## Archive for September, 2009

### Surface Area of Revolution

September 27, 2009

Suppose $y=f(x)$ is some function you’d like to rotate around the $x$-axis (over some interval $I$), and that you’d like to calculate the surface area of the resulting surface. Dig up your favorite calculus reference, and you’ll find something like the formula:

$\int_I 2\pi f(x)\sqrt{1+(f'(x))^2}\ dx$.

If you read through the proof, you’ll see that this formula comes from breaking up the surface into lots of little end bits of cones. The way I explained those shapes to my class this semester was: take an ice cream cone and dip the “top” (where the ice cream goes) into chocolate. The bit that’s covered in chocolate is then the sort of shape we’re considering, on each little interval. These little bits give us a reasonable approximations on each interval, and we take a limit and a sum and it becomes the above integral.

Recently, another of the calc instructors asked: Why don’t we approximate each little bit with just cylinders? Well, the first several answers he got (there were a few of us talking about this) were “you get the wrong answer”, essentially. “The cylinder and cone have different surface areas”. Of course they do, but in the limit, why does it matter? Don’t those cylinders get really close to approximating the length of the curve?

Let’s start getting some formulas going. Consider the line $y=mx$ on the interval $[a,a+h]$. For convenience, take $m,a>0$. Also consider the horizontal line $y=ma$ on the same interval. Rotating the line $y=mx$ gives us the cone, and rotating the line $y=ma$ gives us the cylinder. Notice that we chose the horizontal line to be the value of the diagonal line at the point $x=a$.

While we’re thinking about surface area, let’s also think about volumes, to see what is different between the two situations. The formulas for the cylinder are easy, the surface area of the cylinder we made in the previous paragraph is just $2\pi mah$, and the volume is $\pi (ma)^2h$. The formulas for the bit of the cone are slightly more complicated, but doable. If you’ve still got your calc reference out and find these formulas, you should be able to get the surface area to be $\pi m\sqrt{m^2+1}(2ah+h^2)$ and volume as $\frac{1}{3}\pi m^2(3a^2h+3ah^2+h^3)$.

Now what we want to do is compare how the formulas for surface area (and volume) compare as $h\to 0$. The way to see how they compare is to take their ratio. In the case of volume, if we put the volume of the cone in the numerator, we can simplify the formula to

$\dfrac{3a^2+3ah+h^3}{3a^2}$

Taking the limit as $h\to 0$ we get 1. That means that the formulas are essentially equivalent in terms of using them to find volume of a solid of revolution. However, in the case of surface area, the quotient boils down to

$\dfrac{\sqrt{m^2+1}(2a+h)}{2a}$

which only goes to 1 in the limit if $m=0$ (i.e., the cone is really a cylinder).

Now, if a student asks about this in calc class, how do we justify this process of taking quotients and looking at the limit? Has anybody ever had a student ask them this question?

### Homotopy Limits and Arrow Categories

September 13, 2009

Ouch, no posts in 2 months? Shame on me. Let’s suppose I’ve been busy.

Today, and other scattered times in the past, I’ve been trying to come to grips with a proof concerning homotopy limits. Suppose $\mathscr{C}'\subseteq \mathscr{C}$ and that $|\mathscr{C}'\downarrow c|\simeq *$ for every $c\in \mathscr{C}$. Suppose that $F$ is a (covariant) functor from $\mathscr{C}$ to, say, spaces. Then the lemma I’ve been thinking about is that $\text{holim}_{\mathscr{C}'}F\simeq \text{holim}_{\mathscr{C}}F$, and in fact the natural map from right to left gives the equivalence.

I’ve been studying Bousfield and Kan‘s proof a little, and trying to make it look like something I can think about. Their $F$ goes to simplicial sets, or so, and they have some assumption about it taking values in the collection of fibrant objects. I’m going to ignore that, because in Top all spaces are fibrant (right?), and because ignoring hypotheses can never lead you astray (right?).

So anyway, I thought I’d share something like an outline of a proof. I’ll write = where there probably should be $\simeq$ instead. And that likely won’t be the worst of my transgressions.

