# Area of countries given by GeoBoundaries does not equal area in database

In[3]:= GeoArea[CountryData["World", "FullPolygon"]]/CountryData["World", "LandArea"]

Out[3]= 0.918839


Why don't the polygons of the World add up to the world's land area? I've read CountryData and the areas of the world but think this is a different issue, since Antarctica is part of the world.

I realize GeoArea[CountryData["World", "FullPolygon"]] doesn't include the part of Antarctica south of 89.9 degrees, but this isn't enough to compensate for the difference.

• I suspect it is because Polygons are just an approximation to the full geo boundary. – David G. Stork Jul 24 '18 at 1:00
• @DavidG.Stork this makes it difficult to do geographical analysis. I assumed Mathematica's curated data, especially using FullPolygon, would be complete. Are you saying there are more complete sources? – user1722 Jul 24 '18 at 17:49
• No FullPolygon will go down to 1 cm of coastline (for instance)... every polygon is an approximation. – David G. Stork Jul 24 '18 at 22:17
• Actually CountryData["World", "FullPolygon"] does not contain Antarctica. That accounts for the missing 8%. Check with GeoArea[Entity["GeographicRegion", "Antarctica"]] / CountryData["World", "LandArea"]. – jose Jul 27 '18 at 20:57
• @jose You are correct and I actually ran into this earlier at github.com/barrycarter/bcapps/blob/master/STACK/bc-equ-dist.m and forgot. The only part I remembered was that Antarctica's polygons stop at 89.9 degrees south. If you make your comment an answer, I'll accept it. – user1722 Jul 27 '18 at 21:07

Antarctica is not included in CountryData["World", "FullPolygon"]. Its area explains the missing 8%:

GeoArea[Entity["GeographicRegion", "Antarctica"]] / CountryData["World", "LandArea"]
(* 0.0838191 *)


You can obtain a polygon that contains Antarctica using the entity

world = Entity["GeographicRegion", "World"];


Then we have

GeoArea[world] / world["LandArea"]
(* 0.98607 *)


and now the 1.4% missing can be attributed to imperfections of the polygons.