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 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\%.

Source : Link , Question Author : Holly , Answer Author : J. W. Tanner

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