Basically, this is a classic physics problem. Very easy. AND, physics was meant to be done with vector calculus. Anyways,, to make this problem easy, we use spherical coordinates. Here is a link to Wolfram, which will help you with understanding the spherical coordinate system (if you don't know already that is):
http://mathworld.wolfram.com/SphericalCoordinates.html
So, the area element in spherical coordinates is given by dA = (r^2)*sin(phi)*dphi*dtheta. SO, what we do is integrate this area element to find the surface area of the submerged portion. What we know that theta ranges from 0 to 2pi, just because of how the coordinate system is set up. But we DON'T know what phi ranges to, so we leave it as a variable (let's call it x).
A = Double Integral((R^2)*sin(phi)*dphi*dtheta, 0<theta<2pi, 0<phi<x)
=
2pi(R^2)[1-cos(x)]
But, this equals 72 percent of the total surface area, so (0.72)4pi(R^2)
Which, solving for [1-cos(x)], gives us [1-cos(x)] = 1.44
We didn't explicitly solve for x because it is more efficient to leave it this way, as you will see later.
Now we know what the ranges are for theta and phi, but we still need to find the radius. So now, we have to solve for the volume.
The volume element in spherical coordinates is dV = (r^2)*sin(phi)*dphi*dtheta*dr. So we integrate over the ranges of the variables. (0<theta<2pi, 0<phi<x, 0<r<R)(note: x is known, but R is unknown)
V = Triple Integral((r^2)*sin(phi)*dphi*dtheta, 0<theta<2pi, 0<phi<x, 0<r<R) = (2/3)pi[1-cos(x)](R^3) = (2.88/3)pi(R^3) = 280000 cm^3
Which, after solving for R and plugging into the equation for the total volume, gives you the final answer:
V = (4/3)pi(R^3) = 388 888.889 = 388889 cm^3
