Solids Lab: Sphere, Cone & Cylinder

Put a sphere of radius 3 inside the smallest cylinder that holds it. Show that the volumes are in the ratio 2:3, that the total surfaces are in the same 2:3, and then say why the same number turns up twice.

1. The tight cylinder is fixed by the sphere: its radius is r and its height is 2r, because the sphere has to touch the curved wall and both flat ends. Nothing is free to choose.

   $$V_{\text{sph}}=\frac{4}{3}\pi r^3,V_{\text{cyl}}=\pi r^2\cdot 2r=2\pi r^3$$

2. Volumes. The sphere is ⁴⁄₃πr³ and the cylinder is πr² × 2r = 2πr³. Divide: (⁴⁄₃) ÷ 2 = ²⁄₃, and r has cancelled, so the ratio holds for every sphere there has ever been. At r = 3 that is 113.10 against 169.65.

   $$\frac{V_{\text{sph}}}{V_{\text{cyl}}}=\frac{4/3}{2}=\frac{2}{3}$$

3. Surfaces. The sphere is 4πr². The cylinder needs three pieces: two flat caps at πr² each, and the curved wall, which unrolls into a rectangle 2πr wide and 2r tall — so 4πr². Total 2πr² + 4πr² = 6πr².

   $$A_{\text{cyl}}=\underset{\text{caps}}{\underbrace{2\pi r^2}}+\underset{\text{wall}}{\underbrace{2\pi r\cdot 2r}}=6\pi r^2$$

4. Divide again: 4πr² ÷ 6πr² = ²⁄₃. The same ratio, from a completely different calculation. At r = 3 that is 113.10 against 169.65 once more.

   $$\frac{A_{\text{sph}}}{A_{\text{cyl}}}=\frac{4\pi r^2}{6\pi r^2}=\frac{2}{3}$$

5. Here is why it is not luck. Look at the curved wall on its own: it is 4πr², which is exactly the sphere's whole surface. That is Archimedes' result — slice the sphere and the cylinder with any two planes square to the cylinder’s axis and the two bands between them have equal area, because going up the sphere the circles shrink at precisely the rate the surface tilts. Add the two caps, 2πr², and 4 becomes 6.

   $$A_{\text{sph}}=4\pi r^2=A_{\text{wall}}\text{(Archimedes)}$$

6. One warning about the panel at these settings. It prints 113.0973 for the sphere's volume and 113.0973 for its surface, and those are not the same kind of quantity — one is cubic units and the other square. They agree only because ⁴⁄₃πr³ = 4πr² needs r = 3. Set r to 4 and they separate at once.

   $$\frac{4}{3}\pi r^3=4\pi r^2⟺r=3\text{(why the panel prints 113.0973 twice)}$$

At equal volume the panel charges the cylinder 14.5% more surface than the sphere and the cone 26.0%. Work out where those two figures come from, and decide whether the fixed-material preset is a second fact or the same one.

1. Start with the sphere. At r = 3 the volume is (4/3)π × 27 = 113.10 and the surface is 4π × 9 = 113.10, which is the target every other shape has to match.

   $$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi \cdot 27=113.10,A_{\text{sph}}=4\pi r^2=113.10$$

2. The cheapest closed cylinder holding that volume has h = 2r, the proportion that minimises surface. Then V = 2πr³ and A = 6πr², so r = ∛18 = 2.6207 and the area comes to 129.46.

   $$\text{Best cylinder: }h=2r_c\Rightarrow V=2\pi r_c^3,A=6\pi r_c^2;r_c=\sqrt[3]{18}=2.6207,A=129.46$$

3. Do the comparison without numbers and r drops out. Matching the volumes forces r_sphere³ = (3/2)r_cyl³, and the area ratio collapses to the cube root of 3/2, which is 1.1447. That is the 14.5%, and it holds at every size.

   $$r_s^3=\frac{3}{2}{r}_c^3⟹\frac{A_{\text{cyl}}}{A_{\text{sph}}}=\frac{6\pi r_c^2}{4\pi {\left(\frac{3}{2}\right)}^{2/3}{r}_c^2}=\sqrt[3]{\frac{3}{2}}=1.1447\Rightarrow +14.5\%$$

4. The cheapest cone has h = √8 r, which makes its slant exactly 3r and its area 4πr². The same cancellation leaves the cube root of 2, which is 1.2599, so 26.0%.

   $$\text{Best cone: }h=\sqrt{8}{r}_n\Rightarrow \ell =3r_n,A=4\pi r_n^2,V=\frac{2\sqrt{2}}{3}\pi r_n^3;\frac{A_{\text{cone}}}{A_{\text{sph}}}=\sqrt[3]{2}=1.2599\Rightarrow +26.0\%$$

5. Two cube roots of small fractions is a strong hint that one quantity is doing the work, and it is Q = 36πV²/A³. It carries no units and does not move when a shape is scaled. The sphere gives 1, the best cylinder ⅔ and the best cone exactly ½.

   $$Q=\frac{36\pi V^2}{A^3}:Q_{\text{sph}}=1,Q_{\text{cyl}}=\frac{2}{3},Q_{\text{cone}}=\frac{1}{2}$$

6. Q settles both presets at once. At fixed volume the surface goes as Q to the power −⅓, which returns 1.1447 and 1.2599. At fixed surface the volume goes as √Q, so the sphere holds √(3/2) = 1.2247 times the cylinder and √2 = 1.4142 times the cone. Those are the 22.5% and 41.4% on the fixed-material cards.

   $$\text{fixed }V:A\propto Q^{-1/3}\Rightarrow 1.1447,1.2599;\text{fixed }A:V\propto Q^{1/2}\Rightarrow \sqrt{\frac{3}{2}}=1.2247,\sqrt{2}=1.4142$$

7. And the cylinder's ⅔ is the ratio this page opened with. The surface-minimising cylinder is h = 2r, which is Archimedes's circumscribing cylinder, and there the sphere holds two thirds of both the volume and the surface. Put that into Q and the squares and cubes leave (3/2)² ÷ (3/2)³ = ⅔.

   $$\frac{V_{\text{sph}}}{V_{\text{cyl}}}=\frac{A_{\text{sph}}}{A_{\text{cyl}}}=\frac{2}{3}⟹Q_{\text{cyl}}=\frac{(3/2{)}^2}{(3/2{)}^3}=\frac{2}{3}$$