Describe the region of R^3, sphere with inequality

[alias]

Homework Statement

Describe the region of R^3 that is represented by:

Homework Equations

x^2 + y^2 + z^2 > 2z

The Attempt at a Solution

I'm not sure what to do with this at, especially at z=0 and z=2

 

[flyingpig]

Maybe you should complete the square

 

[alias]

x^2 + y^2 + (z-1)^2 > 1
a sphere with centre (0,0,1) and radius >1?

 

[SammyS]

Seems like that could be many spheres.

 

[alias]

would it be:
x^2 + y^2 + (z-1)^2 > 1

infinite number of spheres with centre (0,0,1) and radius >1?

 

[SammyS]

Where do you find any point (x, y) that satisfies x^2 + y^2 + (z-1)^2 > 1 , relative to the set of points that satisfy x^2 + y^2 + (z-1)^2 = 1 ?

 

[alias]

Where do you find any point (x, y) that satisfies x^2 + y^2 + (z-1)^2 > 1 , relative to the set of points that satisfy x^2 + y^2 + (z-1)^2 = 1 ?

I'm lost now. I am fairly sure I have the centre correct at (0,0,1), the only thing I can think of regarding your previous post is:
(x,y) = (x>1, y>1) or
(x,y) = (x>=1, y>=1). But this seems incorrect.

 

[SammyS]

Yes, the equation x² + y² + (z-1)² = 1 describes a sphere of radius 1 with center at (x, y, z) = (0, 0, 1) .

Pick any point that is outside of the sphere. What do you get for x² + y² + (z-1)² for that point?

Pick any point that is inside of the sphere. What do you get for x² + y² + (z-1)² for this point?

 

[alias]

for the equation, a point inside the sphere would be (1/2, 0, 1)
.5^2 + 0^2 + (1-1)^2 = .25

outside the the sphere would be (1,1,1)
1^2 + 1^2 + (1-1)^2 = 2

 

[SammyS]

for the equation, a point inside the sphere would be (1/2, 0, 1)
.5^2 + 0^2 + (1-1)^2 = .25

outside the the sphere would be (1,1,1)
1^2 + 1^2 + (1-1)^2 = 2

Notice that .25 < 1, so for point (1/2, 0, 1), x² + y² +(z-1)² < 1 .


Similarly, 2 > 1, , so for point (1, 1, 1), x² + y² +(z-1)² > 1 .

Does that give you any ideas?

BTW, what does \sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2} represent ?

 

[alias]

\sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}
is the distance between two points, (0, 0, -1) and any point (x,y,z)
but I still don't understand the constraint on x, y, and/or z.

 

[alias]

\sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}
is the distance between two points, (0, 0, -1) and any point (x,y,z)
but I still don't understand the constraint on x, y, and/or z.

or the inequality of the radius

 

[HallsofIvy]

I'm afraid SammyS rather confused things when he asked about "(x, y)" in post #6.

alias, In post #9, you said, "outside the the sphere would be (1,1,1)". Yes, that is a point in this set but any point on that set lies in this set. The only problem with your post #5, "infinite number of spheres with centre (0,0,1) and radius >1?", is not enough. The set of all spheres with integer radius, greater than 1, satisfies that but does not include points that satisfy, for example, x^2+ y^2= \frac{9}{4}.

 

[SammyS]

I'm afraid SammyS rather confused things when he asked about "(x, y)" in post #6.
...

Yes. TYPO !

Sorry about that.

I meant to type " point (x, y, z) = ... "

 

[SammyS]

\sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}
is the distance between two points, (0, 0, -1) and any point (x,y,z)
but I still don't understand the constraint on x, y, and/or z.

Actually, it's the distance between two points, (0, 0, 1) and any point (x,y,z). In other words, it's the distance from the center of your sphere to any point (x,y,z).

If \sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}&gt;1\,, where is the point (x, y, z) located in relation to the sphere ?

 

[alias]

The distance between (0,0,1) and (x,y,z) is greater than 1. I'm not sure about the relation to the sphere though.

 

[SammyS]

What is the distance from (0,0,1) to any point on the surface of the sphere ?

 

[alias]

The distance from (0,0,1) to any point on the surface of the sphere is 1.
How does this relate to the inequality with the radius?

 

[SammyS]

The distance between (0,0,1) and (x,y,z) is greater than 1. I'm not sure about the relation to the sphere though.

The distance from (0,0,1) to any point on the surface of the sphere is 1.
How does this relate to the inequality with the radius?

If the distance from (0,0,1) is 1, the point (x,y,z) is on the surface of the sphere. If that distance is greater than 1 doesn't the point lies outside the sphere?

If (x)^2 + (y)^2 + (z-1)^2&gt;1\,, what does that say about \sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}\,?

 

[alias]

If the distance from (0,0,1) is 1, the point (x,y,z) is on the surface of the sphere. If that distance is greater than 1 doesn't the point lies outside the sphere?

If (x)^2 + (y)^2 + (z-1)^2&gt;1\,, what does that say about \sqrt{(x-0)^2 + (y-0)^2 + (z-1)^2}\,?

Yes, any point (x,y,z) that has a distance greater than 1 from (0,0,1) lies outside the the sphere. So the original eqn, x^2 + y^2 + (z-1)^2 > 1 consists of all points outside the outside the sphere with with centre (0,0,1) and radius 1? Would that not be all spheres with radius >1 and the same centre?

 

[SammyS]

Well, the point (x,y,z) = (1,1,1) satisfies your inequality (It's not an equation.), but that single point is not a sphere.

 

[alias]

Then the region of R^3 that is represented by:
(x-0)^2 + (y-0)^2 + (z-1)^2 > 1 is all spheres with centre(0,0,1) and a radius greater than 1?

 

[SammyS]

I suppose you could say it that way. I think it's simpler and clearer to say it's the region external to the sphere of radius 1, that's centered at (0,0,1) .

 

[alias]

Thanks a lot for your help. You have a lot of patience!

 

[SammyS]

You're more than welcome.

Sorry about that typo earlier.