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I am currently thinking about computing the surface area of surfaces in $\mathbb{R}^3$ through the lens of a traditional multivariable course, and I'm a bit confused about how parameterizations are meant to work. As a toy example, let's say I want to find the surface area of the sphere $x^2 + y^2 + z^2 = 1$. I can use the parameterization given by

$$r(u,v) = \langle \cos(u)\sin(v), \sin(u)\sin(v), \cos(v)\rangle$$

and set up the corresponding integral

$$\int_0^{\pi}\int_0^{2\pi} \sin(v)~dudv$$

where $\sin(v) = \|r_u \times r_v\|$. This is all fine and good. However, if I want to think about my parameterization landing in $\mathbb{R}^3$ from the perspective of spherical coordinates, I can use the parameterization

$$r(u,v) = \langle 1, u, v \rangle.$$

In this case, the function $\|r_u\times r_v\|$ is just equal to 1. If I take this at face value, I should think that I've done something wrong since this would give me $4\pi^2$ as the surface area, not $4\pi$. However, If I close my eyes and throw in the Jacobian for spherical coordinates as it is along the sphere, then I get the "correct" integral.

Why should I use the Jacobian here? I'm not doing a triple integral, and there's no volume form running around here. I'm strictly integrating a function $f(u,v)$ over a region in the $uv$-plane. What gives?

TheAssistant
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  • (P.S. If you need to use some differential topology to answer this question, that's perfectly fine.) – TheAssistant Jul 15 '22 at 00:01
  • Not sure I understand the question: If the angle brackets denote Cartesian coordinates, then $r(u, v) = \langle 1, u, v\rangle$ is a parametrization of a plane, not part of a sphere. If instead $\langle 1, u, v\rangle$ connotes values substituted into spherical coordinates, then $|r_u\times r_v| = \sin v$ as expected. – Andrew D. Hwang Jul 15 '22 at 01:44
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    @Andrew Here the angle brackets are just denoting a vector in $\mathbb{R}^3$, irrespective of the coordinate system. – TheAssistant Jul 15 '22 at 01:51
  • I see: So you're really asking why a length/area/volume element is not just the product of coordinate differentials, but includes the Jacobian factor? (From the wording I understood something different.) – Andrew D. Hwang Jul 15 '22 at 21:23
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    I think I'm confused about why the Jacobian comes into play, even though the object/region I'm integrating over isn't "in" $\mathbb{R}^3$ or "in" spherical coordinates. There is no three-dimensional volume form here, because the integral is in the $uv$-plane. Right...? – TheAssistant Jul 15 '22 at 22:18
  • Possibly of interest: What does it mean to multiply differentials? – Andrew D. Hwang Jul 16 '22 at 11:33

1 Answers1

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The reason is because the term $r^2sin(\theta)d\theta d\phi$ is literally what you call $r_u \wedge r_v$$dudv$. Namely, the term $r\sin(\theta)d\phi$ is the increment of arclength along the $\phi$ direction and $rd\theta$ is the arclength increment along the $\theta$ direction, and to find the incremental area you just multiply them (or wedge them) together resulting in $r^2sin(\theta)d\theta d\phi$ (you can do that, since they form a pointwise orthogonal frame/basis). That makes it clear why the Jacobian appears in computing the area and it should also give you an idea what the Jacobian factor means geometrically. To find the incremental volume in spherical coordinates you just wedge along the increment in the $r$ direction resulting in $r^2sin(\theta)d\theta d\phi dr$, hence why it appears in computing the area, because computing the volume just adds a factor of $dr$ to it. My advice to you: Revise the spherical coordinates from a geometric perspective, it will make all these formulas more clear.

Leonid
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