# Need help with volumes of spheres with holes.

1. Dec 3, 2006

### timm3r

1. The problem statement, all variables and given/known data

if you guys have trouble reading that, it says,

Suppose you make napkin rings by drilling holes with different diameters through two wooden balls (which also have different diameters). You discover that both napkin rings have the same height h, as shown in the figure.
a) Guess which ring has more wood in it.
b) Check your guess: use cylindrical shells to compute the volume of a napkin ring created by drilling a hole with radius r through the center of a sphere of radius R and express the answers in terms of h.

2. Relevant equations
I know how to find the volume of this using discs, which the equation would be
integral[-R, R] sqrt[R^2 - X^2] - integral[-r, r] sqrt[R^2 - x^2]

3. The attempt at a solution
i have no idea how to answer this in terms of h, can someone please help me put this in terms of h.

2. Dec 3, 2006

### StatusX

There is a pretty simple relation between h, the radius of the sphere, and the radius of the hole that follows from pythagoras' theorem.

3. Dec 3, 2006

### timm3r

so if we use the pythagoras theorem, 2r would be the relation between the radius of the hole, what would be the relation of the radius of the sphere?

4. Dec 3, 2006

### Grogs

timm3r,

Draw a triangle with one vertex at the center of the sphere, the 2nd at the top wall of the cylinder, and the 3rd straight down from the second so that it forms a 90 degree angle with the first two. What is the length of the 3 sides of the triangle? That should give you your relationship between R, r, and h.

5. Dec 4, 2006

### timm3r

im still stuck on this problem, can someone make it more dumby proof for me?

6. Dec 4, 2006

### timm3r

ok i still don't get it, this is what i did

i still don't see how there is a relation between the radius of the circles.

7. Dec 4, 2006

### StatusX

It's hard to make it any clearer. One side of that triangle you drew is h/2, one is the radius of the sphere, and one is the radius of the hole. It's a right triangle, so you can use pythagoras' law.

8. Dec 4, 2006

### timm3r

ok did you see my picture? so if we go back to the c^2=a^2+b^2, a would be h/2, b would be the radius of the the hole, and c would be the radius of the sphere right?

9. Dec 5, 2006

### HallsofIvy

Staff Emeritus
How about writing that in the terms you are given: If r is the radius of the hole and R is the radius of the sphere, then $r^2+ (h/2)^2= R^2$. Since h, and so h/2, is the same in both spheres, we have R^2- r^2 the same in both spheres.