Dirichlet's Integral to find Volume with gamma function values Evaluate volume $\iiint_D dx dy dz$ over a domain $D$ where $D$ is the region bounded by $x\ge 0$, $y\ge 0$ and $z\ge 0$ and ellipsoid $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} = 1$.
My approach:
I used the RMS$\ge$AM inequality (since $x,y,z\ge 0$ and $a,b,c>0$ without loss of generality) to arrive at $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\le\sqrt{3}$
and then did a substitution $X=\frac{x}{a}$, $Y=\frac{y}{b}$ and $Z=\frac{z}{c}$
And I got answer to be
$\frac{abc(\cdot\Gamma(1)\Gamma(1)\Gamma(1)\cdot(\sqrt{3})^3)}{\Gamma(4)}$ ie $abc\sqrt{3}/2$
My friend's approach:
Direct substitution of $X=\frac{x^2}{a^2}$, $Y=\frac{y^2}{b^2}$ and $Z=\frac{z^2}{c^2}$ and his answer was
$(\frac{abc}{8})\frac{(\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})\Gamma(\frac{1}{2}))}{\Gamma(\frac{5}{2})}$ ie $\frac{abc\cdot\pi}{6}$
Professor says mine is wrong and friend's method is right but I am not satisfied without the reason.

Did you just replace the ellipsoid $(\frac{x}{a})^2+(\frac{y}{b})^2+(\frac{z}{c})^2\le 1$ with the octahedron $|\frac{x}{a}|+|\frac{y}{b}|+|\frac{z}{c}|\le\sqrt{3}$ ?
No wonder you are getting a different result for its volume. The RMS$\ge$AM
inequality guarantees that the ellipsoid is inscribed in the octahedron, i.e. fully contained in it - but of course, they are not the same bodies.
