Consider two planes, which are parallel, intersecting a sphere of radius $1$, such that the volume between the planes is half the volume of the sphere. Then, compute the minimum distance between the two planes.

Here is what I've done so far.

I tried to use the formula $V=\frac{1}{3} \pi h^2 (3R-h)$, coupled with Pythagoras' theorem and a few routine formulas based on the fact the planes are symmetric about the centre of the sphere.

However, I don't know why, but I don't feel happy with this solution. I know that segments of spheres have close connections to integrals (i.e. volume of revolutions and polar coordinates). Is there a slick not-heavily computational answer with integrals. Please bear in mind that I have limited knowledge of polar coordinates. Thanks for your help.


3 Answers 3


If the volume of the sphere between the two planes is one half of the total volume of the sphere, then the volume on each side of either plane not sandwiched between is also one half of the total volume.

Consider one of the two planes at a distance $0 < x_0 < 1$ from the center of the sphere. Place the unit sphere's center at the origin of a Cartesian coordinate system and assume without loss of generality that the plane has the equation $x = x_0$. Then we can use the method of disks, to see that the radius of a representative disk on the interval $x \in [x_0, 1]$ is $r(x) = \sqrt{1 - x^2}$, and thus its differential volume is $$dV = \pi r^2(x) \, dx = \pi (1-x^2) \, dx.$$ Therefore the total volume of one "cap" as a function of the distance of the plane from the origin is $$V(x_0) = \int_{x=x_0}^1 \pi (1-x^2) \, dx = \pi \left( \frac{2}{3} - x_0 + \frac{x_0^3}{3} \right).$$ It follows by symmetry that for two planes whose distances are $x_0$, $x_1$ away from the origin, the total volume of the "caps" is simply $$V(x_0) + V(x_1).$$ Subject to the constraint that this sum is half the total volume of the sphere, which is $\frac{4}{3} \pi$, we obtain the relationship $$2-3x_0 + x_0^3 - 3x_1 + x_1^3 = 0.$$ The distance between the planes is $x_0 + x_1$, and treating $x_1$ as an implicit function of $x_0$, we find that the volume is minimized for a critical point satisfying $$0 = 1 + x_1'(x_0) = 1 + \frac{1 - x_0^2}{x_1^2 - 1},$$ or equivalently, $$x_1^2 = x_0^2.$$ Thus $x_1 = x_0$ (as both must be between $0$ and $1$), and the desired separation between the planes satisfies $$V(x_0) = \frac{\pi}{3},$$ which corresponds to the unique root of $$x_0^3 - 3x_0 + 1 = 0$$ in the interval $(0,1)$. This has a value of approximately $x_0 \approx 0.347296$.

  • $\begingroup$ Is the answer not equal to $2 x_0 \approx 0.69495$? Thanks for your help. I did the same method, but proved the formula I stated using a volume of revolution. $\endgroup$
    – YDP
    Sep 8, 2015 at 22:03
  • 1
    $\begingroup$ @Kinhu Sorry, in my rush to finish the answer I was not clear in the last part. The value of $x_0$ is indeed around $0.347$ but the separation between the two planes is twice this amount, as you stated in your comment. $\endgroup$
    – heropup
    Sep 8, 2015 at 22:45

Volume of a Sphere

If we rework this equation with known variables.

$$V = \int_{1-h}^1 \pi(1-y^2)dy = \bigl[\pi(y-\frac{1}{3}y^3)\bigr]_{1-h}^1 = \frac{\pi}{3}\left(3h - 1 + (1-h)^3\right)$$

$$ \begin{align} \frac{\pi}{3}\left(3h - 1 + (1-h)^3\right) &= \frac{\pi}{3}\\ 3h + (1-h)^3 &= 2 &2|1-h|<1 \end{align} $$ Therefore, through the use of a C.A.S., because reasons. $$h = 1- \frac{1-i\sqrt3}{2^\frac{2}{3}\sqrt[3]{1+i\sqrt3}} - \frac{(1+i\sqrt3)^\frac{4}{3}}{2\sqrt[3]2} \approx 0.6527036447$$

Therefore the point is at $1-h \approx 0.3472963553$.

Bah, heropup beat me to submission! xD. Gg

And the distance between the points would then be $2(1-h) \approx 0.6945927107$


I came across this question recently thinking about how to slice a watermelon into 4 equal volume parts with parallel slices. I'm glad to see that I'm not the only one thinking about this kind of problem.

If you simplify the question to focus on a hemisphere with radius 1, then you want to find the height $r$ from the base of the hemisphere to the cut such that the two remaining pieces have equal volume.

Using an integration under a surface of rotation, much as in the other answers, one finds that $r$ is the solution to the equation $$ r^3 -3r+1 =0 $$ I knew there must be a closed-form solution, but not being all that familiar with cubics with non-integer solutions (and being stubborn enough to want to solve it myself with no outside help), it took me a few days to crack this.

It was first clear that there were three (likely irrational) solutions, around 0.34, 1.5 and -1.8, and the first was the $r$ needed. But this is unsatisfying, so about three days tinkering with $x^3+ax+b=0$ and some variants finally led me to the solutions in closed form: $$ r = 2\cos(\tfrac{2\pi}{9}) $$ $$ r = -\cos(\tfrac{2\pi}{9}) + \sqrt{3}\sin(\tfrac{2\pi}{9}) $$ $$ r = -\cos(\tfrac{2\pi}{9})-\sqrt{3}\sin(\tfrac{2\pi}{9}) $$ The middle one is the culprit. Who knew that slicing watermelons could be so complicated?


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