# Describing a region using spherical coordinates

## 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.

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:

haruspex
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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.
x = ρsinφcosθ
y = ρsinφcosθ
One of those should be sin θ.
ρ=√(2ρ2sin2φ)/cosφ
Where does the 2 come from? Can you not simplify this?

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?

haruspex
Homework Helper
Gold Member
2020 Award
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?

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

haruspex