Sphere in Swimming Pool

Forget the sphere. Instead just imagine a water column that extends from the surface to the depth of the sphere such that its upper end is a cylinder of radius R, and the lower end has a spherical end (like the lower end of a test tube) with radius R,

BASICALLY A HEMISPHERE ATTACHED TO CYLINDER OF HEIGHT 2R and RADIUS R

Clearly this is in equilibrium , because this is a water column in a water body, so its weight is equal to the upper force exerted by lower part of water, or

$F=W=mg=\frac { 2 }{ 3 } \rho \pi { R }^{ 3 }g+\rho \pi { R }^{ 2 }g(2R)\quad =\frac { 8 }{ 3 } \rho \pi { R }^{ 3 }g$ .

or answer is 14 !

Nice innovative problem !

let me tell a good method ..... $F_{upward}$ - $F_{Down}$ = buoyant force , since buoyant force can be easily calculate from it's volume, thew answer you get will provide a = 8 , b= 3 , c= 3 , easy enough ! :)

Archimedes' Principle can be used. In this scenario, there are 2 forces: upward and downward force, both exerted by the fluid. However, upward > downward, so there's a buoyant force (Fup - Fdown = Fbuoy = weight of water displaced). The forces vary throughout the sphere's surface, so here's the trick: split the sphere in half such that we only consider the bottom half of the sphere (force exerted on bottom of sphere won't change in both cases). Since the top of the bottom hemisphere is a circle, we know Fdown right away, and can use it to find Fup.

well i did it with integration!.(and got d right answer!) step 1: take a small ring element which subtends an angle (theta) at the centre next write the expressesion for force at that small area and multiply it by cos(theta) because by symmetry the sine theta components cancels out you will get something like this 2πR^2 ρg[h sin(theta)*cos(theta) + Rsin(theta) * cos^2(theta)]• d(theta) (dA = 2πR^2 *sin(theta)•d(theta) NOW INTEGRATE,PUTTING LIMITS FROM(0,π/2)