Imagine you're immerging a shpere, with a diameter d, in a liquid. What will be V(x), where:

x is th height of the immerged part of the sphere

V(x) the volume of the part that is immerged

 

for x=d, V(x)=4/3πd^3 and for x=d/2, V(x)=4/6πd^3

It´s often helpful to draw a sketch for such questions. I assume you´re talking about a scenario I drew in the sketch atttached below.

 

The key to calculating volumes is integration. In this case you can -losely speaking- add up all the circle slides lying in the water to get the answer.

So [math] V(x) = \int_0^x \pi r(h)^2 dh [/math]

r(h) here denotes the radius of the circle at height h (where h=0 is at the bottom and h=d is at the top). Now the trick you can use is that r(h)² = h*(d-h). Integrate.

For the volumn of a sphere, I have always used 4.188 (4/3 pi) times the radius cubed.

 

Works for me.....

 

Oops, sorry I didn't notice the part about it being only partially immersed in the liquid.

 

This site seems to have it worked out....

http://www.monolithic.com/plan_design/calcs/