Find the volume below $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$ in the first quadrant

Question

I understand that we have to use transformation $$x = u^2, y = v^2, z = w^2$$ but I cannot figure out the limits.

I just need a rough sketch of how to approach this. Could anyone give me some ideas?

Answer 1

We can still do it directly. For the set up, it should be:

$V = \displaystyle \int_{0}^1 \int_{0}^{(1-\sqrt{x})^2} \int_{0}^{(1-\sqrt{x}-\sqrt{y})^2} 1 dzdydx$

Answer 2

The region in $xyz$-space is the set of points which satisfy $x,y,z \ge 0$ and $\sqrt{x}+\sqrt{y}+\sqrt{z} \le 1$.

Under the given transformation, the inequality $\sqrt{x}+\sqrt{y}+\sqrt{z} \le 1$ becomes $u+v+w \le 1$.

Since we need our transformation to be 1:1, we need to restrict $u,v,w \ge 0$.

The region in the $uvw$-space is the set of points which satisfy $u,v,w \ge 0$ and $u+v+w \le 1$.

Can you find the bounds for this region? Also, don't forget the Jacobian when integrating.