Volume Of Intersection Between Square Pyramid And Sphere

[EquationOfMotion]

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.

 

[etotheipi]

What is a square cone?

 

[EquationOfMotion]

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.

 

[DaveE]

Must be a square equilateral pyramid with all edges = 2r.

 

[fresh_42]

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!

 

[EquationOfMotion]

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.

 

[fresh_42]

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.

 

[etotheipi]

Arghh, this does look quite tricky. Because the cap is weird... sketched crudely:

 

[fresh_42]

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?)

 

[etotheipi]

(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!

 

[fresh_42]

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!

 

[haruspex]

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.

 

[haruspex]

@EquationOfMotion , were you able to follow my method in post #12 to get the answer or do you need more detail?

 

[haruspex]

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 $\frac 13\Omega r^3$.

 

[TSny]

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.

 

[haruspex]

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.