Finding volume of a modified sphere using integrals

Click For Summary

Discussion Overview

The discussion revolves around calculating the volume of a modified sphere, specifically an olive shape, after it has undergone a process of capping and punching. Participants explore the application of calculus, particularly integrals, to determine how the original diameter of the olive affects the final volume of the pitted olive.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • A participant describes a project involving an olive modeled as a sphere that is modified by cutting caps and punching a hole, seeking help with setting up the necessary integrals.
  • Another participant suggests that the geometry of spheres and spherical caps is well-known and provides a link to a relevant resource, implying that calculus may not be necessary.
  • One participant proposes a method to calculate the volume of the caps cut off the sphere using integrals, detailing the geometric relationships involved and suggesting a formula for the volume of the caps.
  • A later reply reiterates the suggestion that calculus may not be needed, but the original poster emphasizes the requirement to use calculus for their project.

Areas of Agreement / Disagreement

Participants express differing views on whether calculus is necessary for the problem, with some advocating for its use while others suggest that geometric reasoning could suffice. The discussion remains unresolved regarding the best approach to take.

Contextual Notes

There are limitations in the assumptions made about the geometry and the specific requirements of the project, as well as the dependence on the definitions of the terms used in the calculations.

Parker Tibbs
Messages
2
Reaction score
0
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.

Thanks in advance!
 

Attachments

  • IMG_1145.JPG
    IMG_1145.JPG
    27.7 KB · Views: 554
Physics news on Phys.org
Parker Tibbs said:
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.

Thanks in advance!
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
 
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}.
 
SteamKing said:
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

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).
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K