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: $$ \left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)^2+\frac{z^2}{c^2}=1. $$
Explanation of Method
With $\alpha=1$ (standard elliptical polar coordinates), $\frac{x^2}{a^2}+\frac{y^2}{b^2}=r^2$, so the surface becomes $z=\pm c\sqrt{1-r^4}$ over the disk $r\leqslant1$. The volume is the integral of the full height $2c\sqrt{1-r^4}$ over that disk.
- Convert to elliptical polar coordinates.
- Set up the volume integral over $r\in[0,1]$, $\varphi\in[0,2\pi]$.
- Evaluate using the substitution $u=r^2$.
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 $\frac{z^2}{c^2}=1-r^4$, giving $z=\pm c\sqrt{1-r^4}$, real for $r\leqslant1$.
Step 2 · Set up the volume integral
The full height at each point is $h(r)=2c\sqrt{1-r^4}$:
$$ V=\int_0^{2\pi}\int_0^12c\sqrt{1-r^4}\cdot abr\,dr\,d\varphi = 4\pi abc\int_0^1r\sqrt{1-r^4}\,dr. $$
Step 3 · Evaluate
Let $u=r^2,\,du=2r\,dr$:
$$ \int_0^1r\sqrt{1-r^4}\,dr = \frac12\int_0^1\sqrt{1-u^2}\,du = \frac12\cdot\frac\pi4 = \frac\pi8. $$
Therefore:
$$ V = 4\pi abc\cdot\frac\pi8 = \frac{\pi^2abc}2. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically: direct evaluation of $4\pi abc\int_0^1r\sqrt{1-r^4}\,dr$ gives exactly $\dfrac{\pi^2abc}2$.
- This surface is a spindle-torus-like shape, with elliptical cross-sections perpendicular to $z$ whose size follows a circular relationship in $(r^2,z)$.
Result
The volume is: