# Thread: What solid is this?

1. ## What solid is this?

If it helps, i know it's on the first octant (everything is positive)
x=[0,1]
y=[0,1]
z>0

It seems like a sphere of radius 2 on the first octant, but then they put $\sqrt{1-y^2}$ which is the projection of a circle on the first quadrant (because x belongs to [0,1]) of radius 1.

So i'ts kind of weird. If it is what i think, they might as well have put a cylinder z=[0,1] and then the rest of the sphere of radius 2 above it, being all values positive (first octant).

2. Originally Posted by AniMuS

If it helps, i know it's on the first octant (everything is positive)
x=[0,1]
y=[0,1]
z>0

It seems like a sphere of radius 2 on the first octant, but then they put $\sqrt{1-y^2}$ which is the projection of a circle on the first quadrant (because x belongs to [0,1]) of radius 1.

So i'ts kind of weird. If it is what i think, they might as well have put a cylinder z=[0,1] and then the rest of the sphere of radius 2 above it, being all values positive (first octant).
So y ranges from 0 to 1 and, for each y, x ranges from x= 0 to $x= \sqrt{1- y^2}$ so $x^2= 1- y^2$ or the upper part of the circle $x^2+ y^2= 1$. Then, for each (x,y), z ranges from z= 0 to $z= \sqrt{4- x^2- y^2}$ so $z^2= 4- x^2- y^2$, the circle $x^2+ y^2+ z^2= 4$.

The figure is one octant of an ellipsoid.

3. How can it be an ellipsoid?

I was thinking about it and:
The sphere is centered on the origin, and has radius 2 which means by the time z reaches $\sqrt{3}$ (counting from 2) the radius is already on 1. From $\sqrt{3}$ until z=0, the sphere has to be "cut" so that the projection on the xy plane doesnt pass the quarter circle of radius, making the part of the solid below z= $\sqrt{3}$ a quarter of a cylinder.

4. Looks like that little piece in there:

5. Nice shawsend =). Thats exactly what i had in mind, thanks! May i ask you how did you do that image?

Cheers

6. Originally Posted by AniMuS
Nice shawsend =). Thats exactly what i had in mind, thanks! May i ask you how did you do that image?

Cheers
Pretty messy . . . but I like making plots. Here's the Mathematica code, and as usual, I've converted it to Raw input form so I can cut and paste it so that you can if you want, just cut and paste this code directly into Mathematica and if you want, convert it back to standard form by selecting it, then choose Cell/Convert To/Standard Form. Also, if you download it to a Mathematica version before ver 7, there may be some downward-compatability issues but not sure.

Code:
plane1 = Graphics3D[Polygon[
{{3, 3, 0}, {-3, 3, 0}, {-3, -3, 0},
{3, -3, 0}}]];
plane2 = ContourPlot3D[y == 0, {x, 0, 1},
{y, 0, 1}, {z, 0, 2},
RegionFunction -> Function[{x, y, z},
0 <= x <= 1 && 0 <= z <=
Sqrt[4 - x^2]], ContourStyle ->
{Opacity[0.2], Green}, Mesh -> None]
plane3 = ContourPlot3D[x^2 + y^2 == 1,
{x, 0, 1}, {y, 0, 1}, {z, 0, 2},
RegionFunction -> Function[{x, y, z},
0 <= x <= 1 && 0 <= y <= 1 &&
0 <= z <= Sqrt[4 - x^2 - y^2]],
ContourStyle -> {Opacity[0.2], Green},
Mesh -> None]
c1trace = ContourPlot3D[
{x^2 + y^2 + z^2 == 4}, {x, -3, 3},
{y, -3, 3}, {z, -3, 3},
ContourStyle -> {LightRed},
Mesh -> None, RegionFunction ->
Function[{x, y, z}, z > 0 &&
0 <= y <= 1 && 0 <= x <=
Sqrt[1 - y^2]]]
c1 = ContourPlot3D[{x^2 + y^2 + z^2 ==
4}, {x, -3, 3}, {y, -3, 3},
{z, -3, 3}, ContourStyle ->
{Opacity[0.4], LightRed},
Mesh -> None]
axes = Graphics3D[
{{Line[{{-3, 0, 0}, {3, 0, 0}}],
Line[{{0, 3, 0}, {0, -3, 0}}]}}];
p1 = ParametricPlot3D[{t, Sqrt[1 - t^2],
0}, {t, 0, 1}];
Show[{c1, c1trace, plane1, p1, axes,
plane2, plane3}, PlotRange ->
{{-2, 2}, {-2, 2}, {-2, 2}},
Boxed -> False]

7. Will do.