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Volume Estimation


Alignments to Content Standards: G-GMD G-GMD.A.3

Task

Charles and Olivia are trying to estimate the volume of water that could be held by the figure shown below, which is 10 feet high and has a circular top of radius 20 feet. Charles proposes they approximate the volume by using a cylinder of radius 20 feet and height 10 feet. Olivia proposes that they instead use a circular cone connecting the top of the tank to the vertex at the bottom.

Paraboloid_1b5b02d3f3e750cb9d4bbc8799c99500

What answers would the two methods predict? Which is likely to be most accurate? What is your best estimate for the volume of the tank?

IM Commentary

This task has the dual purpose of having students apply geometric volume formulas, and to have them reason about modeling with geometric figures. Students are presented with a surface, though not by that name, and asked to consider the process of modeling its volume by that of a cylinder or a cone. The task also provides ample opportunity for open-ended reasoning, as the lack of more explicit information about the surface makes a standard volume formula impossible. (Indeed, such formulas do not typically appear until late in the calculus sequence).

As such, the task gives students the opportunity to model a large range of practice standards. For example, students may be unaccustomed to making volume arguments without explicit formulas, in which case making sense and persevering in solving problems (MP 1), modeling (MP 4), and using available tools/formulas (MP 5), all come into play. Indeed, students might reason intuitively about the relationship between the cone, cylinder, and the surface, or might develop their own more rigorous techniques. For example, a student might compare the areas in a given cross-section, reducing the problem to a comparison of the area under a line and under a quadratic-like curve. A diagram of the comparison is below:

Paraboloid_1_47257a86199ee8418428559f53c93aa1Paraboloid_1_47257a86199ee8418428559f53c93aa1

The shaded region in the left diagram shows the difference in area between the cylinder and paraboloid in such a cross-section, and the shaded region in the right diagram shows the analogous difference between the paraboloid and the cone.

Incidentally, we note that the actual surface represented by the picture is a so-called paraboloid, for which calculus provides volume formulas via integration. In fact, the numbers in the problem were chosen so that after intuiting that the cone gave the more accurate approximation, a reasonable rough guess of $2000\pi$ is exactly the correct answer. For the curious calculophiles, the exact answer can be computed as a surface of revolution, or as a double or triple integral, e.g.: $$ V=\int_0^{2\pi}\int_0^{10} \int_0^{20} r\, dr\,dz\,d\theta=2000\pi. $$

Solution

As valid reasonings are varied and far-ranging, this solution provides only the volume calculations of the cone and cylinder, and some brief discussion of the comparison.

Both the volumes of the cone and the cylinder can be computed solely from the known radius of $r=20$ feet and the height of $h=10$ feet. The volume of the cylinder is

$$ V(\text{cylinder}) = \pi r^2h=4000\pi\,\text{ft}^3, $$

and the volume of the cone is $\frac{1}{3}$ of that:

$$V(\text{cone}) = \frac{1}{3}\,\pi r^2h=\frac{4000}{3}\pi\,\text{ft}^3,$$

It is intuitively clear to see that the volume of the cone is an under-estimate, since the cone would fit completely in the given tank, and the volume of the cylinder is an over-estimate, since the tank would fit completely inside of the cylinder. We conclude that

$$ \frac{4000}{3}\pi\,\text{ft}^3 < V(\text{tank}) < 4000\pi\,\text{ft}^3. $$

(Note that $\frac{4000}{3}=1333.\overline{3}$). A reasonable estimate the for the volume of the tank thus might be to average the two to get $$ V(\text{tank})\approx 2666\pi \text{ft}^3. $$

It is true and possibly intuitive, but difficult to show, that in fact the cone does a significantly better job at approximating the tank than does the cylinder. As a consequence, a better approximation might be to choose an estimate closer to the volume of the cone than the cylinder. An estimate closer to $$ V(\text{tank})\approx 2000\pi \text{ft}^3 $$

might therefore be a little more reasonable. As it happens, this estimate turns out to be exactly correct, assuming the walls of the tank are quadratic curves.