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So basically I wanted to know, what percentage of the Northern hemisphere's surface area lies below a given latitude $\phi$?

So my idea was to first find the volume of the Earth up to that latitude and then differentiate the answer with respect to the radius (if this is invalid, please let me know; it seems to be valid- for example $\frac{4}{3}πr^3$ becomes $4πr^2$ when differentiated)

Think of chopping off the top bit of the Earth at a given latitude and chopping at the equator and then measuring the volume of that part of the truncated object.

I created a running variable $h$ for height above equatorial plane and realised the value varies from $0$ to $R*sin(\phi)$ Then for the volume we get $\int_0^{R*sin(\phi)} π(R^2-h^2)dh= πR^3*sin(\phi)-\frac{πR^3*sin(\phi)^3}{3}$

Differentiating for the surface area then gives $3πR^2*sin(\phi)-πR^2*sin(\phi)^3$

Comparing this with $2πR^2$ should give the correct proportion. But it doesn't seem to be giving the right results. For example wikipedia says on their page https://en.m.wikipedia.org/wiki/49th_parallel_north#:~:text=The%2049th%20parallel%20north%20is,the%2048th%20and%2049th%20parallels. that slightly less than one eight of the world is above that latitude while my estimate would yield around 8%

Where have I gone wrong? Is it just the bulge of the Earth's equator that somehow accounts for the error?

aman
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1 Answers1

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In general, you cannot differentiate volume of a region to find its surface area. For sphere, $x^2 + y^2 + z^2 = R^2$, the surface area element in cartesian coordinates is given by,

$ds = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial x}\right)^2} ~ dx ~ dy$

Note $z = \sqrt{R^2 - x^2 - y^2}$ for the half sphere above xy-plane.

But given you have the angle $\phi$ above xy-plane, it is easier to do this in spherical coordinates. Also note that this is for a spherical surface and will of course not be accurate for earth which is not a true spherical shape.

Surface area element in spherical coordinates is $R^2 \sin \varphi ~ d\varphi ~ d\theta$ and integral to find required surface area is,

$ \displaystyle \int_0^{2\pi} \int_0^{\pi/2 - \phi} R^2 \sin \varphi ~ d\varphi ~ d\theta = 2 \pi R^2 (1 - \sin \phi)$

Math Lover
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  • Thanks, I'm still getting the hang of applying integral calculus to 3D so this was helpful. Is there any intuitive way to determine when differentiating the volume gives the surface area? – aman Oct 24 '21 at 06:41
  • Sorry for a late reply. Here is a link you will find interesting that tries to answer your question: https://projecteuclid.org/journalArticle/Download?urlid=10.1216%2Frmjm%2F1181068766 – Math Lover Oct 24 '21 at 18:41