Why is the slope a part of the area of a surface of revolution?

Question

I have previously learned about the volume of a solid of revolution about the x axis and that equation makes sense to me since it's taking the integral between points a and b of the area of a circle at each point.

$$ \int_{a}^{b} \pi [f(x)]²dx = \int_{a}^{b}Area(x)dx $$

But now when I've come across the formula for the area of a solid of revolution I can't quite understand why the slope factor is there. From looking at the volume of a solid of revolution I would assume counting the surface area is just the same as counting the volume except the area is replaced by the circumference. And looking at the equation that seems to be true, but the slope of the function on a small interval has also been added as a factor.

$$ \int_{a}^{b} 2\pi f(x)\sqrt{1+[f'(x)]²}dx $$

From the picture of it in the book it takes the arch length $$ ds = \sqrt{dx²+dy²} $$ of a small dx. But taking this ds and the integral from a to b aren't we counting every part twice since the integral already goes over every dx?

Answer 1

Note that $$ds = \sqrt{(dx)²+(dy)²} = \sqrt{1+\frac {(dy)²}{(dx)^2}}dx $$

We integrate $$\int_{a}^{b} 2\pi f(x)ds =\int_{a}^{b} 2\pi f(x)\sqrt{1+[f'(x)]²}dx$$

Thus in the integral there is only one copy of $dx$

Answer 2

If you integrate the circumference with respect to $x$, you are in essence finding a sort of $x$-component of the surface area. When you integrate the circumference, you should so it in the direction of the curve so you are moving with the curve. The correct integral, if you are starting at $x=a$ and ending at $x=b$ is

$$ \int_{s(a)}^{s(b)} 2 \pi f(x) ds $$

Where the variable $s$ goes along the curve (think of the value of $s$ as the length of the curve from $x=0$ to your current $x$-value). Hence if we want to make everything in terms of $x$, we must use the relation that the length of the curve has with $x$:

$$ ds = \sqrt{dx^2+dy^2} $$

Substituting in gives,

$$ \int_{s(a)}^{s(b)} 2 \pi f(x) ds = \int_{s(a)}^{s(b)} 2 \pi f(x) \sqrt{dx^2+dy^2}=\int_{a}^{b} 2 \pi f(x) \sqrt{1+(\dfrac{dy}{dx})^2} dx, $$

Answer 3

Consider taking the area of an ordinary well-behaved function $f(x)$, as shown below. As is usual, we consider the surface area of an infinitesimally thin slice of the surface of revolution.

This surface is equivalent in area to—i.e., can be rolled out into—a trapezoid whose width is $2\pi f(x)$.

And what is the height of this trapezoid? It is $ds$, where $ds^2 = dx^2 + dy^2$, by the Pythagorean theorem.

Note that the slope of the curve has a negligible effect on the volume of the infinitesimal slice; the difference is essentially the volume of a thin ring around the slice (with a wedge-shaped cross section). However, the effect on the area of the slice is substantial. It essentially acts to "boost" the height of the trapezoid. That is why $\frac{dy}{dx}$ shows up in the integrand for area, but not for volume.

From here on out, the rest is simple algebra. We integrate

\begin{align} \int_{x=a}^b 2 \pi f(x) \, ds & = \int_{x=a}^b 2 \pi f(x) \sqrt{ds^2} \\ & = \int_{x=a}^b 2 \pi f(x) \sqrt{dx^2+dy^2} \\ & = \int_{x=a}^b 2 \pi f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \end{align}

It might seem as though we have to take the slope into account when identifying the width (the base) of the equivalent trapezoid, but that is not so, because the difference in the top and bottom sides of that trapezoid (and therefore the impact on the area) are infinitesimally different.

Answer 4

Let's step back from the calculus for a little while and just do some plain old geometry.

