After class on Monday, Cyrus asked me a great question.
You know how volume is the integral of cross-sectional area? Why isn't surface area the integral of cross-sectional circumference?
Let's talk about this. Suppose is a function on , and (just to be safe), suppose that is differentiable on and its derivative is continuous on , too. Use the graph of to create a solid of revolution by revolving the graph around the -axis. In class, we derived an integral formula for the surface area, , of this solid:
We arrived at this formula using a cut-up-and-add-up approach using conical frustrums to approximate a slice of the surface. Summing the surface areas of those frustrums, applying the Mean Value Theorem, and taking a limit led to the above definite integral.
Below you can see a SageMath representation of how a conical frustrum approximates a curving surface on a slice of the domain of the function that generates the solid.
Cyrus's question may appear to be inspired by what we learned about volume and is a natural extension of that knowledge. Or it may appear to be inspired by noticing when is small, the frustrum is basically a circle, so it 'appears' that the surface area of the frustrum is essentially the circumference. Or it may appear to be inspired by wondering if we could use cylinders to approximate the surface of a slice of the solid and get a simpler formula for surface area. Regardless of how the very reasonable question arose, it's a question that's worth exporing
If it's true that the surface area is just the integral of cross-sectional circumferences, then we could write
There are two problems with this.
The first problem with this approach reveals itself when look at the definite integral as representing the area between the -axis and the graph of the integrand. If the above integral were the surface area of a solid of revolution, then that surface area could be written as
which is a constant multiple of the area under the curve on . It would be remarkable if the surface area of a solid of revolution were equal to a multiple of the area betweent he axis of rotation and the generating curve! Unfortunately, it's not true. To see why, let's do a thought experiment that starts easy and then gets complicated.
Consider and on . The plots of these are shown below with in blue and in red. By inspection, it is clear to anyone who knows Calculus that
It follows that any positive constant multiple of the integrals also obeys the inequality. If it were true that the surface area of a solid of revolution were a constant multiple of the area under its generating curve, then the surface area of the solid of revolution generated by on would be greater than the surface area of the solid of revolution generated by on for any whose graph sat below the graph of on .
This certainly seems true for the functions and that we've graphed. We haven't thought much about surface areas of three dimensional solids, so our intuition about this is weak. To build our intuition about surface areas, consider this statement:
the closer a part of the graph is to the solid's axis of rotation, the less surface area that part of the graph generates.
Even though the arc length of on is greater than that of , because the graph gets closer to the -axis faster than , when rotated about the -axis, its graph generates less surface area than the nearest part of . We can modify slightly to 'fix' this. That is, we can modify to increase the surface area of the solid it generates. We'll do this by taking a page out of Nature's playbook.
We're going to build an example function whose graph lies above the -axis andbelow the graph of , and that generates a solid of revolution whose surface area is clearly greater than the surface area of the solid generated by .
When Nature wants to make something bigger but has limited resources or space to work in, it braches or bends. For example, the human cortex (or grey matter) is a sheet of tissue consisting of the neurons that allow a human to think, and the amount of grey matter in a human brain depends on the size of the sheet of tissue that can fit near the top fo the skull. To get more of that sheet in there, Nature decided to fold it intricately so the surface area of the sheet is huge even though the volume in the skull in small. We see similar phenomena in the human intestines (more surface area means more efficient digestion), lungs (more surface area means more efficient oxygen transport into the blood), and nalas passages (more surface area means a greater chance the mucous membranes grab garbage from inhales air before it gets to the lungs).
So let's add ripples to so the surface area of the solid it generates gets bigger.
Our favorite way to add ripples to a function is by adding one a sinusoidal function. Consider , a sinusoidal function with period . A slight modification of this function give which has period . This means that will have equally spaced maxima and minima on .
Define the function Like , this function passes through and , and will, on average, decrease between these points, having oscillations along the way.
Below is a plot of both and with . The plot shows that isn't the function we're looking for because it does not satisfy for all in . We need to get the oscillations 'under control' so the red graph never crosses the blue graph.
We want to limiting the amplitude of the oscillations so they are small near and and not so big in between that the graphs cross. We'll do this by multiplying by the function built from pictured below.
We will narrow the function so its values get closer to zero faster. Let for some . This compresses the graph of horizontally. We can compress the graph vertically be multiplying its output by a constant.
Below of a plot of (in blue), (in green), and (in red).
Now we use to control the amplitude of the sinusoidal function that will give us our ripples. Before doing that, there one more tweak we need to make.
The function is symmetric with respect to the -axis. We would like to shift it right so its symmetry occurs somewhere in the middle of the interval , so we use the function , which is symmetric about . Therefore, the modified sinusoidal function is
A quick plot of shows the amplitude of our sinusoids is still too great, so we will compress them vertically by multiuplying by a constant between 0 and 1. Guessing a checking suggests works. (We can always make it smaller later, if we need to.)
If you're not immediately convined that our 'rippled' curve generates a solid with a greater surface area than that generated by , you can increase to add more ripples. You can even shift to be centered at to make the ripples start further away from the axis or rotation.
Here's a plot of a version of with more ripples that start closer to .
We have created a convincing example of two functions with that generate solids with surface areas measure in the opposite order. This shows that it's not possible that the surface area of a solid of revolution can be calculated as the integral of its cross-sectional circumferences.
This counterexample shows that we can't use cross-sectional circumferences to calculate the surface area of a solid of revolution, but it does not explain why this doesn't work.
So understand why surface area is not the integral of cross-sectional area, we first need to agree that the definite integral we derived does work. Surface area of the solid generated by rotating on about the -axis is If it were also true that then we could equate these formula and write
We can see that changes in are important when calculating surface area by writing using Leibniz notation. Notice
What if we approached calculating arc length in a similar manner, but approximating curve segments by horizontal lines instead of secant line segments. This approach to calculating arc length is illustrated below, where we have an arc (in blue) subdivided into four subarcs. On each subarc, we approximate the subarc with (1) a secant line (in green) and (2) a horizontal line (in red).
In our alternative formulation of approximating the length of a subarc using a horiztonal line segment, only the length of the line segments matter, not their distance from the -axis, so the arc length approximation becomes
This shows that it's important to account for how the output of the function changes.
Something analogous is happening when we calculuate the surface area of a solid of revolution. We can't assume that surface area is the integral of cross-sectional circumference because that does not take into account how the function is getting closer to or farther from the axis of rotation. Only a horizontal cylinder has a surface area that equals the integral of its cross-sectional circumference.