Geometrically speaking, it seems to me that if you have for example $y^2=8x$ revolved around the x-axis, taking the limit of the sum of $n$ surfaces of cylinders as $n$ approaches infinity should give you the surface area of that surface of revolution.

This is how the author initially derives the formula for finding the volume of solids of revolution. Take a rectangle under the curve over $\Delta x$ and revolve it around the axis to get an approximation of the volume of the solid over that interval. Add up those rectangles over $n$ changes in $x$ and take the limit as $n$ approaches infinity, which is the integral of the function that gives you the $y$ value (radius of that approximating cylinder) for each $x$ value.

Following the same principle, why wouldn't we be able to take those same cylinders, but instead of taking their volume, taking their surface area and take the limit as the number of those cylinders approaches zero?

In other words, in this case each $y$ value is given by $y = \sqrt{8x}$, which is the radius of that cylinder of height $\Delta x$ and an approximation of the surface area over that interval.

Why doesn't that work? Why do we need to deal with arc length? I don't understand why it doesn't work in this case, it seems to me that you're still getting a better and better approximation of surface area as those cylinders get smaller and smaller, eventually getting the exact surface area with the limit as their number goes to infinity.

PS: I saw this Surface area of a solid of revolution: Why does not $ \int_{b}^{a} 2\pi \,f(x) \,dx $ work? but it's still not making sense visually/geometrically.


5 Answers 5


We have an option to cut the solid of revolution (obtained by revolution of $y = f(x)$ between $x = a$ and $x = b$) into multiple slices in the following manner:

  • each slice is a cylinder
  • each slice is a section of cone cut by two parallel planes (a frustum of a cone)

Let the desired slicing be done via partition $$P = \{x_{0}, x_{1}, x_{2}, \dots, x_{n}\}$$ of $[a, b]$. We will apply both the approaches mentioned earlier to calculate the surface area as well as volume of the solid of revolution.

First we deal with volume which has an easier analysis. If we slice the solid as cylinders then the approximation of volume is given by $$V(P) = \pi\sum_{i = 1}^{n}\{f(x_{i})\}^{2}(x_{i} - x_{i - 1})\tag{1}$$ which is a Riemann sum for the integral $\pi\int_{a}^{b}\{f(x)\}^{2}\,dx$ and this is the desired volume.

If we slice the solid into frustums of cone we get the approximation of volume as $$V(P) = \frac{\pi}{3}\sum_{i = 1}^{n}\left[\{f(x_{i - 1})\}^{2} + f(x_{i - 1})f(x_{i}) + \{f(x_{i})\}^{2}\right](x_{i} - x_{i -1})\tag{2}$$ which is split into 3 terms and each term is a Riemann sum for $(\pi/3)\int_{a}^{b}\{f(x)\}^{2}\,dx$ so that the desired volume is again $\pi\int_{a}^{b}\{f(x)\}^{2}\,dx$

Let's now come to surface area of the solid of revolution. If we slice the solid into cylinders then the surface area is approximated by $$S(P) = 2\pi\sum_{i = 1}^{n}f(x_{i})(x_{i} - x_{i - 1})\tag{3}$$ which tends to $2\pi\int_{a}^{b}f(x)\, dx$.

If we slice the solid into frustums we get the approximation for surface area as $$S(P) = \pi\sum_{i = 1}^{n}\{f(x_{i - 1}) + f(x_{i})\}\sqrt{(x_{i} - x_{i - 1})^{2} + ((f(x_{i}) - f(x_{i - 1}))^{2}}\tag{4}$$ which can be simplified by the use of mean value theorem as $$S(P) = \pi\sum_{i = 1}^{n}\{f(x_{i - 1}) + f(x_{i})\}\sqrt{1 + \{f'(t_{i})\}^{2}}\cdot(x_{i} - x_{i - 1})\tag{5}$$ for some points $t_{i} \in (x_{i - 1}, x_{i})$. This can be split into two sums each of which is a Riemann sum for $\pi\int_{a}^{b}f(x)\sqrt{1 + \{f'(x)\}^{2}}\,dx$ so that the desired surface area is $2\pi\int_{a}^{b}f(x)\sqrt{1 + \{f'(x)\}^{2}}\,dx$.

We see that in case of volume both the approaches give the same answer. But in case of surface area the answers obtained by both the methods are different. Further note that out of the two answers we can easily verify which one is correct by using $y = x, a = 0, b = 1$ so that the solid of revolution is a circular cone. This verification shows that the technique used in equation $(4), (5)$ gives the correct surface area.

