Following on from my last post On errors in math and code, here’s an example (adapted from Khan Academy) of how an error can be caught with a small amount of symbolic prompting along the way towards the final numeric answer.
For the purpose of reading this post, it doesn’t matter so much whether you understand all the steps, more that taken together, the whole thing is complex enough that you can see how an error is possible. Even a simpler problem could be enough for this.
The circle of radius 4 above has the following equation:
where 16 is 
The green lines perpendicular to the x-axis each form the base of an equilateral triangle. For a given x value, the line extends the same amount, y, above and below the x-axis, so has length 2y, as shown below.
If you imagine many such triangles closely packed together over the circular base, the 3D shape at left below emerges with (a) The solid, showing a few equilateral triangles extending above the x-y plane. Note that, incidentally, in the example below, the circle’s radius is 1 not 4 as in our example (not important for how to approach a solution to the problem).
source: https://i.stack.imgur.com/e7WX9.png
The problem: how to compute the volume of this 3 dimensional solid? The answer requires the use of integral calculus, which allows the area or volume of non-trivial functions to be computed.
To compute the volume of the 3D shape presented in the problem, we need to follow these steps:
- Determine the triangle height by taking half of the yellow equilateral triangle, giving a right angle triangle, allowing us to apply the Pythagorean Theorem to determine the height. This is already shown in the yellow triangle above as the square root of 3 but we will show how to calculate it given a hypotenuse of length 2y and a base of half that (y).
- Determine the area of a triangle first in terms of y, then x since we will be summing areas over a range on the x axis.
- Calculate the volume of the 3D shape by summing the area of an “infinite number” of triangles by computing the definite integral over the appropriate x axis range.
Along the way I will show an error and how a sanity check could serve as a correction before proceeding to calculate the final answer.
Triangle height using the Pythagorean Theorem:
Triangle area in terms of y:
Triangle area in terms of x (from circle equation), substituting for y squared:
Volume by integration over area of triangles of width dx (delta x), where dx becomes very small, so there are “infinitely many” triangles.
This is almost right, but the bounds are wrong. Look at the circle at the start of this post. It spans -4 to 4 on the x axis, not 0 to 4. The volume should therefore be twice that calculated above. We get that from what follows, with the correct integration bounds.
But if just one half of the circle’s base is used for determining the volume, the math is simpler. Compare the last two solutions: in the first there are fewer numbers in the calculation, so less opportunity for error. If one half of the volume is computed, the whole volume is simply twice this, and that was what was initially in my mind; I simply forgot to multiply by two! Taking this into account gives:
So, a simple error (wrong bounds or not multiplying by two) was introduced in the very first step of the integration but did not become tangible until we started substituting numbers.
A graphing calculator application like Desmos first confirms the result of using the erroneous bounds and then the equivalence of the last two approaches:
After all that, the main point I’m making is that guidance during learning can be given along the way to a solution (in the textbook or online resource where the problem appears) by showing multiple choices for the equation itself, not the numeric answer, such as the first and last above along with some incorrect ones. This may be enough for someone to realise that a mistake has been made before proceeding to substitute numbers.
Online learning resources often use the multiple choice approach or the “just give the numeric answer” approach, e.g. to three decimal places or in terms of 
An argument against this is that it is better to allow failure and provide step-by-step guidance in the event of an incorrect answer, something Khan Academy excels at. Of course, this doesn’t guarantee that the same mistake won’t be made across similar problems.
Leave a Reply