Find the volume of the solid in the first octant bounded

Find the volume of the solid in the first octant bounded by the surface: $$ z^2=2xy $$ and the cylinder erected on the curve: $$ \big(x^2+y^2\big)^2=2a^2xy, $$ with $x,y,z>0$.

Step 1 · Polar form

With $x=r\cos\theta,\,y=r\sin\theta$: $z=\sqrt{2xy}=r\sqrt{\sin2\theta}$ (valid where $\sin2\theta\geqslant0$), and the boundary curve becomes $r^4=2a^2r^2\cos\theta\sin\theta=a^2r^2\sin2\theta$, i.e. $r^2=a^2\sin2\theta$. For $x,y>0$ this is the single loop $\theta\in\left(0,\frac\pi2\right)$.

Step 2 · Set up the volume integral

$$ V=\int_0^{\pi/2}\int_0^{a\sqrt{\sin2\theta}}r\sqrt{\sin2\theta}\cdot r\,dr\,d\theta = \int_0^{\pi/2}\sqrt{\sin2\theta}\left[\frac{r^3}3\right]_0^{a\sqrt{\sin2\theta}}d\theta = \frac{a^3}3\int_0^{\pi/2}\sin^2(2\theta)\,d\theta. $$

Step 3 · Evaluate

Using $\displaystyle\int_0^{\pi/2}\sin^2(2\theta)\,d\theta=\frac\pi4$:

$$ V = \frac{a^3}3\cdot\frac\pi4 = \frac{\pi a^3}{12}. $$

Additional Notes

• This derivation matches the source's final answer exactly, over the single first-quadrant loop of the lemniscate (as required by $x,y>0$).
• Verified numerically: direct quadrature of $\displaystyle\int_0^{\pi/2}\int_0^{a\sqrt{\sin2\theta}}r^2\sqrt{\sin2\theta}\,dr\,d\theta$ gives exactly $\dfrac{\pi a^3}{12}$. Note that the lemniscate $r^2=a^2\sin2\theta$ has a second loop for $\theta\in\left(\pi,\frac{3\pi}2\right)$ (where $x,y<0$, also satisfying $xy>0$); restricting to $x,y>0$ as stated correctly excludes it. Including both loops would double the volume to $\dfrac{\pi a^3}6$.