How deep is the liquid in a half-full hemisphere?

I have a baking recipe that calls for 1/2 tsp of vanilla extract, but I only have a 1 tsp measuring spoon available, since the dishwasher is running. The measuring spoon is very nearly a perfect hemisphere.

My question is, to what depth (as a percentage of hemisphere radius) must I fill my teaspoon with vanilla such that it contains precisely 1/2 tsp of vanilla? Due to the shape, I obviously have to fill it more than halfway, but how much more?

(I nearly posted this in the Cooking forum, but I have a feeling the answer will involve more math knowledge than baking knowledge.)

Solution 1:

Assuming the spoon is a hemisphere with radius $R$,

let $x$ be the height from the bottom of the spoon, and let $h$ range from $0$ to $x$.

The radius $r$ of the circle at height $h$ satisfies $r^2=R^2-(R-h)^2=2hR-h^2$.

The volume of liquid in the spoon when it is filled to height $x$ is $$\int_0^x\pi r^2 dh=\int_0^x\pi(2hR-h^2)dh=\pi Rh^2-\frac13\pi h^3\mid_0^x=\pi Rx^2-\frac13\pi x^3.$$

(As a check, when the spoon is full, $x=R$ and the volume is $\frac23\pi R^3,$ that of a hemisphere.)

The spoon is half full when $\pi Rx^2-\frac13\pi x^3=\frac13\pi R^3;$ i.e., $3Rx^2-x^3=R^3;$

i.e., $a^3-3a^2+1=0$, where $a=x/R$.

The only physically meaningful solution of this cubic equation is $a\approx 65\%.$

Solution 2:

It may be surprising that the problem actually admits an analytic solution.

A spherical cap is the difference between two overlapping cones, one with a spherical bottom and the other with a flat bottom, i.e.

$$ V = \frac{2\pi}{3}r^2h - \frac{\pi}{3}(2rh-h^2)(r-h) =\frac{\pi}{3}(3rh^2-h^3)$$

which, with half of the semisphere volume $V=\frac{2\pi}{3}r^3$, becomes

$$\left(\frac rh \right)^3 - 3\frac rh+1=0$$

Let $\frac rh = 2\cos x$ and compare with $4\cos^3 x -3\cos x -\cos 3x=0$ to obtain $x=40^\circ$. Thus, the depth $h$ as a fraction of the radius $r$ is

$$\frac hr = \frac{1}{2}\sec 40^\circ$$

Solution 3:

It makes things a bit simpler if we turn your measuring spoon upside down, and model it as the set of points $\{(x,y,z):x^2+y^2+z^2=1, z\ge 0\}$. The area of a cross-section at height $z$ is then $\pi(1-z^2)$, so the volume of the spoon between the planes $z=0$ and $z=h$ is

$$\pi\int_0^h(1-z^2)dz = \pi\left(h-\frac13h^3\right)$$

The volume of the hemisphere is $\frac23\pi$, and we want the integral to be equal to half this, i.e. $$\pi\left(h-\frac13h^3\right)=\frac{\pi}{3}$$ or $$h^3-3h+1=0$$ This cubic equation doesn't factorize nicely, so we ask Wolfram Alpha what it thinks. The relevant root is $h\approx 0.34730$. Remember that we turned the spoon upside down, so you should fill it to a height of $1-h=0.65270$, or $65.27\%$.

Solution 4:

Without loss of generality we assume the radius of the sphere to be $1$

The volume of the liquid is found by an integral $$V= \int _{-1}^{-1+h} \pi (1-y^2 )dy$$

and you want the volume of the liquid to be half of the hemisphere which is $\pi/3$

After evaluating the integral and solving the equation I have found $$h=0.65270365$$ That is a little bit more than half as expected.

Solution 5:

Alternative: use two teaspoons.

Use water as you develop your skill. Fill tsp A, and pour into tsp B until the contents appear equal. Each now contains half a tsp. And now you know what half a tsp looks like in practice.

And you don't have to calculate cosines against thumb-sized hardware.