I'm sorry maybe it's obvious but English is not my first language. I just want to know what is asked in this question:

The area of a circle (in square inches) is numerically larger than its circumference (in inches). What is the smallest possible integral area of the circle, in square inches?

Specifically I don't understand what integral area means. I'm familiar with integration and how you can calculate the area between the curve and the axes but what should the integral area of a circle mean?

  • 2
    $\begingroup$ @EricDuminil That's why they specify the units. It allows you to uniquely extract dimensionless numbers (a number of square inches for the area, a number of inches for the circumference), and you can compare these dimensionless numbers. This question may instill the dangerous habit of thinking that lengths and areas are comparable, and in that sense it may be a bad question, but as asked, with units given, there is no room for misinterpretation. $\endgroup$
    – Wouter
    Mar 27 '18 at 11:46
  • $\begingroup$ @Wouter, you're right, it's possible to compare units/inch**2 and length/unit. It's pretty much useless though since it depends on the choice of units, and it's simply wrong in physics. I'll delete my comment. $\endgroup$ Mar 27 '18 at 12:12
  • $\begingroup$ I edited the title because the previous title was completely nondescript. $\endgroup$ Mar 27 '18 at 21:28
  • 1
    $\begingroup$ @CameronWilliams: Unfortunately your edit incorporates into the title exactly the information that Denis-George Mih was missing, and thus makes him look like a complete moron. I'll edit again to fix that. $\endgroup$
    – celtschk
    Mar 28 '18 at 1:38
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    $\begingroup$ @celtschk fair! I hadn't thought of that. $\endgroup$ Mar 28 '18 at 4:47

Most likely, "integral" means "is an integer". The area of this circle, expressed in square inches, is an integer.

  • 21
    $\begingroup$ This is an ambiguity frequently seen. "Integral" is the adjective both for "integral" and for "integer". I once went to a lecture with a title about "integral equations", thinking it was equations involving $\int$. But no, it was about equations with coefficients in $\mathbb Z$. $\endgroup$
    – GEdgar
    Mar 27 '18 at 11:12
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    $\begingroup$ @GEdgar Well at least you didn't get a lecture on the importance of equations to the modern way of life.;) $\endgroup$
    – DRF
    Mar 27 '18 at 12:11
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    $\begingroup$ There's a special place in heaven for those who can write a math question that's understandable to even a first language reader. In this case, "nearest integer" or similar, would have been my first choice of wording. As I help high school students, I realize that not everyone knows what a ferris wheel is (for trig problems) or is familiar with a standard 52-card deck of playing cards (for probability). To be fair, the student this is targeted to is a much lower level than calculus. But still. $\endgroup$ Mar 27 '18 at 13:04
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    $\begingroup$ @JoeTaxpayer Using "nearest" would be inappropriate here since this would indicate you are allowed to round down (which you're not). It would be better to write "the smallest possible area in $\mathbb{Z}$". $\endgroup$
    – fabian
    Mar 27 '18 at 14:06
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    $\begingroup$ What typo? Nothing here to see... $\endgroup$ Mar 27 '18 at 16:00

As Wouter points out, "integral" almost certainly means "integer valued". That is, the area of the circle is an integer. We can now answer the question, which I likely would have written as

The area of a circle (in square inches) is numerically larger than its circumference (in inches). What is the smallest integer which could be the area of the circle (in square inches)?

To fix notation, suppose that we have a circle with area and circumference given by

$$ \text{Area} = A \text{ in}^2 \qquad\text{and}\qquad \text{Circumference} = C \text{ in}. $$

The first sentence tells us that $A > C$. The second sentence asks us to find the smallest integer value of $A$ possible. From general theory, we know that if $r$ inches is the radius of the circle, then $$ A = \pi r^2 \qquad\text{and}\qquad C = 2\pi r. $$ This implies that $$ A = \frac{C^2}{4\pi}. $$ Since we need $A > C$ (and we can assume that $C > 0$), it follows that $$ \frac{C^2}{4\pi} > C \implies C > 4\pi \implies A = \frac{C^2}{4\pi} > \frac{(4\pi)^2}{4\pi} = 4\pi, $$ since $C > 4\pi > 1$ implies that $C^2 > (4\pi)^2$. But then (1) $A$ has to be an integer and (2) $A$ must be bigger than $4\pi$, so we round up to obtain $$ A = \lceil 4\pi \rceil = 13 $$ (since $4\pi \approx 12.566$; thanks Google!). That is, the smallest integer which could be the area of the circle is 13 square inches.


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