Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the lemniscate $(x^2+y^2)^2=a^2(x^2-y^2)$ about the $x$-axis.

Step 1: Convert to polar coordinates

\[\rho^2=a^2\cos2\varphi \implies \rho=a\sqrt{\cos2\varphi\,},\qquad \varphi\in\left[-\frac\pi4,\frac\pi4\right]\ \text{(right lobe)}\]

Step 2: Differentiate

\[\rho'=-\frac{a\sin2\varphi}{\sqrt{\cos2\varphi}}\]

Step 3: Simplify $\rho^2+(\rho')^2$

\[\rho^2+(\rho')^2=a^2\cos2\varphi+\frac{a^2\sin^22\varphi}{\cos2\varphi}=\frac{a^2(\cos^22\varphi+\sin^22\varphi)}{\cos2\varphi}=\frac{a^2}{\cos2\varphi}\]

\[\sqrt{\rho^2+(\rho')^2}=\frac a{\sqrt{\cos2\varphi}}\]

Step 4: Set up the integral

By symmetry, integrate the upper half of the right lobe ($\varphi\in[0,\pi/4]$) and double:

\[S=2\cdot2\pi\int_0^{\pi/4}a\sqrt{\cos2\varphi}\cdot\sin\varphi\cdot\frac a{\sqrt{\cos2\varphi}}\,d\varphi = 4\pi a^2\int_0^{\pi/4}\sin\varphi\,d\varphi\]

(the $\sqrt{\cos2\varphi}$ factors cancel exactly, since the polar radius and the arc-length correction offset one another.)

Step 5: Evaluate

\[S=4\pi a^2\left[-\cos\varphi\right]_0^{\pi/4}=4\pi a^2\left(1-\frac{\sqrt2}2\right)=2\pi a^2(2-\sqrt2)\]

\[\boxed{S=2\pi a^2(2-\sqrt2)\approx3.68a^2}\]