Problem 6

A cone of revolution has an inscribed sphere tangent to its base and sloping surface. A cylinder is circumscribed about the sphere, with its base in the base of the cone. If the volumes of the cone and cylinder are

V_1

and

V_2

, respectively: (a) prove that

V_1\ne V_2

; (b) find the smallest possible value of

V_1/V_2

, and in this case construct the half-angle of the cone.

Step 2 of 4: Compute the volume ratio

Previous step Next step

V_2=2\pi r^3,\quad \dfrac{V_1}{V_2}=\dfrac{(1+s)^3}{6s(1-s^2)},\quad s=\sin\theta

Detailed analysis

The circumscribed cylinder has radius $r$ and height $2r$ , so $V_2=2\pi r^3$ . With $s=\sin\theta$ and $\cos^2\theta=1-s^2$ , substituting $H$ and $R$ into $V_1=\tfrac13\pi R^2H$ gives $V_1/V_2=(1+s)^3/[6s(1-s^2)]$ , for $0<s<1$ .