# Why is area of a surface of revolution integral $2\pi y~ds$? not '$dx$'? [duplicate]

For me, intuitively, integral $2\pi y~dx$ make more sense. I know intuition can not be proof, but by far, most part of math I've learned does match with my intuition. So, I think this one should 'make sense' as well. Probably I didn't understand the way surface area is measured. It will be great if any one could tell me how 'integral $2\pi y~dx$' is wrong. (By the way, how to use mathematical symbols in texts?)

• latex-project.org for adding symbols in text. – user223391 Apr 12 '15 at 22:15
• What is ${}{}s$? – GFauxPas Apr 12 '15 at 22:18
• Thanks for the guidelines. s hereby refers to arc length funciton. – Jinyoung Sung Apr 12 '15 at 22:46
• @JinyoungSung: Welcome to Math.SE! This question has an answer that may address your uncertainty. – Andrew D. Hwang Apr 12 '15 at 22:57

Here's some intuition:

Draw a diagram of some element of a graph of $y = f(x)$ you are rotating. If the line element described by $dx$ or $ds$ is not parallel to the $x$ axis, then notice that $ds > dx$; that is, the length of the line element is longer than just $dx$ because there is a component in the $dy$ direction; i.e., $ds^2 = dx^2 + dy^2$.

So if were to calculate the integral

$$\int_a^b 2\pi f(x) \ dx$$

that would be a lower bound on the actual surface area. For continuous $f$ on the interval $[a,b]$, this integral is equal to the actual surface area if and only if $f$ is a constant. That is, if we were calculating the surface area of the curved part of a cylinder. If $f(x) = r > 0$, then the surface are is indeed

$$\int_a^b 2\pi f(x) \ dx = \int_a^b 2\pi r \ dx = 2\pi rL \ \ \ \text{ where } L = b - a$$

For every other $f$, we need $ds$:

$$\int 2\pi f(x) \ ds$$

As $\displaystyle \left( { ds \over dx } \right)^2 = 1 + \left( { dy \over dx } \right)^2$, this last integral is normally written as

$$\int_a^b 2\pi f(x) \ \sqrt{1 + \left( { dy \over dx } \right)^2} \ dx$$

• Thank you very much for replying. In fact, this is my first time to use this forum. – Jinyoung Sung Apr 12 '15 at 22:33
• Yeah, I do understand that ds>dx, but I still don't get why ds is used. I think using ds is liking intentionally streching the curved line and multiply the f(x) which is actually dependent on x, so it doesn't make sense. – Jinyoung Sung Apr 12 '15 at 22:35
• You say only when f(x)=c integral 2*pi*y dx is correct, but I think(intuitively) the integral makes sense even when f(x) is not a constant... – Jinyoung Sung Apr 12 '15 at 22:36
• If that bothers you, were can rewrite to surface area integral as $$\int_{s_1}^{s_2} 2\pi f(x(s)) \ ds$$ The important conceptual point about using $ds$ is that we remember the length of the line segment we are rotating has that length and not length $dx$. That is why writing $\int_a^b 2\pi f(x) \ dx$ for surface area of rotation is wrong in general. If you are not convinced about that point, do a simple calculation, a cone $f(x) = mx$ on $x \in [0,1]$ and see how one integral gives a good answer and the other rubbish. – Simon S Apr 12 '15 at 22:38
• For example, when calculating the surface area of y=x^(1/2) from 1<=x<=4. We could approximate the surface area by slicing it into 4 pieces with same interval. When we could take right end as we do in approximating area and volume and arc length. Then right end are (1,1), (2,sqrt(2)),(3,sqrt(3)),(4,2). And I think when we slice the revolution into infinitely many pieces, we get the surface area... – Jinyoung Sung Apr 12 '15 at 22:44

OP, if I'm guessing right, the intuition point that you are puzzling over is "how come we can get away with integrating $\pi y^2 dx$ for volume (there too, we are approximating an infinitesimal frustrum by an infinitesimal cylinder) but we are not able to approximate the surface area of the infinitesimal frustrum $2 \pi y ds$ by that of the infinitesimal cylinder $2 \pi y dx$?"

To understand the difference consider the error of the approximation in each case as a fraction of the thing you are trying to calculate. The difference arises when the function is not a constant ($y' \neq 0$). In the case of volume, that error is negligible (because $\pi y^2 dx$ does capture most of the volume) while in the case of the surface area calculation, that error is of the same order as the surface area itself.

