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 paraboloid: $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac zc $$ and the curve: $$ \frac{x^4}{a^4}+\frac{y^4}{b^4}=\frac{x^2}{a^2}+\frac{y^2}{b^2},\qquad z=0. $$
Explanation of Method
In standard elliptical polar coordinates, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=r^2$, and the boundary curve becomes $r^2=\dfrac1{\cos^4\varphi+\sin^4\varphi}$. Since $z=cr^2$, the volume integral separates into an elementary radial part and an angular part that reduces to a definite integral evaluated numerically to a clean closed form.
- Convert the boundary curve to elliptical polar form.
- Set up the volume integral with $z=cr^2$.
- Evaluate the radial part, then the angular integral.
Step 1 · Boundary curve in polar form
With $x=ar\cos\varphi,\,y=br\sin\varphi$ (Jacobian $J=abr$): $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=r^2$, and the boundary curve $r^4(\cos^4\varphi+\sin^4\varphi)=r^2$ gives $r^2=\dfrac1{\cos^4\varphi+\sin^4\varphi}$ (for $r\neq0$).
Step 2 · Set up the volume integral
Since $z=c\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=cr^2$:
$$ V=\int_0^{2\pi}\int_0^{r(\varphi)}cr^2\cdot abr\,dr\,d\varphi = \frac{abc}4\int_0^{2\pi}\frac{d\varphi}{\big(\cos^4\varphi+\sin^4\varphi\big)^2}. $$
Step 3 · Evaluate the angular integral
High-precision numerical evaluation of $\displaystyle J=\int_0^{2\pi}\frac{d\varphi}{(\cos^4\varphi+\sin^4\varphi)^2}$ gives exactly $J=3\sqrt2\,\pi$ (confirmed to 50 significant digits), so:
$$ V = \frac{abc}4\cdot3\sqrt2\,\pi = \frac{3\sqrt2\,\pi\,abc}4. $$
Additional Notes
- A note on the source computation: the source's derivation reached the same Step 2 setup, but then made an error evaluating the angular integral $\int_0^{2\pi}\frac{d\varphi}{(\cos^4\varphi+\sin^4\varphi)^2}$: it first claimed the value was $\frac{4\pi}3$, then second-guessed itself mid-derivation and "corrected" it to $\frac{8\pi}3$, arriving at a final answer of $\frac{2\pi abc}3$. Neither $\frac{4\pi}3\approx4.19$ nor $\frac{8\pi}3\approx8.38$ matches the true value of the integral. High-precision numerical evaluation gives $J\approx13.3286488\ldots=3\sqrt2\pi$ exactly (confirmed to 50 digits), and a fully independent direct Cartesian double integral over the true boundary curve (e.g. with $a=1.1,b=0.9,c=1.7$) gives $\approx5.60803$, matching the corrected closed form $\frac{3\sqrt2\pi\,abc}4$ at those parameters exactly — not the source's $\frac{2\pi abc}3\approx3.52$.
Result
The volume is: