I want to calculate the surface area of a sphere with radius $R$.

As I was calculating the norm of the cross product of partial derivatives of the surface parametrisation, I stumbled to a problem.

I use the spherical coordinates as the parametrisation for the sphere.

Let $x(\phi,\theta) = R\cdot \sin(\phi)\cos(\theta)$

$y(\phi,\theta) = R\cdot \sin(\phi)\sin(\theta)$

$z(\phi,\theta) = R\cdot \cos(\phi)$

$|| r_\phi \times r_\theta || = || R\cdot \sin(\phi) (x,y,z) || = |R\sin(\phi)| \cdot||(x,y,z)||$

$\sin(\phi)$ is a function of $\phi$, so why are we able to treat it as a constant and factor it from the norm?

  • $\begingroup$ It is not a constant, it is a number. Since it is a number not a vector it can be factored out from the norm. $\endgroup$ – Paul Nov 2 '16 at 10:29
  • $\begingroup$ Hint: The norm $||(x,y,z)|| = R$ for all points on the sphere of radius $R$. (I'm not sure how you're computing $||r_{\phi} \times r_{\theta}||$. If you compute it algebraically in spherical coordinates, you find that it equals $R^2 \sin \phi$ directly. It seems like your calculation is based on a geometric argument of some sort.) $\endgroup$ – Michael Joyce Nov 2 '16 at 13:13
  • $\begingroup$ i calculate the cross product of the partial derivatives of the chosen parametrisation. here $(x,y,z)$ is simply $(R\cdot \sin(\phi)\cos(\theta), R\cdot \sin(\phi)\sin(\theta), R\cdot \cos(\phi))$ $\endgroup$ – Little Rookie Nov 3 '16 at 7:22
  • $\begingroup$ $\left\| a(x,y,z) \right\|=\left| a \right|\left\| (x,y,z) \right\|$ where a is a scalar quantity. It does not matter if a is a variable scalar, just so long as it is not a vector. $\endgroup$ – Paul Nov 3 '16 at 14:52

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