Day: March 9, 2010

A Super Specific Multivariable Calculus Question

Hi all,

I have a question about multivariable calculus, that I need some help with. My kids and I are both slightly stumped about this.

The question we are asked — in a section thrillingly titled, replete with semicolon, “Parametric Surfaces; Surface Area” — is to find the surface area of “The portion of the sphere x^2+y^2+z^2=16 between the planes z=1 and z=2.”

In class, the formula we derived for surface area for any parametric surface \vec{p}(u,v) is

S=\underset{R}{\int\int} \left\Vert \frac {\partial\vec{p}}{\partial u}\times\frac{\partial \vec{p}}{\partial v}\right\Vert dA.

We solved this by converting the (top part) of the sphere to a parametric surface:

x=r\cos(\theta)
y=r\sin(\theta)
z=(16-r^2)^{1/2}

Then we defined \vec{p}=<r\cos(\theta),r\sin(\theta),(16-r^2)^{1/2}> (where \theta ranged between 0 and 2\pi and r ranged between \sqrt{12} and \sqrt{15}. (Those limits for r come from the fact that we want the surface area of the sphere between z=1 and z=2 — which correspond to r=\sqrt{15} and r=\sqrt{12} respectively.) [1]

So I calculate \frac{\partial \vec{p}}{\partial r}=<\cos \theta, \sin \theta, -r(16-r^2)^{-1/2}> and \frac{\partial \vec{p}}{\partial \theta}=<-r\sin \theta, r\cos \theta, 0>.

So to use our surface area formula above, we need to find \left\Vert \frac {\partial\vec{p}}{\partial r}\times\frac{\partial \vec{p}}{\partial \theta}\right\Vert. Calculating that out, we get it to equal \frac{4r}{\sqrt{16-r^2}}. Phew, now we have something we can plug into the surface area formula for that “norm of the cross product” thingie.

Here’s where the question comes in. We know

S=\underset{R}{\int\int} \left\Vert \frac {\partial\vec{p}}{\partial r}\times\frac{\partial \vec{p}}{\partial \theta}\right\Vert dA=\underset{R}{\int\int} \frac{4r}{\sqrt{16-r^2}} dA.

Why is it that when we finally evaluate this beast, dA is not equal to our normal area element for polar, namely r dr d\theta? For the answer to come out right, we need to let dA equal to simply d r d\theta.

WHY? Why don’t we plug in the normal polar area element?

Here’s my thinking. Even though we usually use dA to represent an area element, in this particular surface area formula, it doesn’t represent anything more than du dv (for whatever parametrization gets made). The reason I think this? When I look at the derivation of the formula, it defines du dv to be dA. Simple as that.

I used to think that dA had a fixed meaning: the area element in a particular coordinate system. However, I’m now thinking that it might mean different things in different equations? Either that or our book is being sloppy.

If anyone can follow what I’ve written here and has any help to proffer, I would be much obliged. It’s a small point — one that won’t really matter in the long run for this course — but both my kids and I would like to have this resolved once and for all.

[1] If you don’t see that, imagine you have this sphere and you make a slice at z=1 and another slice at z=2. You want the surface area of that little curved “ring” — and if you find the shadow of that ring on the x-y plane, you’ll get two concentric circles with radius \sqrt{12} and \sqrt{15}. That’s the region R that you will be integrating over.