Help calculating this volume

Question

Being $x,y,z \geq 0$ , calculate the volume bounded by $\sqrt{x} + \sqrt{2y} + \sqrt{3z} = 1$

Thanks in advance for the help

Answer 1

As suggested, use $x=u^2,\ 2y=v^2\ \mbox{and}\ 3z=w^2$. Then your volume is $$ V = \int \int \int \frac43 u v w \; du \;dv \;dw $$ with $u,v,w > 0$ and $u + v + w \le 1$. The boundaries can be implemented by

$$ V = \int_0^{1} \int_0^{1-u} \int_0^{1-u-v} \frac43 u v w \; du \;dv \;dw $$ so $$ V = \frac46 \int_0^{1} \int_0^{1-u} u v (1-u-v)^2 \; du \;dv $$

$$ V = \frac46 \frac{1}{12} \int_0^{1} u (u-1)^4 \; du = \frac46 \frac{1}{12} \frac{1}{30} = \frac{1}{540} \simeq 0.00185 $$