# Volume of $E=\{5x^8 \le y \le 7x^8; 2y^5 \le z \le 3y^5; z^7 \le x \le 6z^7\}$.

I am stuck with the following problem:

Compute volume of $E=\{5x^8 \le y \le 7x^8; 2y^5 \le z \le 3y^5; z^7 \le x \le 6z^7\}$.

My progress:

It is easy to see that $x,y,z \ge 0$. Therefore I can add all inequalities and after simple algebra get the following: $$5x^8 + 2y^5+z^7 \le x+y+z \le 7x^8 + 3y^5+6x^7$$ Update: as it was pointed out by A.Γ. adding inequalities does not work because different sets can give the same inequalities. So, it looks like I should solve or at least get some kind of information for integration directly from the system somehow.

Any hint?

Thanks a lot for your help!

• The first idea is wrong for sure - the single inequality is true for large $x,y,z$ and gives en unbounded region while $E$ is bounded.
– A.Γ.
May 15 '17 at 20:58
• Thanks for your comment, of course you are right, I updated my ideas. May 16 '17 at 4:43
• No, when you add inequalities you lose information. Look - if you add instead another inequalities $E'=\{ 5x^8 \le \color{red}x \le 7x^8; 2y^5 \le \color{red}y \le 3y^5; z^7 \le \color{red}z \le 6z^7\}$ you get exactly the same sum as before. Do you think that $E=E'$?
– A.Γ.
May 16 '17 at 7:08
• I think that $E \neq E'$. Ok, adding them does not work. So I should somehow solve this system to obtain borders, right? May 16 '17 at 7:11

Hint: the set is given by $$\begin{cases} 5\le\frac{y}{x^8}\le 7,\\ 2\le\frac{z}{y^5}\le 3,\\ 1\le\frac{x}{z^7}\le 6 \end{cases}\quad\rightarrow\quad\text{substitution}\quad\rightarrow\quad \begin{cases} 5\le u\le 7,\\ 2\le v\le 3,\\ 1\le w\le 6 \end{cases}.$$
• @Hedgehog $y=ux^8$, $z=vy^5$, $x=wz^7$ $\Rightarrow$ $$x=wz^7=wv^7y^{5\cdot 7}=wv^7u^{5\cdot 7}x^{5\cdot 7\cdot 8}\quad\Rightarrow\quad x=...$$