Show set (which is a subset of R^n) is bounded

[Berrius]

Homework Statement

Show that $D = { (x,y,z) \in \mathbb{R}^{3} | 7x^2+2y^2 \leq 6, x^3+y \leq z \leq x^2y+5y^3}$ is bounded.

Homework Equations

Definition of bounded:$D \subseteq \mathbb{R}^{n}$ is called bounded if there exists a M > 0 such that $D \subseteq \{x \in \mathbb{R}^{n} | ||x|| \leq M\}$

The Attempt at a Solution

I have to find a M such that $D \subseteq \{(x,y,z) \in \mathbb{R}^{3} | x^2 + y^2 + z^2 \leq M\}$. I thought of just picking a very high M, say 999999. But how do I show it works?

 

[Office_Shredder]

It's often easier to show that each coordinate is bounded, say x,y and z are all smaller than 10 maybe. Then ||(x,y,z)|| < ||(10,10,10)|| = M

 

[Berrius]

So if I say something like: $7x^2+2y^2 \leq 6 \Rightarrow y^2 \leq 3 \lt 4 \Rightarrow y \lt 2$
and $7x^2+2y^2 \leq 6 \Rightarrow x^2 \leq \frac{6}{7} \lt 1 \Rightarrow x \lt 1$
and $z \leq x^2y+5y^3 \Rightarrow z \lt (2+5*8)=42$
So choose M = 4+1+42^2 = 1769. And this M will do.

 

[Office_Shredder]

Careful, M=||(1,2,42)|| is the square root of the number you put up. As long as M is larger than 1 that won't matter, but if the norm happens to be smaller than 1 failing to take the square root can give a value of M that doesn't work