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 is because our book does this section on double integrals of polar regions, and they basically say that to integrate over a region , you will basically have to integrate . And they show a derivation of where that comes from. So I just got in the habit of sticking for polar. I always that that was 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 come out of that.

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

Sam

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.

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 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 and , 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 (that’s intentionally odd-looking), where for me here, is an area element of the box you’re mapping from, which is also .

(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.

So, in your case, the transformation is occurring when you compute .

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

and

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

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:

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

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

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

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