Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The first question that comes into my mind here is whether any cylinder that touches(at 4 pts) the circumference of the sphere and does not go out of it, has equal volume?

Second, how do i mathematically limit the volume of the cylinder to be less than that of a sphere? Squeeze theorem?

Please help, thanks!

share|cite|improve this question
Are there spheres of non-constant radius? :) – Thomas Andrews Nov 27 '12 at 15:31
This is basically the same problem, but with a cone instead of a cylinder. See if you can apply the same ideas.… – apnorton Nov 27 '12 at 15:38
You can see that not all such cylinders have equal volume, just by considering the extreme case of when two of the points are very close together. You get either a long thin rod, or a big flat pancake, and the volume of each of these tends to zero as the points become closer together. So the maximum volume must lie somewhere between these two extremes. – TonyK Nov 27 '12 at 15:58
up vote 2 down vote accepted

Let $R$ be the radius of the sphere and let $h$ be the height of the cylinder centered on the center of the sphere. By the Pythagorean theorem, the radius of the cylinder is given by $$ r^2 = R^2 - \left(\frac{h}{2}\right)^2. $$

The volume of the cylinder is hence $$ \begin{align} V &= \pi r^2 h\\ &= \pi \left(h R^2 - \frac{h^3}{4}\right). \end{align} $$

Differentiating with respect to $h$ and equating to $0$ to find extrema gives $$ \frac{dV}{dh}=\pi \left(R^2 - \frac{3h^2}{4}\right) = 0\\ \therefore h_0 = \frac{2R}{\sqrt{3}} $$

The second derivative of the volume with respect to $h$ is negative if $h>0$ such that the volume is maximal at $h = h_0$. Substituting gives $$ V_{max}=\frac{4 \pi R^3}{3\sqrt{3}}. $$

share|cite|improve this answer
nice solution , just one correction. When applying Pythagorean theorem, the exact way would be $r^2+(h/2)^2=R^2$ where $r$ is the radius of small cylinder and $R$ is the radius of big cylinder and $h$ is the height of cylinder. – ashmantak Mar 28 '14 at 10:26
Thanks for the catch. That is now fixed. – Till Hoffmann Mar 28 '14 at 11:17

Hint: In the context of a calculus course, I think you are first expected to argue informally that such a maximal cylinder must have axis that goes through the center of the circle, and that without loss of generality that axis is the $z$-axis.

So now suppose that the cylinder meets the $x$-$y$ plane in a circle of radius $t$. Find the height of the cylinder in terms of $t$, and hence the volume. Now use the ordinary tools to maximize.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.