MHB Is $D_4$ a bounded region in 3-dimensional space?

[mathmari]

Hey!

I want to draw the following spaces:

1. $D_1=\{(x,y,z)\mid 0\leq x\leq 1, \ 1-x\leq y\leq 1, \ x\leq z\leq 1\}$
   
   By the inequalities $0\leq x\leq 1, \ 1-x\leq y\leq 1$ we get a triangle on the $xy$-plane with vertices $(0,1)$, $(1,0)$ and $(1,1)$.
   
   So, we have that triangle at the plane $z=0$. Do we shift that triangle to the plane $z=1$ because of the inequality $x\leq z\leq 1$, and so we get all the points between the triangle at $z=0$ till the triangle at $z=1$ ?
   
   $$$$
   
2. $D_2=\left \{(x,y,z)\mid \frac{1}{2}\leq z\leq 1, \ x^2+y^2+z^2\leq 1\right \}$
   
   $x^2+y^2+z^2\leq 1$ is a sphere. From the inequality $\frac{1}{2}\leq z\leq 1$ we take only one part of the sphere, or not?
   
   $$$$
   
3. $D_3=\{(x,y,z)\mid 0\leq x\leq \pi , \ 0\leq y \leq 1, \ 0\leq z\leq x\}$
   
   We draw the line $z=x$ on the $xz$-plane for $0\leq x\leq \pi$. Since $0\leq y\leq 1$, we get the following parallelogramm:
   
   View attachment 7396
   
   Is this correct?
   
   $$$$
   
4. $D_4$ is defined by the paraboloid with equation $z=2x^2+y^2$ and the cylinder with the equation $z=8-y^2$.
   
   The paraboloid and the cylinder intersect at $2x^2+y^2=8-y^2 \Rightarrow x^2=4-y^2 \Rightarrow x=\pm \sqrt{4-y^2}$, therefore we get $-\sqrt{4-y^2}\leq x\leq \sqrt{4-y^2}$.
   It must hold that $4-y^2\geq 0 \Rightarrow y^2\leq 4 \Rightarrow -2\leq y \leq 2$.
   We have that $2x^2+y^2\leq 2\sqrt{4-y^2}^2+y^2=8-y^2 \Rightarrow 2x^2+y^2\leq z\leq 8-y^2$.
   
   Therefore the space $D_4$ is the set $\{(x,y,z)\mid -\sqrt{4-y^2}\leq x\leq \sqrt{4-y^2}, \ -2\leq y \leq 2, \ 2x^2+y^2\leq z\leq 8-y^2\}$, or not?

 

[I like Serena]

Hey mathmari! (Smile)

Picking 2 of them for now.

2. $D_2=\left \{(x,y,z)\mid \frac{1}{2}\leq z\leq 1, \ x^2+y^2+z^2\leq 1\right \}$

$x^2+y^2+z^2\leq 1$ is a sphere. From the inequality $\frac{1}{2}\leq z\leq 1$ we take only one part of the sphere, or not?

Yep. The top part where $z\ge \frac 12$. (Nod)

3. $D_3=\{(x,y,z)\mid 0\leq x\leq \pi , \ 0\leq y \leq 1, \ 0\leq z\leq x\}$

We draw the line $z=x$ on the $xz$-plane for $0\leq x\leq \pi$. Since $0\leq y\leq 1$, we get the following parallelogramm:

Is this correct?

Wouldn't $(\pi, 0, 0)$ be in $D_3$ as well? (Wondering)

 

[mathmari]

Yep. The top part where $z\ge \frac 12$. (Nod)

So, will it be in the following form?

View attachment 7399

Wouldn't $(\pi, 0, 0)$ be in $D_3$ as well? (Wondering)

So, is $D_3$ the space between these three planes (red, green, purple) :

View attachment 7398

?

 

[I like Serena]

So, will it be in the following form?

Erm... I don't recognize part of a sphere in there.
It should have a circle at $z=\frac 12$, and it should curve into the z-direction to the point (0,0,1), where it is closed.

So, is $D_3$ the space between these three planes (red, green, purple)?

