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.)
 A: 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. 
A: 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 $a\approx 65\%.$
A: 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
$$\frac hr = \frac{1}{2}\sec 40^\circ$$
A: 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.
A: Note about eyeballing: Your eye's reference is the surface of the spoon, so when you eyeball you may actually be measuring along the arc from the bottom of the spoon to its top edge. 
That is, your eye may be watching the red curve, not the blue line:

Using the 65.27% from other answers, the depth measured along the red curve is $$ \frac{\arccos(1 - 0.6527)} {90\deg }\approx 77.42\%$$
So to the eye, the "depth" of a half-full spoon may look like more like three quarters than two thirds.
A: 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 $65.27\%$.
