Find the surface area of (x²+y²+z²)²=a²(x²+y²)
Find the surface area of $(x^2+y^2+z^2)^2=a^2(x^2+y^2)$.
Explanation of Method
In spherical coordinates this surface reduces to $r=a\sin\varphi$ ($\varphi\in[0,\pi]$), independent of $\theta$ — a torus of revolution obtained by rotating the circle $\left(\rho-\frac a2\right)^2+z^2=\left(\frac a2\right)^2$ (in cylindrical $\rho,z$) about the $z$-axis. Since the generating circle passes through the origin, this is a degenerate ("horn") torus. Its area can be found either directly via the surface-of-revolution formula on the generating circle, or via the spherical-coordinates area element for $r=r(\varphi)$ — both are carried out here as a cross-check.
- Convert to spherical coordinates and recognize the torus.
- Compute the area via the surface-of-revolution formula on the generating circle.
- Verify independently via the spherical-coordinates area element.
Step 1 · Spherical coordinates
With $r^2=x^2+y^2+z^2$ and $x^2+y^2=r^2\sin^2\varphi$: $r^4=a^2r^2\sin^2\varphi\Rightarrow r=a\sin\varphi$ ($\varphi\in[0,\pi]$). In cylindrical coordinates ($\rho=\sqrt{x^2+y^2}$): $(\rho^2+z^2)^2=a^2\rho^2\Rightarrow\rho^2+z^2=a\rho\Rightarrow\left(\rho-\frac a2\right)^2+z^2=\left(\frac a2\right)^2$ — a circle of radius $\frac a2$ centered at $\left(\frac a2,0\right)$, i.e. a horn torus.
Step 2 · Surface-of-revolution formula
Parametrize the generating circle as $\rho(t)=\frac a2(1+\cos t),\,z(t)=\frac a2\sin t$, $t\in[0,2\pi]$, so $\sqrt{\rho'(t)^2+z'(t)^2}=\frac a2$. Then:
$$ S = 2\pi\int_0^{2\pi}\rho(t)\cdot\frac a2\,dt = \frac{\pi a^2}2\int_0^{2\pi}(1+\cos t)\,dt = \frac{\pi a^2}2\cdot2\pi = \pi^2a^2. $$
Step 3 · Cross-check via spherical coordinates
With $r=a\sin\varphi$, $\dfrac{dr}{d\varphi}=a\cos\varphi$, so $r^2+\left(\dfrac{dr}{d\varphi}\right)^2=a^2$, giving area element $r\sqrt{r^2+r_\varphi^2}\,\sin\varphi\,d\varphi\,d\theta=a^2\sin^2\varphi\,d\varphi\,d\theta$, and:
$$ S = \int_0^{2\pi}\int_0^\pi a^2\sin^2\varphi\,d\varphi\,d\theta = 2\pi a^2\cdot\frac\pi2 = \pi^2a^2, $$
confirming Step 2.
Additional Notes
- This derivation matches the source exactly, including its own internal cross-check between the two methods.
- Verified numerically (e.g. $a=1.9$): direct double integration of $a^2\sin^2\varphi$ over $\varphi\in[0,\pi],\,\theta\in[0,2\pi]$ gives $\approx35.6293$, matching $\pi^2a^2$ exactly.
Result
The surface area is: