Find the volume of the solid bounded by the surface

Find the volume of the solid bounded by the surface: $$ \frac{x^n}{a^n}+\frac{y^n}{b^n}+\frac{z^n}{c^n}=1,\qquad x=0,\ y=0,\ z=0,\qquad(n>0). $$

Step 1 · Rescale to unit variables

Let $u=\frac xa,\,v=\frac yb,\,w=\frac zc$ (Jacobian $abc$). The region becomes $D=\{u,v,w\geqslant0:\ u^n+v^n+w^n\leqslant1\}$, and:

$$ V = abc\iiint_D du\,dv\,dw. $$

Step 2 · Dirichlet's formula

The classical Dirichlet integral for the volume of the region $x_1,\ldots,x_k\geqslant0,\ \sum x_i^{p_i}\leqslant1$ is $\displaystyle\prod_i\frac{\Gamma\!\left(1+\frac1{p_i}\right)}{\Gamma\!\left(1+\sum_i\frac1{p_i}\right)}$. With $k=3$ and $p_1=p_2=p_3=n$:

$$ \iiint_Ddu\,dv\,dw = \frac{\Gamma\!\left(1+\frac1n\right)^3}{\Gamma\!\left(1+\frac3n\right)}. $$

Therefore:

$$ V = \frac{abc\,\Gamma\!\left(1+\frac1n\right)^3}{\Gamma\!\left(1+\frac3n\right)}. $$

Step 3 · Check against known cases

For $n=1$ (tetrahedron): $\dfrac{\Gamma(2)^3}{\Gamma(4)}=\dfrac1{3!}=\dfrac16$, giving $V=\dfrac{abc}6$ — correct. For $n=2$ (ellipsoid octant): $\dfrac{\Gamma(3/2)^3}{\Gamma(5/2)}=\dfrac{(\sqrt\pi/2)^3}{3\sqrt\pi/4}=\dfrac\pi6$, giving $V=\dfrac{\pi abc}6$ — exactly one octant of the full ellipsoid volume $\dfrac{4\pi abc}3$, as required.

Additional Notes

• A note on the source computation: the source reached the same Dirichlet-formula setup, but included a spurious extra factor of $n^2$ in the denominator, $V=\dfrac{abc\,\Gamma(1+1/n)^3}{n^2\,\Gamma(1+3/n)}$. This passes the $n=1$ check only because $n^2=1$ there, but fails the source's own $n=2$ check: with the erroneous $n^2$ factor, $n=2$ gives $\dfrac{\pi abc}{24}$, not the correct one-octant-of-ellipsoid volume $\dfrac{\pi abc}6$ (a factor of $4$ off) — a discrepancy the source itself flagged without resolving.
• The corrected formula (without the $n^2$ factor) was verified by Monte Carlo simulation for $n=1,2,3,4$, matching in every case; the source's version matches only at $n=1$.