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.

9 comments

samjshah says:

March 9, 2010 at 3:37 am

And although you’re welcome to solve it and give me the answer, we don’t need the answer. We solved it already using a different method. We just wanted to know why we aren’t using $r dr d\theta$ for $dA$ .

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Peeter Joot says:

March 9, 2010 at 4:29 am

Why do you expect the area element to be $r dr d\theta$?

Think about it geometrically. On the surface, if you make the surface element small enough, you can form a parallogram by the span of the two differential elements

$\frac{\partial \mathbf{p}}{\partial \theta} d\theta$

and

$\frac{\partial \mathbf{p}}{\partial \phi} d\phi$

Here I’ve assumed spherical polar coordinates, with a point on the surface parameterized by:

$\mathbf{p} = (\sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta)$

you can really use whatever parameterization you want, but the area of the pair of differential elements that bound the parallelogram will be of the form:

$dA = \left\vert \frac{\partial \mathbf{p}}{\partial \theta} d\theta \times \frac{\partial \mathbf{p}}{\partial \phi} d\phi \right\vert$

That’s what you want to integrate, in this case, you’ll end up with

$\int_{\phi=0}^\pi \int \delta\theta r^2 \sin\theta d\theta d\phi$

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Allison says:

March 9, 2010 at 4:32 am

You are right to suspect that the notation dA is not fixed to a particular coordinate system; I would actually say that it should NEVER be equal to $r dr d \theta$. If you want (and if I am recalling correctly), you can calculate the area of a circle or some other convenient region in the plane as if it were a parametric surface; then $\left\Vert \frac {\partial\vec{p}}{\partial r}\times\frac{\partial \vec{p}}{\partial \theta}\right\Vert$ should be equal to r. So the dA you are familiar with from polar coordinates is really a specialized version of the whole integrand you have above; it’s just that in this case it is easier to memorize the norm of the cross product than to repeatedly calculate it.

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Peeter Joot says:

March 9, 2010 at 4:35 am

Why do you expect the area element to be $r dr d\theta$ ?

Think about it geometrically. On the surface, if you make the surface element small enough, you can form a parallogram by the span of the two differential elements

$\frac{\partial \mathbf{p}}{\partial \theta} d\theta$

and

$\frac{\partial \mathbf{p}}{\partial \phi} d\phi$

Here I’ve assumed spherical polar coordinates, with a point on the surface parameterized by:

$\mathbf{p} = (\sin\theta \cos\phi, \sin\theta \sin\phi, \cos\theta)$

you can really use whatever parameterization you want, but the area of the pair of differential elements that bound the parallelogram will be of the form:

$dA = \left\vert \frac{\partial \mathbf{p}}{\partial \theta} d\theta \times \frac{\partial \mathbf{p}}{\partial \phi} d\phi \right\vert$

This differential form $dA$ is really the “area element”, although it happens to contain a $d\theta d\phi$ because there is neccessarily such a product if those are our two parameterization variables.

It is this $dA$ that you want to integrate, in this case, you’ll end up with

$\int_{\phi=0}^\pi \int_{\delta\theta} r^2 \sin\theta d\theta d\phi$

Reply
Adam Glesser says:

March 9, 2010 at 6:29 am

To put Peeter’s comment a different way: Heuristically, the answer is that you are already “putting in the $r$ “. Recall that the reason you get $r\ dr\ d\theta$ when you switch from rectangular to polar coordinates is that $r$ is the norm of the Jacobian of the transformation. This is essentially a multiplier converting the rectangles you start with to the parallelograms you finish with. Note that in the special case of a transformation from $\mathbb{R}^2 \to \mathbb{R}^2$ , the norm of the Jacobian equals $||\vec{\rho}_u \times \vec{\rho}_v|| (where$ latex \vec{\rho}$ is the vector function describing the transformation).

So, in your case, the transformation is occurring when you compute $||\vec{\rho}_u \times \vec{\rho}_v||$ .

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DavidC says:

March 9, 2010 at 8:26 pm

Probably you and your students would probably not be confused if you were making this calculation for a surface other than a sphere, right? Then you would have ‘du dv’ (or whatever) and not think that ‘dA’ ought to be something else. So the theorem is applying the same way here, but the notation is similar to other things, so it gets confusing.

Here’s my story:

For me, I think ‘dA’ is always an area element for a particular surface, sometimes just a plane, sometimes something more complicated. Parametrizing a surface is about mapping a plane to that surface via some ‘chart.’ And then you relate an area element of the one surface to one for the other, so here $\left\Vert \frac {\partial\vec{p}}{\partial u}\times\frac{\partial \vec{p}}{\partial v}\right\Vert dA$ is an area element for your parametrized surface, in terms of the variables in your parametrization and your area element (dA) for a plane.

But sometimes we parametrize a part of a plane with another part of a plane! (Weird!) Polar coordinates do this. (It would help here if I could draw a box whose sides are measured by $r$ and $\theta$ , and an arrow mapping that to some kind of pie slice. But you can imagine the picture instead…) An area element for the pie slice is $r dA$ (that’s intentionally odd-looking), where for me here, $dA$ is an area element of the box you’re mapping from, which is also $dr d\theta$ .

(I don’t think I’m saying anything really different from anyone else, but maybe more different ways of saying the same thing is helpful.

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Joshua Zucker says:

March 9, 2010 at 9:03 pm

I also think the answer here is super-awesome: the correspondence between each slice of the sphere and a corresponding piece of the cylinder that encloses it!

It’s easy to see that geometrically, too: the cylinder circle has radius R instead of r for the sphere slice, but the sphere’s surface curves in, and if you look at the triangles you can see that they’re similar, so the amount of area you lose with the smaller r is exactly balanced by the area you gain by going up diagonally instead of vertically.

I hope that makes some sense.

And to echo what everyone has said above, here you are treating r and theta like arbitrary parameters to map the region you are integrating over, and the norm of the cross product part takes care of the area element as well as the size of the surface. You’d only need to put the r dr dtheta in if you calculated that cross product part in cartesian coordinates and then wanted to transform the dx dy over.

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samjshah says:

March 10, 2010 at 3:32 am

Hi all,

Wow, I thought my gibberish wouldn’t make sense. But you guys totally got on it, and what you say is exactly the direction I was thinking. Thank you so much. My kids will be so happy to hear that.

Here’s where I was coming from. The reason I thought it might have to be $r dr d\theta$ is because our book does this section on double integrals of polar regions, and they basically say that to integrate $f(r,\theta)$ over a region $R$ , you will basically have to integrate $\int_{\alpha}^{\beta} \int_{r_1(\theta)}^{r_2(\theta)} f(r,\theta) r dr d\theta$ . And they show a derivation of where that $r dr d\theta$ comes from. So I just got in the habit of sticking $r dr d\theta$ for polar. I always that that was $dA$ for polar and the book intimated that too.

Jacobians will be coming up very soon, and yes, that does also help clarify my conclusion. You can see the $r$ come out of that.

Again, thank you everyone. You’re awesome. My kids also will think so too.

Sam

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