Written by Colin+ in calculus, integration.

What is the volume above a plane, and inside a sphere of radius $r$, such that the radius of the circle where the two intersect is $R \sin(\alpha)$? What is this spherical sector's curved surface area?

I've lost the precise wording of the question that drove a small cabal of MathsJammers to distraction the other weekend, but that was the gist of it. It wasn't so much that we couldn't *do* the question (I mean, come on now. I'd hardly be blogging about it in that case, I'd be working on it still!), just that the two approaches we had – using Cartesian and polar coordinates – were giving wildly different answers, and some of my beloved heuristics were falling up short. So, having figured them out, I thought it'd make sense to write them up.

Of the four problems, this one felt the least difficult to me. Splitting the segment into disks of height $\delta x$ and radius $\sqrt{R^2 – x^2}$ gives a volume of $\pi \int (R^2 – x^2) \d x$, between limits of $R\cos(\alpha)$ and $R$.

The integral works out to $\pi \left[ R^2 x – \frac{1}{3}x^3\right]_{R\cos(\alpha)}^{R} = \frac{\pi R^3}{3} \left[ 2 – 3\cos(\alpha) + \cos^3(\alpha) \right]$. This fits: if $\cos(\alpha)=1$, the integral vanishes; if it's 0, we get the volume of a hemisphere, and if it's -1, we get the volume of a sphere.

Aside: we can even factorise it if we really want to: $V = \frac{\pi R^3}{3} \left(1-\cos(\alpha)\right)^2\left(2-\cos(\alpha)\right)$.

Instead of slicing the volume into disks of equal (perpendicular) height, let's slice the surface into bands of equal (slant) width. Each band has a width of $R\delta \theta$ and a length of $2\pi R\sin(\theta)$, and we're integrating between $\theta = 0$ at the pole and $\theta = \alpha$. (The hardest bit with this one – for me – is getting the limits right!)

$\int_{0}^{\alpha} 2\pi R^2 \sin(\theta) \d \theta = 2\pi R^2 \left[-\cos(\theta) \right]_0^\alpha$, which becomes $2\pi R^2 \left(1-\cos(\alpha)\right)$. Again, this fits for the key values of $0$, $\frac12 \pi$ and $\pi$.

There's no law that says we have to do the volume of revolution with fixed-height disks; we could do it with fixed-angle disks like the strips above. We just need to be a bit more careful.

Each disk has a radius of $R \sin(\theta)$, as before. Its height, though, is $R\sin(\theta) \delta \theta$ – to see this, a disk near the equator of a globe ($\theta \approx \frac{1}{2} \pi$) is almost a cylinder; near the pole, where $\theta \approx 0$, it's practically flat.

We need to calculate $\pi \int_0^\alpha R^3 \sin^3(\theta) \d \theta$. That looks awful until you remember you can rewrite $\sin^3(\theta)$ as $(1-\cos^2(\theta))\sin(\theta)$, which will integrate nicely.

The integral becomes $-\frac{1}{3}\pi R^3 \left[ 3\cos(\theta) – \cos^3(\theta)\right]_0^\alpha$, or $\frac{1}{3}\pi R^3 \left[ 2 – 3\cos(\alpha) + \cos^3(\alpha)]\right]$, as in the first section. I'd say that was a significantly tougher integral, though, and the set-up needs a certain amount of nous.

Over here are Flying Colours Towers, I've just made a horrified face. We're going to split the surface area into strips of equal vertical height, as we did with the original surface area calculation. However, to get the area of each strip, we need the slant height.

We *could* look that up, but the Mathematical Ninja would look scornfully at us. So instead, let's look at a right-angle triangle with legs of length $\delta x$ and $\delta y$. The hypotenuse – which is the slant height – is $\sqrt{(\delta x)^2 + (\delta y)^2}$, or $\delta x \sqrt{1 + \left(\frac{\delta y}{\delta x}\right)^2}$. In the limits, those $\delta$s straighten up nicely^{1}. What's $y$ here? It's the $y$ in $x^2 + y^2 = R^2$, so $2x + 2y \dydx = 0$ and $\dydx = -\frac{x}{y} = -\frac{x}{\sqrt{R^2 – x^2}}$. How lovely.

The width of each strip is $2\pi \sqrt{R^2 – x^2}$. This is really shaping up to be elegant, don't you think?

Our integral is $ \int_{R\cos(\alpha)}^R 2\pi \sqrt{R^2 – x^2} \cdot \sqrt{1 + \frac{x^2}{R^2 – x^2}} \d x$. That tidies up to $ 2\pi \int_{R\cos(\alpha)}^R \sqrt{\left(R^2 – x^2\right) + \left(x^2\right)} \d x$, or $2\pi R \int_{R\cos(\alpha)}^R dx$.

Well, that de-escalated quickly. We end up with $2\pi R \left[ R – R\cos(\alpha)\right] = 2\pi R^2 (1 – \cos(\alpha))$, as before. Although the final integral is beautifully simple, there's a lot of confusion in getting to it.

Why – you might ask – is it so much easier to work out the volume using Cartesian coordinates, while polars are favourite for the surface area?

I think it simply comes down to picking the coordinate system that works most simply for the shapes you're dealing with. In the case of making a volume out of disks, the slant height is much less significant than the vertical height; if you set up the integral in a way that keeps the height – $\delta x$ – constant, you make less work for yourself than by working with constant arc-lengths.

The converse is true for the surface area: the slant height is key, and that is kept constant in polar coordinates but not in Cartesians.

The reasonable question, though, isn't the one that's bugging me. That one, I'll answer another time.

* Many thanks to @realityminus3 and @ajk44 for discussions that led to this post.

- Don't tell @realityminus3! [↩]