Question 1011899
{{{R}}}= radius of the cylinder,
{{{h}}}= height of the cylinder.
A cross-section of the sphere, with the cylinder
(sliced along the middle) looks like this:
{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
green(triangle(0,-0.8,0.6,-0.8,0,0)),
green(rectangle(0,-0.8,0.6,0.8)),
green(rectangle(0,-0.8,0.07,-0.73)),
rectangle(-0.6,-0.8,0.6,0.8),
circle(0,0,1),circle(0,0,0.02),
locate(0.25,-0.37,green(1)),
locate(0.02,0.45,green(h/2)),
locate(0.02,-0.35,green(h/2)),
locate(0.2,-0.7,R),locate(-0.22,-0.7,R)
)}}} Applying the Pythagorean theorem, {{{(h/2)^2+R^2=1^2}}}-->{{{(h/2)^2=1-R^2}}}-->{{{h/2=sqrt(1-R^2)}}}-->{{{h=2sqrt(1-R^2)}}} .
Since for a cylinder, {{{Volume=pi*R^2*h}}} ,
{{{highlight(Volume=2pi*R^2*sqrt(1-R^2))}}}