Using generalized polar coordinates x=ar cos^()alpha phi
Using generalized polar coordinates $x=ar\cos^\alpha\varphi,\ y=br\sin^\alpha\varphi$ (Jacobian $J=ab\alpha r\cos^{\alpha-1}\varphi\sin^{\alpha-1}\varphi$), find the volume of the solid bounded by the surface: $$ \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2+\frac{z^4}{c^4}=1. $$
Explanation of Method
With standard elliptical polar coordinates, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=r^2$, and $z=\pm c(1-r^4)^{1/4}$. The radial integral, via $u=r^4$, reduces to a Beta function whose Gamma-function values simplify using the reflection formula.
- Convert to elliptical polar coordinates.
- Set up the volume integral with full height $2c(1-r^4)^{1/4}$.
- Reduce the radial integral to a Beta function and simplify via the Gamma reflection formula.
Step 1 · Elliptical polar coordinates
Let $x=ar\cos\varphi,\,y=br\sin\varphi$ (Jacobian $J=abr$). Then $\frac{x^2}{a^2}+\frac{y^2}{b^2}=r^2$, so $z=\pm c(1-r^4)^{1/4}$, real for $r\leqslant1$.
Step 2 · Set up the volume integral
$$ V=\int_0^{2\pi}\int_0^12c(1-r^4)^{1/4}\cdot abr\,dr\,d\varphi = 4\pi abc\int_0^1r(1-r^4)^{1/4}\,dr. $$
Step 3 · Reduce to a Beta function
Let $u=r^4$, so $r\,dr=\frac14u^{-1/2}\,du$:
$$ \int_0^1r(1-r^4)^{1/4}\,dr = \frac14\int_0^1u^{-1/2}(1-u)^{1/4}\,du = \frac14B\left(\frac12,\frac54\right) = \frac14\cdot\frac{\Gamma(1/2)\,\Gamma(5/4)}{\Gamma(7/4)}. $$
Using $\Gamma(1/2)=\sqrt\pi$, $\Gamma(5/4)=\frac14\Gamma(1/4)$, $\Gamma(7/4)=\frac34\Gamma(3/4)$, and the reflection formula $\Gamma(1/4)\Gamma(3/4)=\dfrac\pi{\sin(\pi/4)}=\pi\sqrt2$:
$$ \frac{\Gamma(1/2)\,\Gamma(5/4)}{\Gamma(7/4)} = \frac{\sqrt\pi\cdot\frac14\Gamma(1/4)}{\frac34\Gamma(3/4)} = \frac{\sqrt\pi}3\cdot\frac{\Gamma(1/4)}{\Gamma(3/4)} = \frac{\sqrt\pi}3\cdot\frac{\Gamma(1/4)^2}{\Gamma(1/4)\Gamma(3/4)} = \frac{\sqrt\pi}3\cdot\frac{\Gamma(1/4)^2}{\pi\sqrt2} = \frac{\Gamma(1/4)^2}{3\sqrt{2\pi}}. $$
So $\displaystyle\int_0^1r(1-r^4)^{1/4}\,dr = \frac{\Gamma(1/4)^2}{12\sqrt{2\pi}}$.
Step 4 · Assemble the volume
$$ V = 4\pi abc\cdot\frac{\Gamma(1/4)^2}{12\sqrt{2\pi}} = \frac{\pi\,abc\,\Gamma(1/4)^2}{3\sqrt{2\pi}} = \frac{\sqrt{2\pi}\,abc\,\Gamma(1/4)^2}6. $$
Additional Notes
- This derivation matches the source's first boxed answer, $V=\dfrac{\sqrt{2\pi}\,abc\,\Gamma(1/4)^2}6$, confirmed by direct numerical double integration (e.g. with $a=1.1,b=0.9,c=1.7$, giving $\approx9.24240$, matching the formula exactly).
- A note on the source's second "equivalent" form: the source additionally boxed an allegedly equivalent alternative, $V=\dfrac{abc\,\Gamma(1/4)^2}{3\sqrt{2\pi}}$. This is not actually equal to the first form — numerically, $\dfrac{\Gamma(1/4)^2}{3\sqrt{2\pi}}\approx1.748$, whereas the correct coefficient $\dfrac{\sqrt{2\pi}\,\Gamma(1/4)^2}6\approx5.492$, a factor of $2\pi$ apart. The first form (matching the direct numerical check above) is the correct one.
Result
The volume is: