# The surface temperature of the earth in spherical coordinates

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.

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

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

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

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

$$0\le\theta\le\pi$$

$$\frac{\delta(x,y)}{\delta(\phi,\theta)}= \begin{vmatrix} r\cos\theta\cos\phi & r\cos\theta\sin\phi \\ -r\sin\theta\sin\phi & r\sin\theta\cos\phi \\ \end{vmatrix}$$ 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?

• And with dS I can calculate the A of a sphare is that right?
– vvv
Oct 3 '19 at 17:43
• @trula: Check your formula. Oct 3 '19 at 17:43
• 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 $\theta=0,\pi$. Oct 3 '19 at 17:44
• Okey but how is the equator and the average temperature?
– vvv
Oct 3 '19 at 17:47

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 $$A=4*\pi*r^2, dS=r^2sin(\theta)*d\phi*d\theta$$