Finding volume of a modified sphere using integrals

In summary: 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^
  • #1
Parker Tibbs
2
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!
 

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  • #2
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
 
  • #3
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 [itex] \sqrt{R^2- d^2/4}[/itex]. At each point at height h above the plane of the cut, a cross section would be a disk of radius [itex]\rho= \sqrt{h^2+ d^2/4}[/itex] so area [itex]\pi (h^2+ d^2/4)[/itex]. 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 [itex]\pi\int_0^{\sqrt{R^2- d^2/4} (h^2R^2- d^2/4}[/itex]. (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) [itex]2\sqrt{R^2- d^2/4}[/itex].
 
  • #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).
 

Related to Finding volume of a modified sphere using integrals

1. How do you find the volume of a modified sphere using integrals?

To find the volume of a modified sphere using integrals, you first need to determine the formula for the modified sphere. This can be done by using the Pythagorean theorem and the equation of a circle. Once you have the formula, you can set up an integral with the appropriate limits and calculate the volume.

2. What is a modified sphere?

A modified sphere is a three-dimensional shape that is formed by rotating a curve around an axis. This curve is typically not a perfect circle, resulting in a shape that is not a perfect sphere. Examples of modified spheres include ellipsoids and oblate or prolate spheroids.

3. Why is it necessary to use integrals to find the volume of a modified sphere?

Integrals are necessary to find the volume of a modified sphere because the shape is not a perfect sphere with a known formula. By using integrals, we can break down the shape into infinitesimally small sections and calculate the volume of each section, then add them together to get the total volume.

4. What are the limits of integration when finding the volume of a modified sphere?

The limits of integration will depend on the specific formula for the modified sphere. In general, the limits will involve the variables used in the formula and the boundaries of the shape. For example, if the modified sphere is bounded by a certain radius, the limits of integration may involve that radius.

5. Can the volume of a modified sphere be found using other methods besides integrals?

Yes, the volume of a modified sphere can also be found using other methods such as using geometric formulas or approximating the volume using numerical methods. However, using integrals is the most precise method as it takes into account the varying shape of the modified sphere.

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