Surfaces of Revolution

Surfaces of Revolution
Measuring the Enigma

Surface area really is tricky. Volume, as we know, can be measured using liquid. An interesting aside, Archimedes originally discovered this principle while taking a bath, and he subsequently ran through the streets naked shouting about it (you can learn all about that story here). Measuring length is a similarly simple process: take a string, match the curve, and the measure the string. Easy. Surface area, however, is different. Certainly, surface area is simple enough to find on prisms, and they have flat sides, but once curves are involved the process becomes more difficult. (the video I showed a part of during the presentation)

Take this torus for example. The surface area of this object could be measured by calculating the surface area of each individual square on the surface on this torus. However, it would not be all that accurate. But what if we called the surface area of these squares as representative rectangles? What if their size was our change in x? If that was the case, we could set up a limit as the size decreases towards 1/∞ that acts on a summation of all the surface areas. This limit would give us an approximation of the surface area (seen below).

However, this equation is only useful as a model of an idea. It is not really feasible, at least at our level of mathematics. For this reason, we will think about surface area in terms of circumferences. 

Just as each representative radius led us to the area when finding the volume of a surface of revolution (below), that same radius begs us to use circumference when finding surface area.

VOLUME                                         SURFACE AREA

  

The surface area method sums up the surface area of the sides of our beloved representative rectangles. For this reason, it is important to pay attention to the shape of these surfaces of revolution as each limit is approached. If the surface of revolution animation above, there are two small disks at either end. Those disks both have their own surface area, so their area must be individually calculated. If we say that r(x) is the radius function, then the Surface Area integral evolves to look more like this:

But this isn't all. Consider, for instance, the function below. This function has both an exterior surface (y=√x) and an interior surface (y=x/10).

Due to this, the integral must change again, along with some of it's notation. Let r_big(x) be the exterior radius, and r_small(x) be the interior radius. I will leave r(a) in the equation, even though r(a)=0 for this particular problem set. If we think about it, we must then sum both the interior and exterior surfaces, along with the area of the two washer at each end. 

***NOTE: there are now washers instead of disks are each end. This change occurred because of the switch from one function being revolved to two functions being revolved. Now we must use washers to find the difference between the larger and smaller circles.***

       LEFT WASHER              RIGHT WASHER              EXTERIOR SA              INTERIOR SA

Of course, this notation makes everything look a bit more complicated that it needs to be, but it gets the point across. So now let's try this method out for the last video.

We are revolving the area between f(x)=√x and g(x)=x/10 on the interval x=[0,4] about the x-axis. We know that there will be no washer on the left side, as both f and g are equal to zero when x=0. So our initial equation looks like this:

But wait, the second surface [with g(x)] is just a cone, isn't it? An angled line? Yes, it is! So we can use the SA formula for the side of the cone (SA=πrl) instead of the crazy integral mess.

Now that is better. One integral, a couple functions appraised at single values. No problem. Slap that into the calculation (good old fnInt), and … SHAZAM!

That whole mess is equal to ~50.600587.

You are a genius. Go do your happy dance, then try the quiz.

The Quiz

Thanks!
Cole

p.s. Duncan values SATs more than a day in O'Brien's classroom. Let us collectively dishonor him.

More math animations here

Quiz answers (be honest, no one is grading you)