Find the surface area of the part of the paraboloid dfrac

Find the surface area of the part of the paraboloid $\dfrac{x^2}a+\dfrac{y^2}b=2z$ lying inside the cylinder $\left(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\right)^2=\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}$.

Step 1 · Surface-area integrand

With $z=\dfrac{x^2}{2a}+\dfrac{y^2}{2b}$: $1+z_x^2+z_y^2=1+\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}$.

Step 2 · Generalized polar coordinates

With $x=ar\cos\theta,\,y=br\sin\theta$ (Jacobian $abr$): $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=r^2$, and the cylinder becomes $r^4=r^2\cos2\theta\iff r^2=\cos2\theta$, real for $\theta\in\left[-\frac\pi4,\frac\pi4\right]\cup\left[\frac{3\pi}4,\frac{5\pi}4\right]$ (two lobes).

Step 3 · Inner and outer integrals

By symmetry of both lobes: $S=2ab\displaystyle\int_{-\pi/4}^{\pi/4}\int_0^{\sqrt{\cos2\theta}}r\sqrt{1+r^2}\,dr\,d\theta$. With $u=1+r^2$ and $1+\cos2\theta=2\cos^2\theta$:

$$ \int_0^{\sqrt{\cos2\theta}}r\sqrt{1+r^2}\,dr = \frac13\Big[(1+\cos2\theta)^{3/2}-1\Big] = \frac13\big(2\sqrt2\cos^3\theta-1\big). $$

So:

$$ S = \frac{4ab}3\int_0^{\pi/4}\big(2\sqrt2\cos^3\theta-1\big)\,d\theta. $$

With $t=\sin\theta$: $\displaystyle\int_0^{\pi/4}\cos^3\theta\,d\theta=\int_0^{1/\sqrt2}(1-t^2)\,dt=\dfrac5{6\sqrt2}$, and $\displaystyle\int_0^{\pi/4}d\theta=\dfrac\pi4$, giving:

$$ S = \frac{4ab}3\left(2\sqrt2\cdot\frac5{6\sqrt2}-\frac\pi4\right) = \frac{4ab}3\cdot\frac{20-3\pi}{12} = \frac{ab(20-3\pi)}9. $$

Additional Notes

• This derivation matches the source exactly. Note the resulting closed form has the same $(20-3\pi)$ numerator as doubleint-224 (a different problem, an $a^2$-scaled lemniscate volume computation), reflecting the same underlying trigonometric integral $\int_0^\pi(1+\sin\varphi)^{3/2}\,d\varphi=\frac{20}3$.
• Verified numerically (e.g. $a=1.3,\,b=2.1$): direct double integration gives $S\approx3.207817$, matching $\dfrac{ab(20-3\pi)}9$ exactly.