<s>[INST] Calculating the center of mass of a lemniscate rotated around the x-axis? This is a problem I have been stuck on a while, it goes as follows:
A lemniscate has the equation $(x^2+y^2)^2 = 4(x^2-y^2)$. Let the part of the curve that lies in the first quadrant rotate around the $x$-axis to create an object. This object has an homogeneous massdistribution with the density $1$. Determine the coordinates of the center of mass $(x_T,y_T,z_T)$, if $x_T = \frac{1}{M}\int _K x dm$, where $M$ is the mass of the object.
I have so far managed to switch to polar coordinates and found the relation $r^2 = 4\cos{2\theta}$ and tried to calculate the volume (which equals the mass in this case due to the density being equal to 1) with the formula $\int _K \pi y^2 dx$. After the switch to polar coordinates, a point on the curve is $(x,y) = (r\cos{\theta},r\sin{\theta})$, which gives $\pi y^2 = \pi r^2\sin^2{\theta} = 4\pi\cos{2\theta}\sin^2{\theta}$.
I then reason that a small change $dx$ in polar coordinates, corresponds to a small change in $\theta$, hence I substitute $dx$ with $\frac{dx}{d\theta} 2\sqrt{\cos{2\theta}}\cos{\theta} \Leftrightarrow dx = -2(\frac{\sin{2\theta}\cos{\theta}}{\sqrt{\cos{2\theta}}}+ \sin{\theta}\sqrt{\cos{2\theta}}) d\theta$.
But when I try and calculate the integral
$\int^2_0 4\pi\cos{2\theta}\sin^2{\theta} \cdot (-2)(\frac{\sin{2\theta}\cos{\theta}}{\sqrt{\cos{2\theta}}}+ \sin{\theta}\sqrt{\cos{2\theta}}) d\theta $,
I get an incorrect volume/mass.
The correct solution is $(x_T,y_T,z_T) = (\frac{2}{3\sqrt{2}\ln{(\sqrt{2}+1)}-2},0,0)$
I would greatly appreciate any tips on how to proceed, thanks in advance!
 [/INST] Because you are rotating something around the $x$-axis, we will have $y=z=0$.
Now, for this case, I prefer to use the traditional coordinates. First we have ro solve for $y$ the equation:
$$(x^2+y^2)^2-4(x^2-y^2)=0\leftrightarrow  \left[t=x^2\right] \leftrightarrow y=\pm\sqrt{\frac{-2x^2-4\pm\sqrt{(-2x^2-4)^2-4(x^4-4x^2)}}{2}}=\pm\sqrt{-x^2-2\pm\sqrt{1+2x^2}}$$
We are interested in the first quadrant, so:
$$y=\sqrt{-x^2-2+\sqrt{1+2x^2}}$$
Now, by definition of mass centre:
$$x_M=\frac{1}{M}\cdot\int_M xdm=\frac{1}{V}\cdot\int_V x dV$$
Where $dV$ is the volume of a disk with radius $y(x)$ and height $dx$.
We have:
$$V=\int_0^2 \pi \left(\sqrt{-x^2-2+\sqrt{1+2x^2}}\right)^2dx=\int_0^2 \pi\left(-x^2-2+\sqrt{1+2x^2}\right)dx=\pi\cdot\left[\ln\left(\frac{\sqrt{2x^2+1}+\sqrt2x}{\sqrt2}\right)+x\sqrt{2x^2+1}-\frac{x^3}{3}-2x+c\right]_0^2=\pi\cdot\left(\frac{\ln(3+2\sqrt2)}{\sqrt2}+6-\frac83-4\right)=\pi\cdot\left(\frac{\ln(3+2\sqrt2)}{\sqrt2}-\frac23 \right)$$
Also:
$$\int_0^2 \pi x\left(\sqrt{-x^2-2+\sqrt{1+2x^2}}\right)^2dx=\int_0^2 \pi x\left(-x^2-2+\sqrt{1+2x^2}\right)dx=\pi\left[\frac{\sqrt{(2x^2+1)^3}}{3}-\frac{x^4}{4}-x^2+c\right]_0^2=1-\frac13=\frac23\pi$$
Finally:
$$x_M=\frac{1}{M}\cdot\int_M xdm=\frac{1}{V}\cdot\int_V x dV=\frac{\frac23\pi}{\pi\cdot\left(\frac{\ln(3+2\sqrt2)}{\sqrt2}-\frac23 \right)}=\frac{2\sqrt2}{3\ln(3+2\sqrt2)-2\sqrt2}=\frac{2}{3\sqrt{2}\ln{(\sqrt{2}+1)}-2}$$
Where the last equality holds because you can pick up from numerator and denominator a $\sqrt2$ term and then notice that:
$$\sqrt{3+2\sqrt2}=\sqrt{\left( 1+\sqrt2 \right)^2}=1+\sqrt2$$
</s>