The question which OP is asking is this:

Why do both the approaches (using cylinders and frustums) give the same result for volume but different results for surface area?

The reason is simple. Both the sums in $(3)$ and $(4)$ are trying to approximate the surface area of the solid, but there is a huge difference between them namely $$\Delta = 2\pi\sum_{i = 1}^{n}f(x_{i})\left[\sqrt{1 + \{f'(t_{i})\}^{2}} - 1\right](x_{i} - x_{i - 1})$$ and this itself is a non-zero sum unless $f'(x)$ is identically zero. So the approximation $(4)$ is trying to take into account some additional surface area which is left out by sum $(3)$ and this additional part is significant unless $f'(x) = 0$ identically. Hence $(4)$ is a better and correct approximation.

In case of volume both the approximations $(1), (2)$ are Riemann sums for the same integral (but are expressed in slightly different ways).

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    $\begingroup$ Thank you, you identified exactly what I was really asking and explained perfectly step by step; this is extremely helpful. $\endgroup$ Commented Mar 11, 2016 at 9:39
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    $\begingroup$ @jeremyradcliff: Glad to know that I was helpful in some way. $\endgroup$
    – Paramanand Singh
    Commented Mar 11, 2016 at 11:10
  • $\begingroup$ Perfect. This clears it for me. $\endgroup$
    – R004
    Commented Jul 7, 2017 at 8:43
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    $\begingroup$ I guess I can intuit this. In the case of volumes, the difference in the differential volumes( of cylinder and frustum ) is very less when compared with each one's differential volume. So, when the differentials are stacked continuously, the entire volume masks( so to speak ) the integrated error. In the case of surfaces, however, the difference in the differentials is comparable to each surface if you imagine it. When we stack these differentials, we see that stacking differential frustum surfaces gets us closest to the true volume. I can imagine this and I think it is correct. $\endgroup$
    – R004
    Commented Jul 7, 2017 at 9:37
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    $\begingroup$ @R004: I think you have very well summarized my rigorous explanation into an intuitive form which may be easier to grasp for many people. Thanks for your comment. $\endgroup$
    – Paramanand Singh
    Commented Jul 7, 2017 at 14:13

In your link, it explains precisely why integrating $y$ doesn't work: Because the length $ds$ of a little piece of the arc is not $dx$, but $\sqrt{(dx)^2+(dy)^2}$, which can be written as $${\sqrt{(dx)^2+(dy)^2}\over dx}\cdot dx = \sqrt{1 + \left(dy\over dx\right)^2} \cdot dx.$$ It's the same reasoning why the length of a hypotenuse is $\sqrt{a^2+b^2}$ and not $a+b$.

Or this "proof" that $pi=4$: Is value of $\pi = 4$?

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    $\begingroup$ Yes, but why does it work for volume, then? That's what I didn't understand in the link. Shouldn't you run into the same problem when integrating to find the volume? What's the difference? $\endgroup$ Commented Mar 10, 2016 at 4:51
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    $\begingroup$ The integral $\int_a^b f(x)\,dx$ is based on dividing an interval into pieces and multiplying the width of each interval by the value of $f$ at some value in that interval, and adding the values together. This is not an exact answer; there is some error in the approximation. The error analysis for arc length turns out to be different from the error analysis for volume. $\endgroup$ Commented Mar 10, 2016 at 4:56
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    $\begingroup$ There is some error in the approximation, but that error goes to zero when taking the limit as the number of those pieces of the interval goes to infinity, at least in the case of volume. Somehow though, this doesn't happen with surface area; I guess as you said the nature of the error is such that the error doesn't go to zero as $n$ goes to infinity. But why is that the case? What is different in the error analysis between volume and surface area? Visually at least, I just can't "see" the difference. $\endgroup$ Commented Mar 10, 2016 at 5:00
  • $\begingroup$ This is the sort of thing covered in Real Analysis. I took it a long time ago and became a combinatorialist, so I don't remember everything from it. Sorry; someone else will have to pick up the thread here. $\endgroup$ Commented Mar 10, 2016 at 5:05
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    $\begingroup$ @jeremyradcliff Here's a question about a similar problem. Why is $\pi\neq 4$? Just because you're "approximating" your surfaces in 3D doesn't mean that you approximate their area. In 2D, approximating lines doesn't guarantee that you're approximating their lengths. $\endgroup$
    – Kitegi
    Commented Mar 10, 2016 at 5:45

I cannot post a comment yet, but I will try to pick up where Carl Heckman left. The problem with the 'rectangle' approximation in this case, is the same as the idea of $\pi = 4$.

