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: $$ \left(\sqrt{\frac xa}+\sqrt{\frac yb}\right)^4+\left(\frac zc\right)^2=1,\qquad x>0,\ y>0,\ z>0. $$

Step 1 · Generalized polar coordinates

Let $x=ar\cos^2\varphi,\,y=br\sin^2\varphi$, Jacobian $J=2abr\cos\varphi\sin\varphi=abr\sin2\varphi$. Then $\sqrt{x/a}+\sqrt{y/b}=\sqrt r(\cos\varphi+\sin\varphi)$ (for $\varphi\in[0,\frac\pi2]$, where both terms are non-negative), so the surface becomes:

$$ r^2(\cos\varphi+\sin\varphi)^4+\frac{z^2}{c^2}=1,\qquad z=c\sqrt{1-r^2(\cos\varphi+\sin\varphi)^4}\ (z>0), $$

real for $r\leqslant\dfrac1{(\cos\varphi+\sin\varphi)^2}$.

Step 2 · Set up the volume integral

$$ V = \int_0^{\pi/2}\int_0^{1/(\cos\varphi+\sin\varphi)^2}c\sqrt{1-r^2(\cos\varphi+\sin\varphi)^4}\cdot abr\sin2\varphi\,dr\,d\varphi. $$

Step 3 · Evaluate the radial integral

With $s=\cos\varphi+\sin\varphi$ and $u=r^2s^4$ (so $r\,dr=\frac{du}{2s^4}$):

$$ \int_0^{1/s^2}r\sqrt{1-r^2s^4}\,dr = \frac1{2s^4}\int_0^1\sqrt{1-u}\,du = \frac1{2s^4}\cdot\frac23 = \frac1{3s^4}. $$

Step 4 · Evaluate the angular integral

$$ V = \frac{abc}3\int_0^{\pi/2}\frac{\sin2\varphi}{(\cos\varphi+\sin\varphi)^4}\,d\varphi. $$

Since $(\cos\varphi+\sin\varphi)^2=1+\sin2\varphi$, this is $\displaystyle\frac{abc}3\int_0^{\pi/2}\frac{\sin2\varphi}{(1+\sin2\varphi)^2}\,d\varphi$. High-precision numerical evaluation of $\displaystyle\int_0^{\pi/2}\frac{\sin2\varphi}{(1+\sin2\varphi)^2}\,d\varphi$ gives exactly $\dfrac13$ (confirmed to 30 significant digits), so:

$$ V = \frac{abc}3\cdot\frac13 = \frac{abc}9. $$

Additional Notes

• A note on the source computation: the source reached the same Step 4 setup, then attempted the substitution $w=\sin2\varphi$ directly on $\displaystyle\int_0^{\pi/2}\frac{\sin2\varphi}{(1+\sin2\varphi)^2}\,d\varphi$, treating $\sin2\varphi\,d\varphi$ as if it equaled $dw$. It does not: $dw=2\cos2\varphi\,d\varphi\neq\sin2\varphi\,d\varphi$, so this substitution is invalid, and the source's resulting answer $\dfrac{abc(2\ln2-1)}6\approx0.0644\,abc$ is incorrect.
• High-precision numerical evaluation of the angular integral (30 significant digits) confirms it equals exactly $\dfrac13$, giving $V=\dfrac{abc}9\approx0.1111\,abc$ — verified independently by a direct nested numerical double integral over $(r,\varphi)$, which matches to 10 significant figures.