# What angle halves the volume of a hemisphere?

Similar to this question, when needing to measure out 1/2 tablespoon, I use my hemispherical tablespoon, but I hold mine at an angle and fill to the lowest edge. At what angle should I tilt so that my tablespoon is only half full?

Convention note: 0 degrees is level where the tablespoon holds a full tablespoon. 90 degrees would be holding the tablespoon sideways where it holds no liquid. In practice I usually hold at a little less than 45 degree angle, i.e. more level than tipped.

• Mark the centre of the hemisphere. The liquid surface should be $\approx (1- 65\%)$, or exactly $2\cos(4\pi/9)$, of the radius from the centre (result from a comment in the linked question). Then $\sin\theta =2\cos(4\pi/9) \approx 1-65\%$. Commented Oct 7, 2023 at 23:03
• Commented Nov 5, 2023 at 7:33

## 1 Answer

If the radius of the hemisphere is $$R$$ then the volume of the tilted spoon is

$$V = \pi ( \dfrac{2}{3} R^3 - a R^2 + \dfrac{1}{3} a^3 )$$

where $$a = R \sin \theta$$

This comes from the formula for the volume of a spherical cap of height $$h$$ in a sphere of radius $$R$$. The formula is listed here. What it says there is that the volume is given by

$$V = \dfrac{\pi h^2}{3} ( 3 R - h)$$

In our problem, $$h = R - a$$. Therefore,

$$V = \dfrac{\pi (R - a)^2}{3} (2 R + a ) = \dfrac{1}{3} \pi ( R^2 - 2 a R + a^2 )(2 R + a)$$

And this gives the formula above.

Now, we want to solve

$$\dfrac{\pi R^3}{3} = \pi ( \dfrac{2}{3} R^3 - a R^2 + \dfrac{1}{3} a^3 )$$

Take $$R = 1$$, then

$$\dfrac{1}{3} a^3 - a + \dfrac{1}{3} = 0$$

i.e.

$$a^3 - 3 a + 1 = 0$$

Solving the general cubic equation is addressed here.

Here, our cubic equation is of the form

$$t^3 + p t + q = 0$$

It is explained here how a trigonometric solution is obtained for this cubic equation. If $$4 p^3 + 27 q^2 \lt 0$$ then there will be three solutions given by

$$t_k = 2 \sqrt{ -\dfrac{p}{3}} \cos \left( \dfrac{1}{3} \cos^{-1} \left( \dfrac{3 q }{2 p} \sqrt{ \dfrac{-3}{p}} \right) - \dfrac{2 \pi k }{3 } \right) \hspace{10pt} \text{ for } k = 0, 1, 2$$

This formula is due to the French mathematician François Viète.

Substituting $$p = -3 , q = 1$$ we first verify that there is indeed three real roots, because $$4p^3 + 27 q^2 \lt 0$$. Then we have

$$t_k = 2 \cos \left( \dfrac{1}{3} \cos^{-1} \left( -\dfrac{1 }{2 } \right) - \dfrac{2 \pi k }{3 } \right) \hspace{10pt} \text{ for } k = 0, 1, 2$$

This simplifies further to

$$t_k = 2 \cos \left( \dfrac{2 \pi}{9} - \dfrac{2 \pi k }{3 } \right) \hspace{10pt} \text{ for } k = 0, 1, 2$$

So the three roots are

$$t_1 = 2 \cos \left( \dfrac{2 \pi}{9} \right) = 1.53208888$$

$$t_2 = 2 \cos \left( \dfrac{4 \pi}{9} \right) = 0.3472963$$

$$t_3 = 2 \cos \left( \dfrac{ 10 \pi }{9 }\right) = -1.87938524$$

The only valid solution is $$t_2$$. Hence, we have

$$a = \sin \theta = 2 \cos \left( \dfrac{4 \pi}{9} \right)$$

From which

$$\theta = \sin^{-1} \left( 2 \cos \left( \dfrac{4 \pi}{9} \right) \right) = \sin^{-1}(0.3472963) = \boxed{20.322^\circ}$$

• Where does the initial volume formula come from? Commented Oct 8, 2023 at 0:55
• @Aeryk Please check my updated solution. Commented Oct 8, 2023 at 1:12