A sphere is inscribed in a cone with height $3$ and base radius $3$. What is the ratio of the volume of the sphere to the volume of the cone?

Answer 1

Let the sphere have radius $r$. At any level $h$ in the cone, the radius of the cross-section is proportional to $h$, because the cross-section of the cone is similar to the base, and the base has radius 3. Thus, if the cross-section radius is $rh/3$, then the cross-section volume at that level is $\pi r^2(h^2/9)$. The volume of the cone is the integral of this expression from 0 to 3.

The analogous volume fraction for the sphere is $(4/3) \pi r^3/(\pi r^2 \cdot 9) = r/6$, so we consider $r=\frac{1}{6}$ so we do not need to compute fractions.

The integral of the cross-sectional area is
$$\int_0^3 \pi r^2\left(\frac{h^2}{9}\right)\,dh=\pi r^2\cdot\left[\frac{h^3}{27}\right]_0^3=\pi r^2\cdot \frac{3^3}{27}=\pi r^2\cdot\frac13.$$

The ratio of the volume of the sphere to the volume of the cone is, therefore, $(4/3)\pi\left(\frac{1}{6}\right)^3\cdot \frac{\pi\left(\frac{1}{6}\right)^2}{\pi \left(\frac{1}{6}\right)^2\cdot\frac13} = \dfrac{4}{3}$.

[asy]
size(200);
import three;

currentprojection=perspective(0,0,10);

// calculate intersection point of plane and cone
triple planeconeintersect (triple P, triple u, pair V) {
real t = (V.x + u.x*P.z)/(u.x);
return P + t*u;
}

// dot product
triple dotprod (triple u, triple v) {
return (u.x*v.x, u.y*v.y, u.z*v.z);
}

//scale the vector to length 1
triple normalizefactor(real f, triple u){
return f*unit(u);
}

triple A=(0,0,0), B=(3,0,0), C=(0,3,0), I=(1,1,3/2);
triple[] O=intersectionpoints(A--I,B--I--C);

draw(C--A--B--C);
draw(C--I--B);
draw(B--I);

draw(O[0]--(C.x,C.y,O[0].z),dashed);
draw(O[0]--(B.x,B.y,O[0].z),dashed);
draw(O[0]--I,dashed);

label("$A$",A,S);
label("$B$",B,E);
label("$C$",C,N);
label("$I$",I,N);

dot("$O$",O[0],NW);
[/asy]

Alternatively, the ratio of the volumes does not depend on the size of the objects involved. Since the ratio of the radii is $1/6$, the ratio of the volumes is $1/6^3 = 1/216$. Thus, the answer is $\boxed{\frac43}$.