Consider the line $y=a x$, for example. Let $a$ become very large, and look at the surface area of the surface of revolution from $x=0$ to $x=1/a$. This should approach the area of a disc in the limit of $a \rightarrow \infty$, since we have a cone where the top angle will become very flat. However, if we were to put cylinders under this line and simply look at the surface area of the boundary of the cylinders, it is clear that we will not get the right answer, since this area will go to $0$ in this limit (the sum of the widths of the cylinders goes to zero).

We see that we somehow have to take into account the sides of the cylinders (perpendicular to the x-axis) as well. If we try to do that, however, we run into the problem Carl Heckman described. For example, look at the surface of revolution of a small part of the line around a point (not in the limit $a \rightarrow \infty$). $y$ does not vary much in a small enough area, so adding the sides of the cylinders as well, we would find that the surface area would be $$2\pi y(\Delta x+\Delta y) = 2\pi y|a+1| \Delta x$$

The error in this case is similar as to when you are trying to find the length of a hypotenuse of a right angle, by adding the two other lengths. Note also that $\sqrt{1+(y')^2} \Delta x$ is the length of the hypotenuse of a very small right triangle, which appears to be the factor we need to get the right surface area.

I hope I managed to make it slightly clearer why multiplying by $\sqrt{1+(y')^2}$ is the right thing to do without rambling on too much.



Why is the formula for cone $ \pi r l $ ( l slant height) and not $ \pi \bar r h$ ( h is cone height)?... for the same reason.


The answers above are sufficient I think. However, I just wanted to add this as another viewpoint, based on a traveling particle, that some may like, and some informal derivations using infinitesimals.

Imagine a particle traveling over time $t$ along the curve: $$(x(t),y(t))$$ Where we have: $$y=f(x)$$ or $$y(t)=f(x(t))$$

The velocity vector of the particle is: $$(\frac{dx}{dt},\frac{dy}{dt})$$ and the speed of the particle: $$v(t)=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}$$ so that the infinitesimal quantity $$dl=v(t)dt=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} \ dt$$ is speed multiplied by infinitesimal time and thus an infinitesimal length.

We can write: $$\int_a^b f(x) \sqrt{((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)(dt)^2}=$$ $$\int_a^b f(x) \sqrt{dx^2+dy^2}=$$ $$\int_a^b f(x) dl$$

Then, just as in the answer by Christopher Carl Heckman, we have (not rigorously): $$\int_a^b f(x) \sqrt{dx^2+dy^2}=$$ $$\int_a^b f(x) \sqrt{dx^2+dy^2}\frac{dx}{dx}=$$ $$\int_a^b f(x) \sqrt{\frac{dx^2+dy^2}{dx^2}}dx=$$ $$\int_a^b f(x) \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}dx=$$ The "average" radius of the frustum is: $$r=y=f(x)$$ and the circumference $$c=2 \pi r=2 \pi y = 2 \pi f(x)$$ so that the surface area becomes: $$S=\int_a^b 2 \pi f(x) \sqrt{1+\bigg(\frac{dy}{dx}\bigg)^2}dx=$$ $$\int_a^b 2 \pi y \sqrt{1+y'^2}dx$$ This makes it pretty clear that just having $$y dx=y\sqrt{1+0} dx$$ instead of $$y \sqrt{1+y'^2}dx$$ ignores the y-component of the velocity of the particle. In other words, if we are not careful then we are just measuring the distance traveled in the $x$-direction instead of the "Euclidian" distance.

The volume is a second order affect as can be seen by starting with the infinitesimal volume of a conical frustum between $y(x)$ and $y(x+dx)$: $$dV=\frac{1}{3}\pi h(R_1^2+R_1R_2+R_2^2)=\frac{1}{3}\pi dx((y(x))^2+(y(x)y(x+dx))+(y(x+dx))^2)=$$ $$\frac{1}{3}\pi dx((y(x)^2+(y(x)(y(x)+y'(x)dx+...)+(y(x)+y'(x)dx+...)^2\approx$$ $$\frac{1}{3}\pi dx(y^2+y^2+y y' dx + y^2 + 2y y' dx + y'^2)=$$ $$\frac{1}{3}\pi dx(3y^2+3y y'dx+y'^2dx^2)=$$ $$\frac{1}{3}\pi dx(3y^2+ higher order terms )\approx$$ $$\pi y^2 dx$$

Finally, the volume is $$V=\int_a^b \pi y^2 dx$$


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