Find the surface area of the part of the surface
Find the surface area of the part of the surface $(x^2+y^2+z^2)^2=a^2(x^2-y^2)$ that projects onto the interior of the curve $(x^2+y^2)^2=a^2(x^2-y^2)$.
Explanation of Method
In spherical coordinates $x=r\sin\varphi\cos\theta,\,y=r\sin\varphi\sin\theta,\,z=r\cos\varphi$, the surface reduces to $r^2=a^2\sin^2\varphi\cos2\theta$ — a "spherical lemniscate," real only for $\cos2\theta\ge0$. Its projection onto $z=0$ (i.e. $\varphi=\frac\pi2$) recovers the plane lemniscate $\rho^2=a^2\cos2\theta$ exactly as expected. Computing the surface-area element for a surface $r=r(\varphi,\theta)$ given in spherical coordinates and substituting this particular $r$ produces a remarkable simplification, collapsing to $dS=a^2\sin^2\varphi\,d\varphi\,d\theta$.
- Convert the surface equation to spherical coordinates.
- Compute the surface-area element $dS$ for $r=r(\varphi,\theta)$ and simplify.
- Integrate over both lobes ($\theta$-ranges where $\cos2\theta\ge0$) and the full polar range $\varphi\in[0,\pi]$.
Step 1 · Spherical coordinates
With $x^2+y^2+z^2=r^2$ and $x^2-y^2=r^2\sin^2\varphi\cos2\theta$, the surface equation $r^4=a^2r^2\sin^2\varphi\cos2\theta$ gives (for $r\ne0$):
$$ r^2 = a^2\sin^2\varphi\cos2\theta, \qquad r=a\sin\varphi\sqrt{\cos2\theta}, $$
real only for $\theta\in\left[-\frac\pi4,\frac\pi4\right]\cup\left[\frac{3\pi}4,\frac{5\pi}4\right]$ (two lobes, matching the plane lemniscate's two loops).
Step 2 · Surface-area element
For $r=r(\varphi,\theta)$, the surface-area element is $dS=\big\|\mathbf r_\varphi\times\mathbf r_\theta\big\|\,d\varphi\,d\theta$. Carrying out this cross product for $r=a\sin\varphi\sqrt{\cos2\theta}$ (verified directly, term by term) simplifies remarkably to:
$$ dS = a^2\sin^2\varphi\,d\varphi\,d\theta. $$
Step 3 · Integrate
For fixed $\theta$ in either lobe, $\varphi$ ranges over the full $[0,\pi]$. By symmetry, the two lobes contribute equally, so:
$$ S = 2\int_{-\pi/4}^{\pi/4}\int_0^\pi a^2\sin^2\varphi\,d\varphi\,d\theta = 2a^2\left[\int_0^\pi\sin^2\varphi\,d\varphi\right]\left[\int_{-\pi/4}^{\pi/4}d\theta\right] = 2a^2\cdot\frac\pi2\cdot\frac\pi2 = \frac{\pi^2a^2}2. $$
Additional Notes
- This derivation matches the source exactly, including the notable identity $dS=a^2\sin^2\varphi\,d\varphi\,d\theta$, which was verified independently by computing the full cross-product magnitude $\|\mathbf r_\varphi\times\mathbf r_\theta\|$ symbolically (a lengthy expression in $\varphi,\theta$) and confirming it collapses exactly to $a^2\sin^2\varphi$ with no remaining $\theta$-dependence.
- Verified numerically: direct evaluation of the double integral over both lobes (e.g. $a=1.4$) gives $\approx9.672212$, matching $\dfrac{\pi^2a^2}2$ exactly.
Result
The surface area is: