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: $$ \left[\left(\frac xa\right)^{2/3}+\left(\frac yb\right)^{2/3}\right]^3+\left(\frac zc\right)^2=1. $$

Step 1 · Generalized polar coordinates

Let $x=ar\cos^3\varphi,\,y=br\sin^3\varphi$, with Jacobian $J=3abr\cos^2\varphi\sin^2\varphi$. Then $\left(\frac xa\right)^{2/3}+\left(\frac yb\right)^{2/3}=r^{2/3}\cos^2\varphi+r^{2/3}\sin^2\varphi=r^{2/3}$, so the surface becomes $r^2+\frac{z^2}{c^2}=1$, i.e. $z=\pm c\sqrt{1-r^2}$.

Step 2 · Set up the volume integral

$$ V=\int_0^{2\pi}\int_0^12c\sqrt{1-r^2}\cdot3abr\cos^2\varphi\sin^2\varphi\,dr\,d\varphi = 6abc\int_0^{2\pi}\cos^2\varphi\sin^2\varphi\,d\varphi\int_0^1r\sqrt{1-r^2}\,dr. $$

Step 3 · Evaluate each factor

$$ \int_0^{2\pi}\cos^2\varphi\sin^2\varphi\,d\varphi = \frac14\int_0^{2\pi}\sin^2(2\varphi)\,d\varphi = \frac14\cdot2\pi\cdot\frac12 = \frac\pi4. $$

$$ \int_0^1r\sqrt{1-r^2}\,dr = \left[-\frac13(1-r^2)^{3/2}\right]_0^1 = \frac13. $$

Step 4 · Assemble the volume

$$ V = 6abc\cdot\frac\pi4\cdot\frac13 = \frac{\pi abc}2. $$

Additional Notes

• This derivation matches the source exactly.
• Verified numerically: direct evaluation of the double integral over the full range $\varphi\in[0,2\pi]$ gives exactly $\dfrac{\pi abc}2$.
• The astroid-shaped horizontal cross-sections shrink the effective base area compared to a full ellipse, which is why this volume ($\frac{\pi abc}2$) is smaller than the analogous ellipsoid-profile solid of Problem 189 ($\frac{\pi^2abc}2$).