# 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.)

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

let $$xx$$ be the height from the bottom of the spoon, and let $$hh$$ range from $$00$$ to $$xx$$.

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

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

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

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

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

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