When heuristics go bad

Last week, I wrote about the volume and outer surface area of a spherical cap using different methods, both of which gave the volume as $V = \frac{\pi}{3}R^3 (1-\cos(\alpha))^2(2-\cos(\alpha))$ and the surface area as $A_o = 2\pi R^2 (1-\cos(\alpha))$.

All very nice; however, one of my most beloved heuristics fails in this case.

If you differentiate the area of a circle ($\pi r^2$) with respect to the radius, you get $2\pi r$ – the circumference. If you differentiate the volume of a sphere ($\frac{4}{3}\pi r^3$), you get $4\pi r^2$, the surface area. In a way that can be nicely extended to many shapes, differentiating an $n$-dimensional shape gives you its $n-1$-dimensional boundary.

Only it doesn't work – or even really come close to working – here. $\diff Vr = \pi R^2 (1-\cos(\alpha))^2(2-\cos(\alpha))$, which isn't $2\pi R^2 (1-\cos(\alpha))$. So what's gone wrong?

Well, it's not a nice shape, is what's gone wrong. The rationale behind the heuristic is that if you increase the radius of a circle a tiny amount, you add on a tiny extra sliver around the circumference – and similarly for the sphere's volume and surface area.

With the spherical cap, though, there's a difference: increasing the radius does add a tiny shell around the outside. However, it also scrapes off a tiny circle on the inside of the shape. The area of that circle is $A_i = \pi R^2 \sin^2(\alpha)$, or $\pi R^2 (1 – \cos^2(\alpha))$.

So, increasing the radius should add on the outer surface area, $A_o = 2 \pi R^2(1- \cos(\alpha))$ and remove the inner surface area, $A_i = \pi R^2 (1 – \cos^2(\alpha))$.

$A_o – A_i = \pi R^2 \left[ \left( 2 – 2\cos(\alpha) \right) – (1 – \cos^2(\alpha)) \right] = \pi R^2 \left(1-\cos(\alpha)\right)^2$. This, though, still isn't correct.

Again, don't tell @realityminus3, but this is a post about heuristics rather than rigour. The sum can be fixed if we think about the relative depths of the two bits of the surface area. If you increase the radius by $\delta r$, the curved shell's depth thickens by $\delta r$. However, the circle's depth only increases by $\delta r \cos(\alpha)$ – so the appropriate sum is $A_o – \cos(\alpha) A_i = \pi R^2 \left[ \left(2 – 2\cos(\alpha)\right) – \cos(\alpha)(1- \cos^2(\alpha))\right] = \pi R^2 \left[ 2 – 3\cos(\alpha)+\cos^3(\alpha) \right]$, as required.

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

Share

Leave a Reply

Your email address will not be published. Required fields are marked *

Sign up for the Sum Comfort newsletter and get a free e-book of mathematical quotations.

No spam ever, obviously.

Where do you teach?

I teach in my home in Abbotsbury Road, Weymouth.

It's a 15-minute walk from Weymouth station, and it's on bus routes 3, 8 and X53. On-road parking is available nearby.

On twitter