# Earth's surface area

Here we are trying to calculate the earth's surface area via geodetic coordinates: $$x=(Rp(\lambda)+h)\sin (\lambda)\cos(\phi)$$
$$y=(Rp(\lambda)+h)\sin (\lambda)\sin(\phi)$$
$$x=((1-e^2)Rp(\lambda)+h)\cos (\lambda)$$
where $$p(\lambda)=\frac 1 {\sqrt{1-e^2\cos ^2 \lambda}}$$

We first compute the flat metric $$ds^2$$ in these coordinates:
$$ds^2=dh^2+(h+(1-e^2)Rp^3)^2d\lambda ^2+\sin ^2\lambda (h+Rp)^2d\phi ^2$$

Then we compute induced metric on the surface by putting $$h=0$$
$$ds_*^2=R^2p^2((1-e^2)^2p^4d\lambda ^2+\sin ^2\lambda d\phi ^2)$$

where the determinant of this metric is $$g_*=R^4p^8(1-e^2)^2\sin ^2\lambda$$

We found that $$A=\int_0^{2\pi}\int_0^{\pi}\sqrt{g_*} d\lambda d\phi=2\pi R^2(1-e^2) [\frac{1}{1-e^2}+\frac{arctanh(e)}{e}]=2\pi R^2[1+ (\frac{1}{e}-e)arctanh(e)]$$

Now from here my professor wrote that
$$e=0 \implies A=4\pi R ^2$$ and $$e\approx 0 \implies A\approx 4\pi R ^2 - \frac 4 3 \pi R^2 e^2 + O(e^4)$$, and I didn't understand how he got these.
Also from here how do we get that for the earth: $$A_{ellipsoid}=5.10\times 10^8 km^2$$ and $$A_{sphere}=5.12\times 10^8 km^2$$

Consider $$Z=1+\left(\frac{1}{e}-e\right) \tanh ^{-1}(e)$$ Now, by Taylor around $$e=0$$ $$\tanh ^{-1}(e)=e+\frac{e^3}{3}+\frac{e^5}{5}+O\left(e^7\right)$$ $$Z=1+\left(\frac{1}{e}-e\right)\left(e+\frac{e^3}{3}+\frac{e^5}{5}+O\left(e^7\right)\right)=2-\frac{2 e^2}{3}-\frac{2 e^4}{15}+O\left(e^6\right)=2-\frac{2 e^2}{3}+O\left(e^4\right)$$
Then $$A=2\pi R^2 \left(2-\frac{2 e^2}{3}+O\left(e^4\right)\right)=4\pi R^2- \frac{4 \pi e^2}{3} R^2+O\left(e^4\right)$$
• @LeylaAlkan. See my edit. We truncate the Taylor series to $O(e^4)$. – Claude Leibovici Apr 19 at 12:34
• @LeylaAlkan. Now, you must use the value of $e$ and apply the formula. Have a look at mathworks.com/help/map/ref/wgs84ellipsoid.html for its value. – Claude Leibovici Apr 19 at 13:39