Now the calculus problem that I have definitely put on exams is: if you fill a hemisphere (of radius $R$) to exactly half the height ($\frac{1}{2} R$), what's the volume of such a thing? Now of course, because the curve of the hemisphere changes with the height, filling to the halfway

*height*point is going to give you less than half the*volume*. Realizing the volume as a stack of disks, the answer is$\int_{R/2}^{R} \pi(R^2 - z^2) \; dz = \frac{2\pi}{3} R^3 - \int_0^{R/2} \pi(R^2 - z^2) \; dz = \frac{2\pi}{3} R^3 - \pi R^2 z + \frac{\pi}{3} z^3 = \frac{2\pi}{3}R^3 - \frac{\pi}{2} R^3 +\frac{\pi}{3\cdot 8} R^3 = \frac{5\pi}{24} R^3.$

Oh. I thought it involved $\sqrt{2} - 1$ or something. Oh well. My memory is not what it used to be, I guess. I think that may be the answer to my volume-of-a-Sno-Cone problem, lol, which is the volume of the segment of spherical cap cut off by a cone. I don't feel like busting out the spherical coordinates today, anyway, to do that problem, lol. Some other time!

Anyway, this is of course, not half a hemisphere's worth, which is $\frac{1}{3} \pi R^3$. To get that, we have to

*solve*for the height. It is easier to measure from the top of the spoon ($xy$-plane) than from the bottom, so we solve$\frac13 R^3 = \int_0^h \pi(R^2 - z^2) \; dz = hR^3-\frac{1}{3} \pi h^3.$

for $h$. Simple, right? Clearing denominators and moving stuff to the other side, and canceling off the pesky $\pi$, we end up with

$h^3 - 3R^3 h + R^3 = 0$.

A nontrivial cubic polynomial. As it happens, this is the case of the

*depressed*cubic (insert many jokes/puns here and catch flack from the sensitivity police for making light of serious, real mental illnesses). Of course, using Mathematician's Privilege (dimensional consistency is such a ... physicist ... thing), we can just set our units to $1$ and be done with it; this gives us the polynomial$h^3 -3h + 1 = 0$.

Now I could have also tried to have my college algebra students solve this one. Rational roots theorem and all that. Nope, not gonna work, throwing $1$ and $-1$ at it. It turns out that this one is a true

*casus irreducibilis*. It being a depressed cubic, I don't have to deal with the Germans, and can skip directly to Italians with fancy cars (perhaps I'll "retcon" my story and say I was cooking a nice*risotto*and needed to use half a hemispherical teacup), the solution to this is$\sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} - 1}} + \sqrt[3]{-\frac{1}{2} - \sqrt{\frac{1}{4} - 1}}$,

$\zeta\sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} - 1}} + \bar\zeta\sqrt[3]{-\frac{1}{2} - \sqrt{\frac{1}{4} - 1}}$,

$\bar \zeta\sqrt[3]{-\frac{1}{2} + \sqrt{\frac{1}{4} - 1}} + \zeta \sqrt[3]{-\frac{1}{2} -\sqrt{\frac{1}{4} - 1}}$, where $\zeta$ is a cube root of unity (please print these formulas out onto posters and hold them up to enforce social distancing). It happens that $\zeta = -\frac{1}{2} + \frac{\sqrt{3}}{2}\mathsf i$, which is exactly what is under the cube root, so we're dealing with

*ninth*roots of unity here! (who would have thought??). The way to see that these formulas work, generally from the depressed cubic, is basically to substitute $x = w - \frac{p}{3w}$, which results in a quadratic for $w^3$. That it's quadratic explains the nearly identical features of the stuff under the cube roots, but for a sign, and that it's $w^3$ that gets solved for is responsible for the cube roots outside, as well as mutiplying by roots of unity.

The cubic $y = x^3 - 3x + 1$. It has three real roots, but they only can be expressed algebraically by the sum of complex numbers! |

So the three solutions are $\sqrt[3]{\zeta} + \sqrt[3]{\bar{\zeta}} = \zeta^{1/3} + \zeta^{-1/3}$, $\zeta^{4/3} + \zeta^{-4/3}$, and $\zeta^{-2/3} + \zeta^{2/3}$. Taking the cube root of a complex number is not a trivial matter. It happens that the imaginary parts all cancel out and you get three real solutions, but this cannot be done without admitting complex numbers in some form (after now consulting with the French). Well, almost, anyway. Let's call in trigonometry to save the day (but that's cheating, because trigonometry really is $e^{\mathsf i x}$). We'll take $\zeta = e^{2\pi\mathsf i/3}$ and the "principal" complex cube root to take minimum argument. Actually this is irrelevant, because the extra multiplications of the $\zeta$s give you the other cube roots anyway, but for definiteness... $\zeta^{1/3}$ then is $e^{2\pi\mathsf i / 9}$ and the

*possible*solutions are$e^{2\pi\mathsf i / 9} + e^{-2\pi\mathsf i / 9} = 2\cos(2\pi/9)$

$e^{8\pi\mathsf i / 9} + e^{-8\pi\mathsf i / 9} = 2\cos(8\pi/9)$, and

$e^{-4\pi\mathsf i / 9} + e^{4\pi\mathsf i / 9} = 2\cos(4\pi/9)$.

Ok, so which one of these solutions is the correct one? The solution has to be between $-1$ and $1$ (in our units; the original units would be between $-R$ and $R$), and as such, only $2 \cos(4\pi/9) \approx 0.3472$ (the middle one in the graph of the full cubic above) works. So to answer the question, you need to fill up the tablespoon measure so that it is 34.72% of the way down

*from the top*. Here's another picture and zoomed in:A cross section of the hemisphere in question, together with a zoomed in view of the cubic curve (green). The vertical line (cyan) drawn through the cubic at its root will intersect the circle (red) where it needs to be filled (here you'd fill everything to the right of this line) |

Again the fact that solving a cubic

*has*to go through the complex numbers is why complex numbers were finally accepted. It is not, as commonly believed, due to trying to solve for the quadratic, since in that case, complex numbers*only*appear when the parabola fails to intersect the $x$-axis, meaning the complex solutions are of the impossible, nonexistent intersection points, so, the various Renaissance mathematicians thought, obviously should not be taken seriously. But the cubic forces it to happen no matter what, because a cubic curve*always*crosses the $x$-axis. So there it is: half a tablespoon? You'll need complex numbers for that. Some more views (the title image sort of is looking from above, so might obscure the fact you must fill it*more*than halfway from the bottom):View from below the tablespoon. The little extra disk sticking out is the fill line. |

Finally, I'll end with more social distancing formulas: the full cubic formula $ax^3 + bx^2 + cx + d = 0$:

For those who are interested, here's a nice video explaining this formula, and why we don't teach it in high school...

It's not taught because the solutions you get may not be easily reducible to something you can recognize, even in simple cases. Conceptually, however, it is a real adventure, and solving it by breaking it into a 2-step process makes it tractable and interesting. Our tablespoon example already skips step 1 ... all that about Germans and Italians.

It's not taught because the solutions you get may not be easily reducible to something you can recognize, even in simple cases. Conceptually, however, it is a real adventure, and solving it by breaking it into a 2-step process makes it tractable and interesting. Our tablespoon example already skips step 1 ... all that about Germans and Italians.

Enjoy and stay safe!

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