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In class, we are learning to find the surface area using the formula $\iint_D \|r_u \times r_v\|dA$ (where $u$ and $v$ are the variables the parameterization uses). The first question was about a sphere with radius R. Instead of writing the parameterization as

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

I wrote it in spherical coordinates as

$r(\theta, \phi) = \langle R, \theta, \phi\rangle$

After finding that $r_\theta = \langle0,1,0\rangle$ and $r_\phi = \langle0,0,1\rangle$, I found that $r_\theta \times r_\phi = \langle1,0,0\rangle$ and $\|\langle1,0,0\rangle\| = 1$. Integrating that (with the proper Jacobian determinant):

$\int_0^{2\pi} \int_0^\pi (1)R^2\sin(\phi) d\phi d\theta$

$-R^2\int_0^{2\pi}(\cos(\phi) |_0^\pi)d\theta$

$-R^2\int_0^{2\pi}-2d\theta$

$-R^2(-4\pi)$

$4\pi R^2$

This is the correct answer, however, my teacher disagrees that I can use spherical coordinates in a two-dimensional integral. I tend to agree with him, but I'd like to know whether this method will always yield the correct answer or if it only works in the case of a sphere.

user170231
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AndrewEllis
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    For starters, tell us what $\langle 1,0,0\rangle$, $\langle 0,1,0\rangle$, etc., mean. The usual formula for cross product is based on the cartesian coordinate system, where $\vec i,\vec j,\vec k$ give a fixed (constant) reference frame. – Ted Shifrin Nov 14 '23 at 16:58
  • @TedShifrin As I said, I only learned this topic today. I don't think I understand it well enough to be able to answer your question. As far as I know, ⟨1,0,0⟩ is a vector that is orthoganal to both partial derivatives, which are the other vectors listed. – AndrewEllis Nov 14 '23 at 18:14
  • I suggest you use Euclidean coordinates for your vectors until you can begin to understand what you're writing. There are, indeed, unit vectors in the $r$, $\theta$, and $\phi$ directions, but first you need to know what that means and second you need to realize that they vary from point to point, so their derivatives are very much nonzero. – Ted Shifrin Nov 14 '23 at 18:19
  • The jacobian in spherical coordinates is not the cross-product of the component vectors as it is in Cartesian coordinates. To understand more deeply, I suggest you draw a picture of your sphere, with a plane tangent to the sphere at some point. What is the image of $d\theta$ and $d\phi$ on the sphere and the area of the rectangle formed by these two vectors? – user317176 Nov 14 '23 at 19:05
  • Does this answer your question? Surface Element in Spherical Coordinates – Kurt G. Nov 14 '23 at 19:19

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