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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?

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    $\begingroup$ it is correct ! $\endgroup$ – abkds Dec 16 '13 at 6:42
  • $\begingroup$ i am interested about bounds of integral,could you tell me please i little detailed about it? $\endgroup$ – dato datuashvili Dec 16 '13 at 6:44
  • $\begingroup$ you can write $z = (1-\sqrt{x}-\sqrt{y})^2$ , ie z has that range $\endgroup$ – abkds Dec 16 '13 at 6:46
  • $\begingroup$ about $x$ and $y$? $\endgroup$ – dato datuashvili Dec 16 '13 at 6:48
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    $\begingroup$ This is a superellipsoid with $A=B=C=1$ and $r=t=\dfrac12$. The formula of its entire volume is in the article. Of course, you'll only need and eighth of that. As to why these shapes are related to the beta and gamma functions, see here. $\endgroup$ – Lucian Dec 16 '13 at 9:57
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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

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