Biology Forums - Study Force

Hello. This is my first post on the forum, I hope I'm posting in the right place.

I have a problem related with cell biology that certainly is very common in practice: estimate how many cells there are inside something only have some slide.

My problem is as follows.

Let D a cylinder of height H=5000, with circular base of radius R=500, which contains cells (spheres) of radius r = 5 floating in a liquid uniformly distributed.

We freeze D and make a slide parallel to the base of D with height h=5 (a slide like a coin of radius R=500 and height h=5). We can count the number of pieces (slides) of cells contained inside the "coin" (of course, each slide of cell inside the "coin" has different radius).

I need to estimate how many cells there are inside of D, but I also have one slide of D.

How can I estimate how many spheres there are in D? What mathematical tools can I apply?
Or maybe I need to change a bit the statement in order to have some practical solution.

Thanks in advance!

You should find the area of the cylinder by using the formula for a circle.

A = pi*r*r

Let me understand, you want to know how many cells, of radius 500, fit on this area?
/me This is a very strange way of calculating cell volume.

A have a cylinder (like a glass of water) with radius R=500 and height H=5000 containing cells. The interior of the cylinder is solid, and then I can cut the cylinder to see its interior.

The cells are spheres with radius r=50, and are floating in the cylinder. So, the cylinder is not full with cells, but I know that the cells are uniformly distributed in all the cylinder.

I want to know how many cells there are in the cylinder, but I only have an slide (like a coin with  radius R=500 and height h=5), so in the slide I have only have parts of cells.

Aha, so what to know how many little circles (the cells) will fit into a larger circle. If so, that's hard to do lol

I believe there is a calculator found here (http://www.engineeringtoolbox.com/smaller-circles-in-larger-circle-d_1849.html) that can do such calculation.