Can somebody confirm my answer if it's right?

1. Jun 20, 2010

kramer733

Question is:

1. True or False (Explain answer): A sphere with center P (Xo,Yo,Zo) and radius r consists of all points (x, y, z) that satisfy the inequality

(x - xo)^2 + (y - yo)^2 + (z - zo)^2 <= r^2

True because if we pick a vector where the "coordinate" of that vector is an element of the natural numbers (for ease) such as (7,23,16) and let that become the radius of the circle. Let the centre of the point become point of origin (0,0,0) and we can plug in the numbers for the inequality. Looking at the centre of the sphere, it's defined in this situation as (0,0,0). Since the inequality is

(x - xo)^2 + (y - yo)^2 + (z - zo)^2 <= r^2

the (Xo ,Yo , Zo) are equal to zero in this case here.

We will then obtain something that equals r^2 because radius = (7,23,16)

To gain magnitude of vector, we use

7^2 + 23^2 + 16^2

After that, we can square root 7^2 + 23^2 + 16^2

Essentially that will give you the magnitude of the radius which is (834)^(0.5)

We can then just move the sphere around the R^3 plane. It should be the same all around right? I HONESTLY DON'T KNOW AT ALL AND AM MOSTLY PULLING IT OUT OF MY ***. To be honest i' don't even know if it works that way with Inequalities =/

Please correct me if i'm wrong.

2. Jun 20, 2010

djeikyb

Let's look at the problem in a slightly different way.

(x-xo)^2+(y-yo)^2+(z-zo)^2 is the distance squared from (xo,yo,zo), right? If you can't see that, try it in two dimensions.

(x-xo)^2+(y-yo)^2 is the distance from (xo, yo).

So, we know that
(x-xo)^2+(y-yo)^2+(z-zo)^2
is the distance squared from the (xo,yo,zo). This point is the center of the sphere, right?

(distance from center of sphere)^2 <= (radius of sphere)^2.

Now, if your distance from the center is less than the radius, are you going to be within the sphere?

3. Jun 20, 2010

djeikyb

(x - xo)^2 + (y - yo)^2 + (z - zo)^2 <= r^2

the (Xo ,Yo , Zo) are equal to zero in this case here.

We will then obtain something that equals r^2 because radius = (7,23,16)

To gain magnitude of vector, we use

7^2 + 23^2 + 16^2

After that, we can square root 7^2 + 23^2 + 16^2

Essentially that will give you the magnitude of the radius which is (834)^(0.5)

We can then just move the sphere around the R^3 plane. It should be the same all around right? I HONESTLY DON'T KNOW AT ALL AND AM MOSTLY PULLING IT OUT OF MY ***. To be honest i' don't even know if it works that way with Inequalities =/

Please correct me if i'm wrong.[/QUOTE]

There is a problem in your logic. Setting the center of the sphere does not change teh inequality to an equality. The rest of the logic is flawed because of this assumption.

However, the location of the sphere in the R^3 space should not affect the rules governing it, right? It wouldn't make sense for a baseball to change size if you change its location.

4. Jun 20, 2010

kramer733

Thanks. So it's still true then right? So really what is the answer?

oops nvm. THANK YOU VERY MUCH,.

5. Jun 20, 2010

Staff: Mentor

Actually, I don't think it's right. A sphere is the set of points that are equidistant from some fixed point (the center), so the equation of the sphere of radius r, centered at (x0, y, z) is
(x - x0)2 + (y - y0)2 + (z - z0)2 = r2. Note the equality.

The inequality you showed in the first post describes a ball, which includes the sphere (the surface of the ball) together with all the points inside the sphere.

The difference between a sphere and a ball in R3 is similar to the difference between a circle and a disc in R2.

6. Jun 20, 2010

djeikyb

Would that not be the difference between a spherical shell and a sphere?

7. Jun 20, 2010

djeikyb

I think that would probably depend on the book used, though I could be wrong. My old calc book is still packed, so I can't compare with that.

8. Jun 20, 2010

Staff: Mentor

The sphere/ball terminology is what I remember from multiple texts. Wikipedia (http://en.wikipedia.org/wiki/Ball_(mathematics [Broken])) says the ball is the inside of a sphere and that it can include the surface (closed ball) or not (open ball).

Last edited by a moderator: May 4, 2017
9. Jun 20, 2010

djeikyb

Hmm. Then I suppose the trick of the question is deciding which convention to use.

That's annoying.

10. Jun 20, 2010

Staff: Mentor

Well, it's kind of a trick question, but I don't think it requires you to choose between two different conventions. A sphere is defined as the set of all points equidistant from a fixed point. It's not defined as all the points less than or equal to r from some fixed point.

11. Jun 20, 2010

djeikyb

Yeah, I think you're right. Found my book, it agrees with you. Still annoying though.