what is the max volume of cylinder inscribed in a sphere?

Question

Give me the general formula for the volume of the cylinder all in terms of the radius of the sphere.

Answer 1

In a sphere of radius R, inscribe a cylinder of radius r, height h.



Draw a diagram of this. Actually, if a 3D diagram is too tricky, you can just draw the 2D vertical cross section that contains the axis of the cylinder. The sphere then shows up as a circle, and the cylinder as a rectangle inscribed in the circle.



Draw a line from the common center, C, of sphere and cylinder to the center, B, of one base of the cylinder, then from there to A, the endpoint of a radius of the cylinder base, which is also on the sphere (circle), then back to C.



ABC is a right triangle with its right angle at B; legs AB=r and BC=h/2; and hypotenuse AC=R. So it follows that:



r^2 = R^2 - (h/2)^2



The volume of the cylinder is

V = π r^2 h

= (π/4)(4R^2 - h^2)h



To maximize V wrt h,

0 = dV/dh = π(R^2 - (3/4)h^2)

h^2 = (4/3)R^2

h = 2R/√3

r^2 = R^2 - (h/2)^2 = R^2 - (1/3)R^2 = (2/3)R^2



V = π r^2 h = (4π / 3√3)R^3



= V[sphere]/√3



To check that it is a max, not a min,

d^2 V/dh^2 = - (3π/2)h^2 < 0 ==> MAX

Answer 2

It's rather difficult to solve this without a diagram but I'll give it a try



let the sphere have a radius R then the volume of the sphere is 4/3 pi R^3



let x be the radius of the cylinder and h is half the height of the cylinder.



I want to use half the height because I will consider in the region where

x >0 , y>0 and it makes the problem easier.



then the volume of cylinder is Vcyl = pi * x^2 * (2h)



with h = sqrt( R^2 -x^2)



then Vcyl =2 pi* x^2 * sqrt(R^2-x^2)



I assume that you know some calculus



find d(Vcyl)/ dx and set to 0 and solve



x= (sqrt(6) /3 ) *R



Vcyl_max = (4pi *R^3 * sqrt(3) ) / 9



the ratio of Vcylinder max to V sphere is sqrt(3) /3 = 0.577 which



seems reasonable to me.



Note: I wrote the answer in a compact way, hope you understand.

Answer 3

So the cylinder has for its axis between the diameters of the sector. Then if 2h<2R is its height, its radius r would be so as that r^2+h^2=R^2 and you desire to maximise hr^2 this is h(R^2-h^2). merely differentiate in h and you're finished. r is the radius of the cylinder. You get r^2+h^2=R^2 via projecting on a airplane which includes the axis of the cylinder and making use of Pythagoras.