Math Help - Three Dimensional Coordinate Systems

1. Three Dimensional Coordinate Systems

Hey everyone,

Just had my first lecture today and was working on some practice problems, but got stuck on this one. All I have been introduced to so far is the equation of a sphere, but I don't really know what else to do with this.

I completed the square on both sides of the inequality by adding 1/4 to each side and made the middle of the inequality a sphere centered at (0,0,1/2).

Here is my full attempt. The inequality represents the region on/between the spheres centered at (0,0,1/2) with radii 1 1/4 and 3 1/4. The region contains all of the points on the sphere with radius 1 1/4 and includes all of the points of the second sphere except the outermost points where r = 3 1/4.

I think this might be right, but I just thought I'd check with you guys.

Thanks!

2. Originally Posted by statelyplump
Hey everyone,

Just had my first lecture today and was working on some practice problems, but got stuck on this one. All I have been introduced to so far is the equation of a sphere, but I don't really know what else to do with this.

I completed the square on both sides of the inequality by adding 1/4 to each side and made the middle of the inequality a sphere centered at (0,0,1/2).

Here is my full attempt. The inequality represents the region on/between the spheres centered at (0,0,1/2) with radii 1 1/4 and 3 1/4. The region contains all of the points on the sphere with radius 1 1/4 and includes all of the points of the second sphere except the outermost points where r = 3 1/4.

I think this might be right, but I just thought I'd check with you guys.

Thanks!
Well, if z=0, you have an annulus in the xy plane.

What do you have if x = 0?

y = 0?

3. Hello, statelyplump!

$\text{The inequality represents the region on/between the spheres centered}$
$\text{at }(0,0,\frac{1}{2})\text{ with radii }1\frac{1}{4}\text{ and }3\frac{1}{4}.$

$\text{The region contains all of the points on the inner sphere}$
$\text{but not the points on the outer sphere.}$

When you completed-the-square, you got $\frac{13}{4}$ and $\frac{5}{4}$ on the "other side".

Recall the general equation: . $(x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 \:=\:r^2$

What you have is the radius-squared.

The radii are: . $\dfrac{\sqrt{13}}{2}$ and $\dfrac{\sqrt{5}}{2}$

4. dwsmith, I think you must have misunderstood the question.

statelyplump, in addition to what Soroban said, you are misusing some words. Strictly, a "sphere" is the surface of a "ball" and, specifically, to say a point is "on a sphere" implies it is on the surface. When you say "includes all of the points of the second sphere except the outermost points", that is, at best, misleading. What you mean is that it "includes all point within the second sphere and outside the first sphere" or "includes all points inside the ball of radius $\frac{\sqrt{13}}{2}$ and on or outside the ball of radius $\frac{\sqrt{5}}{2}$.

5. Thanks a lot everyone! I knew the wording was off, so thank you for helping out with that. I didn't even know there was a term for the inside of a sphere and that was exactly what I wanted to say.

6. Originally Posted by HallsofIvy
dwsmith, I think you must have misunderstood the question.

statelyplump, in addition to what Soroban said, you are misusing some words. Strictly, a "sphere" is the surface of a "ball" and, specifically, to say a point is "on a sphere" implies it is on the surface. When you say "includes all of the points of the second sphere except the outermost points", that is, at best, misleading. What you mean is that it "includes all point within the second sphere and outside the first sphere" or "includes all points inside the ball of radius $\frac{\sqrt{13}}{2}$ and on or outside the ball of radius $\frac{\sqrt{5}}{2}$.
By isolating each plane, I can easilty see it is a paraboloid. We have a circular annulus in the xy, parabola in the xz, and parabola in yz. To me, that is the easiest method of determine the region.