Consider the bi-complex $A_{i,j}$ (I guess it’s a bi-cosimplicial space, or so) where

$\displaystyle A_{i,j}=\prod_{\substack{c_0'\leftarrow\cdots\leftarrow c_i'\\c_j\leftarrow c_0'\\c_0\leftarrow \cdots\leftarrow c_j}} F(c_0),$

where the $c_k$ represent objects of $\mathscr{C}$, the $c_k'$ come from $\mathscr{C}'$, and arrows lie where they should (in particular, between primed elements, the arrows are in $\mathscr{C}'$). I’m going to get tired of that indexing, so I’ll let $\underline{c_j}$ represent a chain of the form $c_0\leftarrow \cdots \leftarrow c_j$, and similarly $\underline{c_i'}$ is $c_0'\leftarrow\cdots\leftarrow c_i'$. In fact, I’m not going to put the subscripts, as the subscript on a $\underline{c_j}$ will always be $j$, and similarly $i$ for the primes. If I write $\underline{c}\leftarrow \underline{c'}$, I mean that $c_j\leftarrow c_0'$. So I can write

$\displaystyle A_{i,j}=\prod_{\underline{c}\leftarrow \underline{c'}}F(c_0),$

letting the ambiguation begin (or was that earlier?).

Next, let’s start thinking about what some homotopy limits are. I should probably be calling things Tot’s, for totalization, but my understanding is that they’re basically holim. Finally, “\holim” is not a built-in latex command. For convenience, I’m going to just write “lim”.

$\lim$ means $\text{holim}$ below.

Ok, so, fix a $j$. In what follows, $Map(X,Y)$ denotes the mapping space, and $N_n(\mathscr{D})$ represents the $n$-th layer in the simplicial nerve of a category $\mathscr{D}$. The geometric realization of $\mathscr{D}$ is denoted $|\mathscr{D}|$. Compute:

$\begin{array}{rl} \lim_i A_{i,j} &= \displaystyle \lim_i \prod_{\underline{c}\leftarrow \underline{c'}} F(c_0) \\ &= \displaystyle \lim_i \prod_{\underline{c}}\prod_{c_j\leftarrow \underline{c'}} F(c_0) \\ &= \displaystyle \prod_{\underline{c}} \lim_i \prod_{c_j\leftarrow \underline{c'}}F(c_0) \\ &= \displaystyle \prod_{\underline{c}} \lim_i Map(N_i(\mathscr{C}'\downarrow c_j),F(c_0)) \\ &= \displaystyle \prod_{\underline{c}} Map(|\mathscr{C}'\downarrow c_j|,F(c_0))\end{array}$

Finally, our assumption that these arrow categories are contractible lets us write this as simply $\prod_c F(c_0)$. Which means that

$\displaystyle \lim_{\mathscr{C}}F=\lim_j \prod_{\underline{c}} F(c_0) = \lim_j \lim_i A_{i,j}$.

That’s nice. Let’s do the homotopy limit the other way, fixing $i$ to begin.

$\begin{array}{rl} \lim_j A_{i,j} &= \displaystyle \lim_j \prod_{\underline{c}\leftarrow \underline{c'}}F(c_0) \\ &= \displaystyle \lim_j \prod_{\underline{c'}}\prod_{\underline{c}\leftarrow c_0'} F(c_0) \\ &= \displaystyle \prod_{\underline{c'}} \lim_j \prod_{\underline{c}\leftarrow c_0'} F(c_0) \end{array}$

Now $\lim_{\mathscr{C}'}F=\lim_i \prod_{\underline{c'}} F(c_0')$, which almost looks close to what we’ve got above. To see that they are equivalent, notice that the arrow category $c_0'\downarrow \mathscr{C}$ has the initial object $c_0'\to c_0'$. This means that

$\displaystyle F(c_0')=\lim_{c_0'\downarrow \mathscr{C}}F=\lim_j \prod_{\underline{c}\leftarrow c_0'}F(c_0).$

We conclude that

$\displaystyle \lim_{\mathscr{C}'} F=\lim_i \lim_j A_{i,j}.$

Finally, apply Fubini’s theorem for homotopy limits, and conclude that

$\displaystyle \lim_{\mathscr{C}}F=\lim_{\mathscr{C}'}F$

Now, to translate to the situation when my categories are internal categories in Top…