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

Question

I have difficulty understanding the difference between calculating the volume of a solid of revolution, and the surface area of a solid of revolution.

When calculating the volume, using disc integration, I've been taught to reason in the following way: divide the solid into several cylinders of infinitesimal width $dx$. Then each such cylinder has radius $f(x)$ , so its volume is $\pi\cdot f(x)^2~dx$ . Hence the total volume of the solid of revolution is $V=\pi\int_a^b f(x)^2~dx$ .

So when calculating the surface area, I figure I should be able to reason thus: as before, divide the solid into several cylinders, except now calculate the area of each one instead of its volume. This should yield $A=2\pi\int_a^bf(x)~dx$. However, this is the wrong answer. In order to get the proper formula, I need to replace $dx$ with the arc length, which is $\sqrt{1+f'(x)^2}~dx$ .

My question is: why is this the case? There's no need to use arc length when calculating volume, so why does it show up when calculating area?

Answer 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.

Answer 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.

Answer 3

The reason is that the width of the "cylinder" is sensible to both changes in $x$ and in $y$. The cylinder is "slanted": it is more like a slice of a cone in the direction perpendicular to the axis of rotation than like a slice of a cylinder.

Consider for instance a curve which oscillates extremely rapidly between two small values. Your method does not work because it accounts for none of the surface area which is (almost) all in the vertical direction.

When calculating a volume, all of this becomes negligible with respect to the volume of the cylinder.