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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 $A_{sphere}=5.12\times 10^8 km^2$

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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 $$A=2\pi R^2 \left(2-\frac{2 e^2}{3}+O\left(e^4\right)\right)=4\pi R^2- \frac{4 \pi e^2}{3} R^2+O\left(e^4\right)$$

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  • $\begingroup$ @LeylaAlkan. See my edit. We truncate the Taylor series to $O(e^4)$. $\endgroup$ – Claude Leibovici Apr 19 at 12:34
  • $\begingroup$ So, what about the last part of the question? $\endgroup$ – Leyla Alkan Apr 19 at 13:32
  • $\begingroup$ @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. $\endgroup$ – Claude Leibovici Apr 19 at 13:39

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