Please correct, Solid Geometry

[Physicsrapper]

The earth's radius R is 6370 km. Find the height of the water, if 10 000 km^3 of water would have flooded a disc shaped earth.

If a disc-shaped world would be covered with water, wouldn't it be a cylinder then?

So, the cylinde formulas:
Volume: π*r^2*h
Surface area: π*r^2

Surface area of the earth: π * 6370^2 = 127476090.9

Volume / Surface area = hight

10 000 = 127476090.9 * h

10 000 / 127476090.9 = h

h = 7.8 cm

In the solution it should be 5.9 cm. What did I wrong?

 

[Quaid]

The earth's radius R is 6370 km. Find the height of the water, if 10 000 km^3 of water would have flooded a disc shaped earth.

If a disc-shaped world would be covered with water, wouldn't it be a cylinder then?

Hello. A disk (aka disc) is the region of a plane bounded by a circle. Water cannot "flood" a 2-dimensional disk, taking on cylindrical shape, but disks can be extended into 3-dimensional space!

In the given context of a planet covered with water, we need to interpret "disc-shaped earth" as a 3-dimensional disk. (The general name for a 3-dimensional disk is "ball"; it's the space inside a sphere.)

Thinking in three dimensions leads to the approaches posted in the other thread for this exercise: write an equation to model the difference in volume between the two spheres as 10000 km^3, with the variable representing the difference in radii between the two spheres (because that difference is the unknown depth of the water).

Rounding the Real solution (converted to cm) gives the previously-posted answer of 2 cm.