of the cone, as we expect.

We have no problem integrating $\displaystyle\int_0^h\pi\left(\frac{r}{h}y\right)^2\,\mathrm dy$ to get $\dfrac{\pi}{3}r^2h$, and we don't have to do arc length.

Now the problem comes when we try to find the surface area of the cone with integration. What you were doing was taking the integral of the function of circumferences of concentric circles. This seems analogous to the area of the circles last time, but it's not.

You have to realize that like last time, we are looking at that slice of the cone, but this time we're adding up the surface area. It's an approximate rectangle, so the area is $2\pi R\,\mathrm dy$. It's just like what you were doing.

The problem arises when you see that with $n$ slices, $n\,\mathrm dy$ should be equal to the slant height $\ell$, not the height $h$ like last time. The slant height is just the arc length of the line segment. $\mathrm dy=\dfrac{\sqrt{1+f'(y)^2}}{n}$ instead of $\dfrac{1}{n}$ last time.

So you just take the integral $\displaystyle\int_0^h2\pi\frac{r}{h}y\,\mathrm dy$, and you replace $\mathrm dy$ with $\sqrt{1+f'(y)^2}\,\mathrm dy$. Then you integrate $\displaystyle\int_0^h2\pi\frac{r}{h}y\sqrt{1+\left(\frac{r}{h}\right)^2}\,\mathrm dy$ and get $\pi r\sqrt{h^2+r^2}$, the right answer.

Essentially, that slice of the cone is not a cylinder. You can pretend it is when integrating for volume, since the shape of the outside does not matter to you then, but you can't ignore it for surface area.