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?
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 $\boxed{\frac43}$.
To find the ratio of the volume of the sphere to the volume of the cone, we need to find the volumes of both the sphere and the cone.
Let's start by finding the volume of the cone. The volume of a cone is given by the formula $V_{\text{cone}} = \frac{1}{3} \pi r^2 h$, where $r$ is the radius of the base of the cone and $h$ is the height of the cone. In this case, we are given that the base radius of the cone is $3$ and the height is $3$, so the volume of the cone is:
$V_{\text{cone}} = \frac{1}{3} \pi (3^2)(3) = \pi(3)(3) = 9\pi.$
Now, let's find the volume of the sphere. The volume of a sphere is given by the formula $V_{\text{sphere}} = \frac{4}{3} \pi r^3$, where $r$ is the radius of the sphere. In this case, the sphere is inscribed in the cone, so the sphere's diameter is equal to the height of the cone, which is $3$. Therefore, the radius of the sphere is $\frac{1}{2} \times 3 = \frac{3}{2}$. Substituting this value into the formula, we get:
$V_{\text{sphere}} = \frac{4}{3} \pi \left(\frac{3}{2}\right)^3 = \frac{4}{3} \pi \times \frac{27}{8} = \frac{9\pi}{2}.$
The ratio of the volume of the sphere to the volume of the cone is then:
$\frac{V_{\text{sphere}}}{V_{\text{cone}}} = \frac{\frac{9\pi}{2}}{9\pi} = \frac{1}{2}.$
Therefore, the ratio of the volume of the sphere to the volume of the cone is $\boxed{\frac{1}{2}}$.
Let's denote the radius of the inscribed sphere as $r$. We know that the cone has a height of $3$ and a base radius of $3$.
Since the sphere is inscribed in the cone, it touches the base of the cone at its center. Therefore, the diameter of the sphere is equal to the base radius of the cone, which is $3$.
Using the diameter, we can find the radius of the sphere: $r = \frac{3}{2} = 1.5$.
The volume of a cone is given by the formula: $V_{\text{cone}} = \frac{1}{3}\pi r_{\text{cone}}^2 h_{\text{cone}}$.
Plugging in the values for the cone, we have: $V_{\text{cone}} = \frac{1}{3}\pi (3^2)(3) = 9\pi$.
The volume of a sphere is given by the formula: $V_{\text{sphere}} = \frac{4}{3}\pi r_{\text{sphere}}^3$.
Plugging in the value for the sphere, we have: $V_{\text{sphere}} = \frac{4}{3}\pi (1.5^3) = 4.5\pi$.
Therefore, the ratio of the volume of the sphere to the volume of the cone is: $\frac{V_{\text{sphere}}}{V_{\text{cone}}} = \frac{4.5\pi}{9\pi} = \boxed{\frac{1}{2}}$.