Volume generated by lemniscate revolving about a tangent at the pole. The lemniscates $r^2 = a^2\cos2\theta$ revolves about a tangent at the pole. What is the volume generated by it ?
Please explain in detail. I found a couple of answers on finding surface areas, volumes but didn't understood it.
Can someone solve this. I'm stuck in the middle with integrations.
 A: The lemniscate with $a=1$ is the locus of points $(x,y)$ for which:
$$ (x^2+y^2)^2 = x^2-y^2 $$
and the tangents in the origin are the lines $y=\pm x$. Let $R$ be the region of the $x\geq 0$ halfplane bounded by the lemniscate. We just need to compute the area $A$ of $R$, the centroid $G$ of $R$, then apply the second Pappus' centroid theorem. By using polar coordinates we have:
$$ A = \frac{1}{2}\int_{-\pi/4}^{\pi/4}r^2\,d\theta = \frac{1}{2}.\tag{1}$$
By symmetry, the centroid of $R$ lies on the $y=0$ line. Its abscissa is given by:
$$\begin{eqnarray*} G_x &=& \frac{\int_{0}^{1}x\,f(x)\,dx}{\int_{0}^{1}f(x)\,dx}=\frac{\int_{0}^{1}x\,f(x)\,dx}{A/2}=4\int_{0}^{1}x\,f(x)\,dx\\&=&2\sqrt{2}\int_{0}^{1}x\sqrt{-1-2x^2+\sqrt{1+8x^2}}\,dx\\&=&\frac{\pi}{4\sqrt{2}}\tag{2}\end{eqnarray*}$$
hence the distance of the centroid of $R$ from a tangent line in the origin is just $\pi/8$, and:
$$ V = 2\cdot 2\pi\cdot\frac{\pi}{8}\cdot A = \frac{\pi^2}{4},\tag{3}$$
so, in the general case:

$$ V = \frac{1}{4}\pi^2 |a|^3.$$