Yep. (Nod)

 

[mathmari]

Erm... I don't recognize part of a sphere in there.
It should have a circle at $z=\frac 12$, and it should curve into the z-direction to the point (0,0,1), where it is closed.

Ah, so should it be as follows?

View attachment 7400

 

[I like Serena]

Ah, so should it be as follows?

That looks more like it. (Nod)

Although it looks more paraboloidical then spherical. (Nerd)

 

[mathmari]

That looks more like it. (Nod)

Although it looks more paraboloidical then spherical. (Nerd)

Oh yes (Blush)

It should be more like that

View attachment 7401

right? Do you have also an idea about the spaces $D_1$ and $D_4$ ?

 

[I like Serena]

Oh yes (Blush)
It should be more like that
right?

Yep.

1. $D_1=\{(x,y,z)\mid 0\leq x\leq 1, \ 1-x\leq y\leq 1, \ x\leq z\leq 1\}$

By the inequalities $0\leq x\leq 1, \ 1-x\leq y\leq 1$ we get a triangle on the $xy$-plane with vertices $(0,1)$, $(1,0)$ and $(1,1)$.

So, we have that triangle at the plane $z=0$. Do we shift that triangle to the plane $z=1$ because of the inequality $x\leq z\leq 1$, and so we get all the points between the triangle at $z=0$ till the triangle at $z=1$ ?

Suppose we pick (1,0) in the xy-plane, then z cannot be 0 can it? (Wondering)

 

[mathmari]

Suppose we pick (1,0) in the xy-plane, then z cannot be 0 can it? (Wondering)

At the $xy$-plane ($z=0$) we have that $x=0$ and $y=1$. So, we get only the point $(0,1,0)$.

At the plane $z=1$ we get the following points:
for $x=0$ : $y=1$ and so $(0,1,1)$
for $x=1$: $y=0$ or $y=1$ and so $(1,0,1)$ and $(1,1,1)$

Do we connect all these points? The result is a pyramid with base the triangle on the plane $z=1$ defined by the points $(0,1,1)$, $(1,0,1)$, $(1,1,1)$ and the top of the pyramid is $(0,1,0)$.

Is this correct? (Wondering)

 

[I like Serena]

At the $xy$-plane ($z=0$) we have that $x=0$ and $y=1$. So, we get only the point $(0,1,0)$.

At the plane $z=1$ we get the following points:
for $x=0$ : $y=1$ and so $(0,1,1)$
for $x=1$: $y=0$ or $y=1$ and so $(1,0,1)$ and $(1,1,1)$

Do we connect all these points? The result is a pyramid with base the triangle on the plane $z=1$ defined by the points $(0,1,1)$, $(1,0,1)$, $(1,1,1)$ and the top of the pyramid is $(0,1,0)$.

Is this correct? (Wondering)

When solving for each of the boundary conditions x=0, y=0, z=0, x=1, y=1, and z=1, I find the same points.
Since the remaining inequalities are linear equations, I believe our space $D_1$ is indeed that pyramid. (Nod)

 

[I like Serena]

4. $D_4$ is defined by the paraboloid with equation $z=2x^2+y^2$ and the cylinder with the equation $z=8-y^2$.

The paraboloid and the cylinder intersect at $2x^2+y^2=8-y^2 \Rightarrow x^2=4-y^2 \Rightarrow x=\pm \sqrt{4-y^2}$, therefore we get $-\sqrt{4-y^2}\leq x\leq \sqrt{4-y^2}$.
It must hold that $4-y^2\geq 0 \Rightarrow y^2\leq 4 \Rightarrow -2\leq y \leq 2$.
We have that $2x^2+y^2\leq 2\sqrt{4-y^2}^2+y^2=8-y^2 \Rightarrow 2x^2+y^2\leq z\leq 8-y^2$.

Therefore the space $D_4$ is the set $\{(x,y,z)\mid -\sqrt{4-y^2}\leq x\leq \sqrt{4-y^2}, \ -2\leq y \leq 2, \ 2x^2+y^2\leq z\leq 8-y^2\}$, or not?

Can we draw this?
Preferably including the curves that follow from the intersections with the coordinate planes?
Then it becomes a bit easier to see if it matches and if it's all correct. (Thinking)