To be more convinced, you can use the formula for the volume of that frustrum of thickness $dx$. It is $$dV=\pi y^2 dx + \pi y dy dx + \pi dy^2 dx$$ the error is $$dE=dV-\pi y^2 dx= \pi y dy dx + \pi dy^2 dx$$ which is tiny in relation to $dV$ (will vanish as we take $dx$ to zero and integrate) in relation to the true V.

The error in the surface area approximation on the other hand, is $$d\epsilon= dA - 2 \pi y dx = 2 \pi y (ds- dx) = 2 \pi y ds (1 - \frac{1}{\sqrt{1+y'^2}})$$which is of the same order as $dA = 2 \pi y ds$ and upon taking the limit of dx to 0 to compute the integral, this results in a deviation from the true surface area.

A bit late but either way...

The surface area depends on $ds$ because $ds$ runs on the surface, unlike $dx$ which runs through the shape.

Consider two points A and B (not mathematical points, just locations), which are separated by a mountain. Person X simply goes through the mountain tunnel to get from A to B. Person S climbs up the mountain, then climbs down the other side of the mountain to get to point B. It's the same intuition behind $ds$ instead of $dx$.

When calculating volume, we want the interior of the shape. $dx$ gives us just that, since it pierces through the shape, just as person X passes through the mountain. However, for surface area, we want the exterior of the shape, so we need to use $ds$, just as person S has to walk on the mountain.

Here are two ways to measure the surface area of an open cone of base radius $r$ and height $h$ (just the area of the curved surface of the cone; we are not including the area of the flat circular base of the cone).

The first way is, you cut the surface of the cone along a straight line from the apex to the base; then you take the surface of the cone as if it were a sheet of paper and lay it flat on a plane.

You can see for yourself how this works by cutting a circular disk out of a sheet of paper; then draw two radii on that disk and cut along each radius. You will then have cut your disk into two pieces. If you then take either piece of paper and join the two straight edges together, it will make a cone. If you let the edges separate and flatten the paper out again, it returns to its part-of-a-disk shape.

So when we cut and flatten the surface of the cone, it forms part of a disk whose radius is the distance from the apex to the base circle. That distance is $\sqrt{r^2 + h^2}.$ The length of the curved arc is the circumference of the base circle of the cone, which is $2\pi r.$ The area of this part of the disk is half the radius times the length of the arc, namely, $$A_1 = \frac12 \sqrt{r^2 + h^2} \left( 2\pi r \right) = \pi r \sqrt{r^2 + h^2}.$$

A second way is your way: make a stack of disks inside the cone and measure the curved surface areas on the outer edges of the disks. The radius of a disk can be expressed as a function of its distance from the apex; call this distance $y$, and then the disk's radius is $ry/h.$ The curved area is $2\pi(ry/h)(\Delta y),$ where $\Delta y$ is the thickness of the disk. If we take the limit as the thickness of the disks approaches zero, we get the integral $$A_2 = \int_0^h \frac{2\pi ry}{h} dy = \frac{2\pi r}{h} \int_0^h y\,dy = \frac{2\pi r}{h} \left( \frac12 h^2\right) = \pi rh.$$ As you should be able to see, $A_2 < A_1.$ (Try plugging in some numbers such as $r=h=1$ if you doubt it.)

To my intuition, rolling parts of paper disks into cones and unrolling them is a far more convincing way to establish the area of a cone than stacking up a bunch of very thin disks. After all, the cone is one continuous surface, which is preserved (except the connection to itself along one cut) when we unroll it, whereas the curved parts of your disks are not connected to one another at all. I conclude that $A_1$ is the correct area and $A_2$ is not.

This is the difference between the correct formula for area and your formula: the correct formula slices the solid of rotation into thin slices, and approximates each slice by a frustum (a horizontal slice of a cone). If the slices are thin enough, the curved surface of the frustum is very, very close to of its part of the surface area of the solid of revolution, whereas with disks we will always have grooves in the surface (unless the solid was a cylinder to begin with). If we only buy enough paint to paint the curved surface of each disk, it won't be enough paint to connect to the next disk below it, even if we smooth out the grooves. If we can paint the frustums, however, we have enough paint.