How Do You Calculate the Volume Between a Cone and a Sphere?

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Rijad Hadzic
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Homework Statement



Find volume of the solid that lies above the cone Φ = π/3 and below the sphere ρ = 4cosΦ

Homework Equations

The Attempt at a Solution



Obviously this is a triple integral. My book tells me that 0 ≤ρ≤ 4cosΦ

but this makes no sense to me.

From the problem, it lies ABOVE the cone Φ = π/3 and below the sphere ρ = 4cosΦ, so wouldn't that imply that ρ is not starting at 0?

What I did was solved ρ = 4cosΦ

arccos(ρ/4) = Φ and set it = to pi/3

arccos(ρ/4) = π/3

ρ/4 = cosπ/3

ρ = 4 * (1/2) = 2

so wouldn't 2≤ρ≤4cosΦ be the bounds? I don't understand how the lower bound can start at 0 when its asking for what's above the cone and below the sphere..
 
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Orodruin said:
No. Also, please define your variables.

I suggest you draw an image of what things look like.

ρ is the distance to the point from the origin
Φ is the angle from the z axis to the point. 0≤Φ≤pi
θ is the angle to the point from the projection in the xy plane.

I did draw an image, it looks like an icecream cone basically. The part I want is the icecream on the top of the cone. I still don't understand why ρ is 0..
 
Hmm I think I may be graphing it wrong. Maybe I have no clue how to graph ρ = 4cosΦ..
 
Rijad Hadzic said:
ρ is the distance to the point from the origin
Φ is the angle from the z axis to the point. 0≤Φ≤pi
θ is the angle to the point from the projection in the xy plane.
Different textbooks treat spherical coordinates in different ways, particularly the ##\theta## and ##\phi## coordinates. According to this wikipedia article (https://en.wikipedia.org/wiki/Spherical_coordinate_system), physics books consider ##\phi## to be the angle in the "x-y" plane, while math textbooks consider ##\phi## to be measured from the positive z-axis.