Cavalieri the Shape Shifter

Calculus Level 4

In the diagram, a right circular cone (with height and radius of the same length) is stabbing through a hemisphere--a half of a sphere--of the same radius as the cone. During the overlap, the cross-section of the hemisphere expands horizontally, which causes it to shape-shift to a right cylinder. In the end, the tip of the cone coincides with the center of the cylinder base. It can be noted that the solid (hemisphere, initially) spreads like a "pile of sand," so its volume (not including the cone) is constant.

Now, define $V_{\text{min}}$ and $V_{\text{max}}$ to be the minimum and maximum volumes, respectively, of the frustum solid (including the cone part) formed in the process described above. When the tip of the cone goes exactly halfway down, what can be said about the volume of this new solid $V_{\text{new}}?$

Note: To answer the problem, compute the volume of the new solid, including the merged portion of the cone.

V_{\text{new}}

is less than the average of

V_{\text{min}}

and

V_{\text{max}}

V_{\text{new}}

is exactly the average of

V_{\text{min}}

and

V_{\text{max}}

V_{\text{new}}

is larger than the average of

V_{\text{min}}

and

V_{\text{max}}

1 solution

Florian Wechslberger
Jan 29, 2019

$\begin{aligned} V_\mathrm{min}&=\frac 23 r^3\pi\\ V_\mathrm{max}&=V_\mathrm{min}+V_\mathrm{cone}(h=r,R=r)=r^3\pi\\ V_\mathrm{average}&=\frac 56 r^3\pi\\\\ V_\mathrm{new}&=V_\mathrm{min}+V_\mathrm{cone}(h=r/2,R=r/2)=\frac 23 r^3\pi+\frac 1{24} r^3\pi=\frac {17}{24} r^3\pi \end{aligned}$

$\begin{aligned} \Rightarrow\,V_\mathrm{new}<\frac {20}{24} r^3\pi=V_\mathrm{average} \end{aligned}$

Cavalieri the Shape Shifter

1 solution

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