Spherical coordinates Domain

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Homework Statement


f(x,y,z) = 1
[itex]x^{2}[/itex] + [itex]y^{2}[/itex] + [itex]z^{2}[/itex] ≤ 4z
z ≥ [itex]\sqrt{x^2 + y^2}[/itex]

Homework Equations





The Attempt at a Solution


How can I know ρ if there is z variable? Do I just square root both 4 and z?

For θ, since it is sphere it would be 0 ≤ θ ≤ 2[itex]\pi[/itex] right?

for [itex]\phi[/itex], is it 0 ≤ [itex]\phi[/itex] ≤ [itex]\pi[/itex]?
 
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  • #2
LCKurtz
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Well [itex]\phi > \frac \pi 2[/itex] is always necessary to get below the xy plane regardless of what quadrant in xy you are in. What is its upper limit?
 
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I changed the question sorry
 
  • #4
LCKurtz
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Homework Statement


f(x,y,z) = 1
[itex]x^{2}[/itex] + [itex]y^{2}[/itex] + [itex]z^{2}[/itex] ≤ 4z
z ≥ [itex]\sqrt{x^2 + y^2}[/itex]

Homework Equations





The Attempt at a Solution


How can I know ρ if there is z variable? Do I just square root both 4 and z?

For θ, since it is sphere it would be 0 ≤ θ ≤ 2[itex]\pi[/itex] right?

for [itex]\phi[/itex], is it 0 ≤ [itex]\phi[/itex] ≤ [itex]\pi[/itex]?
Your first step is to draw a picture of your two surfaces. For your first equation, take the z onto the other side and complete the square. Then you should recognize both surfaces and you will hopefully be able to see the correct limits.

Also, it is not hard to convert your first equation to spherical coordinates. Try it.
 
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So I must complete the square or I don't have to?
 
  • #6
LCKurtz
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So I must complete the square or I don't have to?
You need to do whatever it takes to understand the picture. I'm guessing you can't draw the first surface without completing the square. And you will need the spherical coordinates version of the first surface in order to figure out [itex]\rho[/itex] on the surface.
 
  • #7
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OHHH ok I went back and reviewed completing the squares.

So it comes about as

-[itex]x^{2}[/itex]-[itex]y^{2}[/itex]-[itex](z-2)^{2}[/itex] = 4

From this I know that

0 ≤ ρ ≤ 2

is z ≥ √x^2+y^2 a cone?
 
  • #8
LCKurtz
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OHHH ok I went back and reviewed completing the squares.

So it comes about as

-[itex]x^{2}[/itex]-[itex]y^{2}[/itex]-[itex](z-2)^{2}[/itex] = 4

From this I know that

0 ≤ ρ ≤ 2

is z ≥ √x^2+y^2 a cone?
You have the signs wrong. Yes the second surface is a cone. But regarding the first surface, please answer these questions:

1. Describe the shape and location of the first surface.
2. What is its spherical equation?
3. How did you get ρ ≤ 2?
 
  • #9
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Since it's there is only Z translation, the sphere would go 2 up from the origin
spherical equation would be... x^2 + y^2 + (z-2)^2 = -4?
ρ is from 0 to 2 because it can't be negative and the sphere of the radius is 2
 
  • #10
LCKurtz
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Since it's there is only Z translation, the sphere would go 2 up from the origin
spherical equation would be... x^2 + y^2 + (z-2)^2 = -4?
ρ is from 0 to 2 because it can't be negative and the sphere of the radius is 2
You can't have positive on one side and negative on the other. Still wrong.
If you have a sphere of radius two moved up 2, then the spherical coordinate ρ is not the radius of the sphere. Draw a picture (as I suggested earlier).

And you ignored my second question without which you will never solve this problem.
 
  • #11
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That's how completing the square came out as. Can I just take out the negative and make it equal to 4?

Why does z translation change the radius of the sphere??

and the spherical equation is [itex]\int\int\int[/itex][itex]ρ^{2}[/itex]sin[itex]\phi[/itex]dρd[itex]\phi[/itex]dθ

is ρ 4?
 
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  • #12
LCKurtz
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That's how completing the square came out as. Can I just take out the negative and make it equal to 4?
No, that isn't how completing the square comes out. Check your arithmetic. I can't correct it if you don't show it but you should be able to find the sign mistakes anyway.
Why does z translation change the radius of the sphere??
It doesn't. When the sphere is not at the origin ρ is not the radius of the sphere. Have you drawn a picture yet?
and the spherical equation is [itex]\int\int\int[/itex][itex]ρ^{2}[/itex]sin[itex]\phi[/itex]dρd[itex]\phi[/itex]dθ
That is the formula for volume once you have the limits, but that is not what I asked you. You need to get the spherical coordinate equation of the sphere or you will never get the limits figured out.
 
  • #13
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ρ is a distance from the origin, so it would be 4.

and I have no idea what spherical coordinate equation you are talking about... I really don't... Would you give me some hint?

x^2 + y^2 + z^2 = 4z

x^2 + y^2 = -z^2 + 4z

-x^2 - y^2 = z^2 - 4z

-x^2 - y^2 - 4 = z^2 - 4z - 4

= (z-2)^2
-x^2 - y^2 - (z - 2)^2 = 4

x^2 + y^2 + (z - 2)^2 = -4
 
  • #14
LCKurtz
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ρ is a distance from the origin, so it would be 4.
ρ is not a constant on this surface.
and I have no idea what spherical coordinate equation you are talking about... I really don't... Would you give me some hint?

x^2 + y^2 + z^2 = 4z

x^2 + y^2 = -z^2 + 4z

-x^2 - y^2 = z^2 - 4z

-x^2 - y^2 - 4 = z^2 - 4z - 4

= (z-2)^2
That isn't equal to (z-2)2 as you will see if you multiply it out. When completing the square you add, not subtract the necessary term.

-x^2 - y^2 - (z - 2)^2 = 4

x^2 + y^2 + (z - 2)^2 = -4
Hint on spherical coordinates: You have x2+y2+z2=4z

What is the left side in spherical coordinates? What is the equation for z in spherical coordinates? (Both questions are easy.) Solve the result for [itex]\rho[/itex].
 

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