Here is a truncated cone:

On the top of the truncated cone is a circular disk of radius $r_1,$ on the bottom is a circular disk of radius $r_2,$ and the height of the truncated cone (that is, the distance between the top and bottom disks) is $h.$

The volume of this shape is $\frac13 \pi h (r_1^2 + r_1 r_2 + r_2^2).$ The lateral area (the area of the darker blue part of the surface that connects the two disks) is $\pi (r_1 + r_2)\sqrt{(r_2 - r_1)^2 + h^2}.$ This does not include the area of the two disks.

The slope of the side is $m = \dfrac h{r_2 - r_1}.$ It follows that $r_1 = r_2 - \dfrac hm.$

Now let's make the slope be some simple value; suppose the slope is $m = 1.$ Further, suppose $h$ is very, very small compared with $r_1$ and $r_2,$ unlike the picture. For example, let's try $r_2 = 20$ and $h = 0.001.$ Therefore $r_1 = 19.999.$

The volume of this example of a truncated cone is $$\frac13 \pi h (r_1^2 + r_1 r_2 + r_2^2) = \frac13 \pi \times 0.001 (19.999^2 + 19.999\times 20 + 20^2) \approx 1.25657.$$

The area is $$\pi (r_1 + r_2)\sqrt{(r_2 - r_1)^2 + h^2} = \pi (19.999 + 20)\sqrt{(20 - 19.999)^2 + 0.001^2} \approx 0.177711.$$

Now let's suppose we had a different slope, for example if the side went straight up. That is, suppose we had a cylinder.

A cylinder has only one radius, not two. But let's try the smaller radius, $r_1 = 19.999.$ That is, take a cylinder of radius $19.999$ and height $0.001.$ Then its volume is $$\pi r^2 h = \pi 19.999^2\times0.001 \approx 1.25651$$ and its lateral area (the area of the curved part between the two end disks, not including the area of the disks) is $$2\pi rh = 2\pi \times 19.999 \times 0.001 \approx 0.125657.$$

Let's see how these compare in relation to the truncated cone. The ratio of the volumes is approximately $$ \frac{1.25651}{1.25657} \approx 0.999952. $$ Not bad. If we mistakenly used the volume of the cylinder when we meant to take the volume of the truncated cone, we'd only be off by a small fraction of a percent.

The ratio of the areas, on the other hand, is approximately $$ \frac{0.125657}{0.177711} \approx 0.707086. $$ Now that is bad. If we used the cylinder instead of the truncated cone, we would underestimate the lateral surface area by more than $29\%.$

But what if we used the larger radius $r_2 = 20$ instead? Then we would have a cylinder of radius $20$ and height $0.001,$ whose volume is $$\pi r^2 h = \pi 20^2 \times 0.001 \approx 1.25664$$ and whose lateral area is $$2\pi rh = 2\pi \times 20 \times 0.001 \approx 0.125664.$$

The ratio of the volumes is approximately $$ \frac{1.25664}{1.25657} \approx 1.000056 $$ while the ratio of the areas is approximately $$ \frac{0.125664}{0.177711} \approx 0.707125. $$

Again, a tiny fraction of a percent error if we use the cylinder instead of the truncated cone--that is, if we don't use the correct slope of the lateral side--to measure volume, and worse than a $29\%$ error when we measure the lateral area.

In the calculus, of course, when we integrate a solid of revolution we don't just make each slice have a thickness of $h = 0.001.$ Instead, we take a limit as we cut the slices finer and finer so $h$ approaches zero. You can try some examples to see what happens if $h$ gets much smaller, for example $r_2 = 20,$ $h = 0.000001,$ $r_1 = 19.999999.$ The errors you get in the volume by using a cylinder instead of the truncate cone get even tinier; the error in the surface area remains just about as bad. Slice it as thin as you like, the error in the surface area will still be worse than $29\%.$ It's hopeless.

Now consider when we get a solid of revolution by stacking up a lot of truncated cones like this. The cones don't all have to have the same slope, but let's suppose they all have some significant slope, that is, the solid isn't just a cylinder. Then in each slice of the solid of revolution, if we ignore the slope of the side (treating the slice like a cylinder), we get a tiny error in the volume but a relatively huge error in the surface area.

And that's why we care so much about the slope when computing area, but ignore it when computing volume.