Volume Of Intersection Between Square Pyramid And Sphere

In summary, the problem involves finding the volume of intersection between a square pyramid with edge lengths 2r and a sphere of radius r, where the sphere is centered at the top vertex of the pyramid. The solution involves setting up the square pyramid and sphere in three planes, finding the angles between these planes, and using a formula to calculate the solid angle formed by the intersection of the planes and the sphere. The final volume is then calculated using this solid angle and the radius of the sphere.
  • #1
EquationOfMotion
22
2
Homework Statement
Find the volume of intersection between a square pyramid with edge lengths 2r and a sphere of radius r where the sphere is centered at the top vertex of the cone
Relevant Equations
I have no idea.
I'm assuming the way to go about it is to integrate in spherical coordinates, but I have no idea what the bounds would be since the bottom edges of the square pyramid are some function of r, theta, and phi.
 
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  • #2
What is a square cone?
 
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  • #3
etotheipi said:
What is a square cone?
Huh maybe it's not called that? Cone with a square base, in this case all sides are the same length so I guess it'd be half an octahedron.

Ok wait it's a square pyramid.
 
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  • #4
Must be a square equilateral pyramid with all edges = 2r.
 
  • #5
EquationOfMotion said:
Homework Statement:: Find the volume of intersection between a square pyramid with edge lengths 2r and a sphere of radius r where the sphere is centered at the top vertex of the cone
Relevant Equations:: I have no idea.

I'm assuming the way to go about it is to integrate in spherical coordinates, but I have no idea what the bounds would be since the bottom edges of the square pyramid are some function of r, theta, and phi.
Is the side length of the square ##2r## or the edges to the top?

Anyway. I would start to draw a picture. First draw a circle of radius ##r## and place the triangle of the pyramid upside down into the circle. Then - depending on which side length it is which is ##2r## - you get the rest of the triangle data and should be able to calculate the area of the two dimensional intersection. After that you only need to calculate the volume of this area rotated.

Mnemonic: If in doubt, draw it out!
 
  • #6
fresh_42 said:
Is the side length of the square ##2r## or the edges to the top?

Anyway. I would start to draw a picture. First draw a circle of radius ##r## and place the triangle of the pyramid upside down into the circle. Then - depending on which side length it is which is ##2r## - you get the rest of the triangle data and should be able to calculate the area of the two dimensional intersection. After that you only need to calculate the volume of this area rotated.

Mnemonic: If in doubt, draw it out!

All edge lengths are 2r - I think it should be half an octahedron.

Issue is that it's not the volume of the area rotated, unless I'm very much mistaken. It's a square pyramid with a cap.
 
  • #7
EquationOfMotion said:
All edge lengths are 2r - I think it should be half an octahedron.

Issue is that it's not the volume of the area rotated, unless I'm very much mistaken. It's a square pyramid with a cap.
Yes, I saw it isn't a rotation immediately after I had written it, sorry. But you can cut it down into 4 equal pieces to use symmetries.
 
  • #8
Arghh, this does look quite tricky. Because the cap is weird... sketched crudely:

1603406468361.png


1603406500918.png
 
  • #9
I think setting up the square with vertices on the axis, and then concentrate on the main quarter/octant would make things easier. It cuts out only one plane. (I'm still struggling with the demo version of my graphics program. What do you use, @etotheipi?)
 
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  • #10
fresh_42 said:
(I'm still struggling with the demo version of my graphics program. What do you use, @etotheipi?)

I find GeoGebra is pretty good, it's also completely free!
 
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  • #11
etotheipi said:
I find GeoGebra is pretty good, it's also completely free!
Da...it. I have that but no icon on the desktop so I had forgotten it. One doesn't get younger ...
Enjoy the time!
 
  • #12
Let the centre of the sphere be O and the base of the pyramid be ABCD. Consider the planes containing OAB, OBC and OCA. These intersect the sphere's surface in three arcs.
Find the angles between the planes, hence the angles between the arcs. A neat formula relates these to the area of the triangular cap the arcs form.
Find the volume between the planes and that cap from the area.
 
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  • #13
@EquationOfMotion , were you able to follow my method in post #12 to get the answer or do you need more detail?
 
  • #14
It's been a week and, disappointingly, the OP seems to have lost interest, so I'll post the full solution.
The angle between two faces of an octahedron is ##2\arctan(\sqrt 2)##.
The arcs where the planes containing OAB, OBC and OCA intersect the sphere therefore form one angle of ##2\arctan(\sqrt 2)## and two of ##\arctan(\sqrt 2)##.
The solid angle that a triangular cap with angles α, β, γ subtends at the centre of the sphere is Ω=α+β+γ-π, so in this case ##\Omega=4\arctan(\sqrt 2)-\pi##.
The volume is ##\frac 13\Omega r^3##.
 
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  • #15
haruspex said:
It's been a week and, disappointingly, the OP seems to have lost interest, so I'll post the full solution.
The angle between two faces of an octahedron is ##2\arctan(\sqrt 2)##.
The arcs where the planes containing OAB, OBC and OCA intersect the sphere therefore form one angle of ##2\arctan(\sqrt 2)## and two of ##\arctan(\sqrt 2)##.
The solid angle that a triangular cap with angles α, β, γ subtends at the centre of the sphere is Ω=α+β+γ-π, so in this case ##\Omega=4\arctan(\sqrt 2)-\pi##.
The volume is ##\frac 13\Omega r^3##.
Very nice. Of course, there are two such triangular caps that contribute to the total volume of interest.
 
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  • #16
TSny said:
Very nice. Of course, there are two such triangular caps that contribute to the total volume of interest.
Right.. I forgot to double when typing it in.
And I could have made it easier because the triangle formula extends to any polygon! Just have to subtract more copies of pi.
 
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Related to Volume Of Intersection Between Square Pyramid And Sphere

1. What is the formula for calculating the volume of intersection between a square pyramid and a sphere?

The formula for calculating the volume of intersection between a square pyramid and a sphere is (1/3)πh(r^2 - (h^2/4)), where h is the height of the pyramid and r is the radius of the sphere.

2. How do you determine the height of the square pyramid for the volume calculation?

The height of the square pyramid can be determined by finding the distance between the center of the sphere and the base of the pyramid. This distance is equal to the radius of the sphere.

3. Can the volume of intersection between a square pyramid and a sphere be negative?

No, the volume of intersection between a square pyramid and a sphere cannot be negative. It represents the amount of space that is shared by both shapes and therefore must always be a positive value.

4. Is there a specific unit for the volume of intersection between a square pyramid and a sphere?

The unit for the volume of intersection between a square pyramid and a sphere will depend on the units used for the height and radius. For example, if the height is measured in meters and the radius in centimeters, the volume will be in cubic meters times centimeters.

5. Can the volume of intersection between a square pyramid and a sphere be greater than the volume of either shape individually?

Yes, it is possible for the volume of intersection between a square pyramid and a sphere to be greater than the volume of either shape individually. This occurs when the pyramid is taller than the radius of the sphere, resulting in a larger shared volume between the two shapes.

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