Find the volume of the first octant region under the surface $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$
I think that the integral should be:
$$\int_{0}^1\int_{0}^{\left(1-\sqrt x\right)^2}\int_{0}^{\left(1-\sqrt x -\sqrt y\right)^2}\,dz\,dy\,dx$$
Could someone tell me if this is correct?
Best Answer
And an even simpler method for finding this integral is replacing $x$ with $u^2$ , $y$ with $v^2$ and $z$ with $w^2$.
Instead of this : $$\int_{0}^1\int_{0}^{\left(1-\sqrt x\right)^2}\int_{0}^{\left(1-\sqrt x -\sqrt y\right)^2}\,dz\,dy\,dx$$ $dx = 2u\,du\\dy=2v\,dv\\dz=2z\,dz$
you will get this :
$$\int_{0}^1\int_{0}^{1-u}\int_{0}^{1-u-v}8uvw\,dw\,dv\,du$$
which i think is easier to visualize