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In the task from Applied numerical analysis,there is given that, a sphere of density $d$ and radius $r$ weights $\frac43 \pi r^3d$. The volume of a spherical segment is $\frac\pi3(3rh^2-h^3)$.

Find the depth to which a sphere of density $0.6$ sinks in water as a fraction of it's radius here is also figure for clarify what is $h$ and what question is about.

enter image description here

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Use the archimedes' principle: – Listing Nov 11 '11 at 6:50
up vote 3 down vote accepted

We do an analysis, and arrive after a while at a cubic equation. This equation is then best tackled using a numerical method.

First we look at the Physics of the situation. The weight of water displaced by the sphere is the same as the weight of the sphere, by a principle that goes back to Archimedes.

Units were not specified, but it seems clear that we are expected to work in grams per cc. Then water can be taken to have density $1$, though the actual density is (slightly) dependent on temperature and pressure. So the weight of the water displaced is, in the appropriate units, equal to $1$ times the volume of the spherical cap. This is equal to the weight of the sphere. You have been given a formula for that. In our case, $d=0.6$.

The spherical cap has volume $$\frac{\pi}{3}(3rh^2-h^3).$$ So we want to solve the equation $$\frac{\pi}{3}(3rh^2-h^3)(1)=\frac{4\pi r^3}{3}(0.6) .$$ We want to find $h$ as a fraction of the radius. Note that since the density of the ball is greater than $0.5$, more than half the sphere will be submerged, unlike what is suggested by the picture!

Let $h=tr$. Substitute for $h$ in the above equation. There is a lot of cancellation, and after we do that we reach $$3t^2-t^3=(4)(0.6), \qquad\text{or equivalently}\qquad t^3-3t^2+2.4=0.$$ There is a formula for the roots of a cubic. However, it is not very useful in this situation. It is best to solve the equation numerically, using your favourite numerical procedure.

Let $f(t)=t^3-3t^2+2.4$. We know the root we are interested in is $>1$, but probably not by much. Note that $f(1.1)=0.101$ and $f(1.2)=-0.192$. So our root is between $1.1$ and $1.2$, probably closer to $1.1$. You should end up with something around $1.134$.

I do not know what method you are now expected to use. Maybe the Newton-Raphson method? If you have trouble setting that up, please leave a message.

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thanks very much @André Nicolas,yes i have missed pi in the second equation,it was my typo,in book there is correctly written,thanks very much – dato datuashvili Nov 11 '11 at 7:38
thanks ones again,it does not matter really,method of finding root is a a lot of in internet and in my book too,so thanks it was great help – dato datuashvili Nov 11 '11 at 7:52

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