# Finding volume of a modified sphere using integrals

1. Nov 26, 2015

### Parker Tibbs

Hello guys, new member here. I've got a calculus project due Tuesday that I could use some help on.

I won't bore you with the all details of the project, but first let's imagine an olive in the shape of a perfect sphere (with a radius always bigger than 6mm) that goes through a set of blades 6mm apart. This chops congruent caps off each side of the olive. Then, a machine takes a punch and punches out the center of the olive. The diameter of the punch always matches the diameter of the circle left behind when the caps were lopped off by the blades.

I need to verify if the diameter of the original olive will affect the volume of the final pitted olive. I was thinking I could set up a double or triple integral to find the volume of a sphere centered at the origin in between the planes z=-3 and z=3 with the bounds on my radius being the radius of the pitted portion and the overall radius of the olive. However, I'm not sure exactly how to set it up. Any ideas?

If you're having trouble picturing this I've attached a sketch to help visualize the problem.

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2. Nov 26, 2015

### SteamKing

Staff Emeritus
You don't need to resort to calculus to analyze this problem unless it's been a while since you've flogged yourself.

The geometry of spheres and spherical caps is well-known:

https://en.wikipedia.org/wiki/Spherical_cap

3. Nov 26, 2015

### HallsofIvy

Do it in two sections. First determine the volume of the two "caps" that are cut off. This is only one "section" because the two caps are obviously the same. If you draw a circle of radius R, representing a cross section of the sphere, then the horizontal line, of length d, representing the diameter of the cap, a line, of length R, from the center of the circle to the point where that horizontal line touches the circle, and the vertical line from the center of the circle perpendicular to the horizontal line, form a right triangle with hypotenuse of length R, one leg of length d/2, and so the other leg (the vertical line) of length $\sqrt{R^2- d^2/4}$. At each point at height h above the plane of the cut, a cross section would be a disk of radius $\rho= \sqrt{h^2+ d^2/4}$ so area $\pi (h^2+ d^2/4)$. To find the volume of each cap, imagine a number of "layers" of discs at height h with thickness of each "dh". Then the volume of the whole cap is $\pi\int_0^{\sqrt{R^2- d^2/4} (h^2R^2- d^2/4}$. (Or you could use the formula SteamKing directed you to!)

The hole connecting the two caps is easier. It is a cylinder of radius d/2 and height (since the sphere and cylinder area symmetric about the xy-plane) $2\sqrt{R^2- d^2/4}$.

4. Nov 26, 2015

### Parker Tibbs

Yeah I realize this, but since this is a calculus project we need to use calculus to complete it (we have to submit a written report with it).