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: $$ \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^k+\frac{z^2}{c^2}=1,\qquad z>0,\ k>0. $$
Explanation of Method
With standard elliptical polar coordinates ($\alpha=1$), $\frac{x^2}{a^2}+\frac{y^2}{b^2}=r^2$, and the (single, $z>0$) surface height becomes $z=c\sqrt{1-r^{2k}}$ over $r\leqslant1$. The radial integral reduces, via $u=r^{2k}$, to a Beta function.
- Convert to elliptical polar coordinates.
- Set up the volume as a single-nappe integral over the disk.
- Reduce the radial integral to a Beta function via $u=r^{2k}$.
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 for $z>0$: $z=c\sqrt{1-r^{2k}}$, real for $r\leqslant1$.
Step 2 · Set up the volume integral
$$ V=\int_0^{2\pi}\int_0^1c\sqrt{1-r^{2k}}\cdot abr\,dr\,d\varphi = 2\pi abc\int_0^1r\sqrt{1-r^{2k}}\,dr. $$
Step 3 · Reduce to a Beta function
Let $u=r^{2k}$, so $r=u^{1/(2k)}$ and $r\,dr=\frac1{2k}u^{1/k-1}\,du$:
$$ \int_0^1r\sqrt{1-r^{2k}}\,dr = \frac1{2k}\int_0^1u^{1/k-1}(1-u)^{1/2}\,du = \frac1{2k}B\left(\frac1k,\frac32\right) = \frac1{2k}\cdot\frac{\Gamma(1/k)\,\Gamma(3/2)}{\Gamma(1/k+3/2)}. $$
Using $\Gamma(3/2)=\dfrac{\sqrt\pi}2$:
$$ \int_0^1r\sqrt{1-r^{2k}}\,dr = \frac{\sqrt\pi}{4k}\cdot\frac{\Gamma(1/k)}{\Gamma(1/k+3/2)}. $$
Step 4 · Assemble the volume
$$ V = 2\pi abc\cdot\frac{\sqrt\pi}{4k}\cdot\frac{\Gamma(1/k)}{\Gamma(1/k+3/2)} = \frac{\pi^{3/2}abc}{2k}\cdot\frac{\Gamma(1/k)}{\Gamma\!\left(\frac1k+\frac32\right)}. $$
Additional Notes
- A note on the source computation: the source's own derivation reached exactly this formula, but then second-guessed itself at the end and replaced it with a different, incorrect "corrected" formula, $\dfrac{\sqrt\pi\,abc}{2k}\cdot\dfrac{\Gamma(1/k)}{\Gamma(1/k+1/2)}$ (mismatched Beta-function parameters). Direct numerical evaluation for several values of $k$ (e.g. $k=1,2,3,\frac12$) confirms the original formula above, $\dfrac{\pi^{3/2}abc}{2k}\cdot\dfrac{\Gamma(1/k)}{\Gamma(1/k+3/2)}$, and refutes the source's late "correction."
- As a consistency check: at $k=2$, this problem's surface is identical to that of Problem 189, but restricted to $z>0$ only. Evaluating the formula above at $k=2$ gives exactly half of Problem 189's volume $\frac{\pi^2abc}2$, i.e. $\frac{\pi^2abc}4$ — confirmed both by the formula and by direct numerical integration.
Result
The volume is: