The surface temperature of the earth is expressed by spherical coordinates:

$\begin{split} T(\phi,\theta)=-30+60\sin\phi \end{split}$

I want to calculate the temperature at the poles (north and south) and the equator and the average temperature.

Already done:

$$\overline T = \frac{1}{A}\iint_{S(R)}T\,dS $$

$x = r\sin \theta\cos\phi$

$y = r\sin\theta\sin\phi$

$0\le\phi \le 2\pi$


$$\frac{\delta(x,y)}{\delta(\phi,\theta)}= \begin{vmatrix} r\cos\theta\cos\phi & r\cos\theta\sin\phi \\ -r\sin\theta\sin\phi & r\sin\theta\cos\phi \\ \end{vmatrix}$$ and that is $r^2\sin\theta\cos\phi$ which is dS(?)

I need to calculate the dS and after that I can calculate the area and after that the temperatures, but how?

How do I proceed from here?

  • $\begingroup$ And with dS I can calculate the A of a sphare is that right? $\endgroup$
    – vvv
    Oct 3 '19 at 17:43
  • $\begingroup$ @trula: Check your formula. $\endgroup$ Oct 3 '19 at 17:43
  • $\begingroup$ Spherical coordinates do not give you a coordinate system at the poles (note that $\phi$ can take any value there), so you have no idea what the temperature will be at the poles. The formula that's given shows that the temperature cannot be continuous at $\theta=0,\pi$. $\endgroup$ Oct 3 '19 at 17:44
  • $\begingroup$ Okey but how is the equator and the average temperature? $\endgroup$
    – vvv
    Oct 3 '19 at 17:47

First your formula, wich says temperature changes going east or west, but not going north or south seams very wrong, check it!

To your calculation, the spere lives in 3d, but you forgot the third coordinate $$z=r*cos(\theta)$$ so your dS is wrong the surface of a sphere you should know as $$A=4*\pi*r^2, dS=r^2sin(\theta)*d\phi*d\theta$$


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