Describing a region using spherical coordinates

Click For Summary

Homework Help Overview

The problem involves describing a solid region in the first octant using spherical coordinates. The region is defined as lying above a half-cone described by the equation z=√(x²+y²) and inside a unit sphere defined by x²+y²+z²=1. The final answer is expected in set-builder notation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the bounds for θ, ρ, and φ in spherical coordinates, with some questioning the relationship between φ and θ. There is an exploration of how to express the limits for ρ based on the geometry of the solid and the equations provided.

Discussion Status

Participants are actively engaging with the problem, clarifying equations and discussing the implications of their simplifications. There is a recognition of potential mistakes in the initial attempts, and some guidance is offered regarding the interpretation of bounds in spherical coordinates.

Contextual Notes

Participants note that the limits for ρ should not reference Cartesian coordinates, indicating a focus on maintaining the integrity of the spherical coordinate system in their descriptions.

Kaleem
Messages
21
Reaction score
0

Homework Statement


Describe using spherical coordinates the solid E in the first octant that lies above the half-cone z=√(x2+y2) but inside x2+y2+z2=1. Your final answer must be written in set-builder notation.

Homework Equations


ρ = x2+y2+z2
x = ρsinφcosθ
y = ρsinφsinθ
z = ρcosφ

The Attempt at a Solution


Since we are in the first octant, θ will go from [0,π/2].
However the problem comes with describing ρ and φ,
Since we are in the first octant I believe that φ will be the same as θ, however for ρ,
I substituted in the relevant equations into both equations that were given.

ρ= ±1 <-- Unit sphere
ρ=√(2ρ2sin2φ)/cosφ <-- Half cone

Would I be able to use the positive bound of the unit sphere as the upper limit and the bound gotten from the cone as the lower limit?
 
Last edited:
Physics news on Phys.org
Kaleem said:
Would I be able to use the positive bound of the unit sphere as the upper limit and the bound gotten from the cone as the lower limit?
Yes.
Kaleem said:
x = ρsinφcosθ
y = ρsinφcosθ
One of those should be sin θ.
Kaleem said:
ρ=√(2ρ2sin2φ)/cosφ
Where does the 2 come from? Can you not simplify this?
 
haruspex said:
Yes.

One of those should be sin θ.

Where does the 2 come from? Can you not simplify this?
You're right, I made a mistake when i factored it out, if i simplify I get 1 = tanφ, which gives me φ= π/4. Which confuses me, would the lower bound for ρ be π/4 or would that just be the upper bound for φ?
 
Kaleem said:
would the lower bound for ρ be π/4 or would that just be the upper bound for φ?
how does tan φ < π/4 turn into a bound on ρ?
Pick a point in the region. Vary ρ up and down until you hit the boundaries of the region. Which boundaries do you hit?
 
haruspex said:
how does tan φ < π/4 turn into a bound on ρ?
Pick a point in the region. Vary ρ up and down until you hit the boundaries of the region. Which boundaries do you hit?
You get z=√(x2+y2) ≤ ρ ≤1
 
Kaleem said:
You get z=√(x2+y2) ≤ ρ
Although that is true, it does not serve as a limit on ρ in polar coordinates. The limits should not mention the Cartesian coordinates.
If the point (ρ, θ,φ) is in the region, is the point (ρ/2, θ, φ) also in the region?
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K