Lemniscate volume by rotation I’ve been trying to create a novel torus-like figure by rotating a lemniscate of r^2=a^2cos(Theta) about the Y axis, with parameters R (distance from midpoint of torus to midpoint of lemniscate cross section) and r (distance from midpoint of lemniscate to endpoint of lemniscate). I’ve likened a traditional torus with a traditional circular cross section to the torus I’m attempting to create by using a lemniscate. 
In my endeavour, I’m trying to derive a formula for the volume of this lemniscate-based torus. To do this, I’ve first looked into ways that a traditional torus’ volume formula has been proven. One method shown below is known as the shell method. The shell method multiplies the height (y) of some rectangle and the distance (x) that rectangle is from the origin. The product is then integrated by the two bounds the circle is constrained to, then multiplied by 2 pi. 
Given what I saw from this proof of a torus’ volume, I tried applying a similar proof to find a formula for the volume of the lemniscate-torus I’m attempting to create. Because the shell method requires I find the height (y) based on the formula of a lemniscate, I took the Cartesian equation of a lemniscate and solved for y. I then simplified the integral but found myself stuck at a dead end. 
Another thing I noticed is that unlike a circle, the distance from the midpoint of a torus to its endpoint isn’t exactly the radius from all lines drawn from the midpoint to some arc. Because a torus relies on constraint r, I didn’t know how I could express the value of r with change in the angle between the line drawn from the midpoint to the arc, and the horizontal x axis. 
I’m wondering if anyone might know where to go from here, or even have any alternative ways to find the equation I’m trying to derive. I’d really appreciate it because I’m at a bit of a dead end and have been staring at my screen for quite some time. 
All the best!
 A: I will follow the same idea as they (you?) did for the torus.
The cartesian equation for the lemniscate is $(x^2+y^2)^2=2a^2(x^2-y^2)$. In particular we get $x=\pm\sqrt2a$ as the extreme points on the $x$-axis by setting $y=0$ in the equation. 
Solving for $y$ and shifting by $R$ gives
$$
y = \pm \sqrt{-a^2-(x-R)^2+\sqrt{a^4+4a^2(x-R)^2}}
$$
Therefore the solid of revolution (the lemniscate "torus") has volume:
$$
4\pi\int_{R-\sqrt2a}^{R+\sqrt2a} x \sqrt{-a^2-(x-R)^2+\sqrt{a^4+4a^2(x-R)^2}} \,dx
$$
$$
= 4\pi\int_{-\sqrt2a}^{\sqrt2a} (u+R) \sqrt{-a^2-u^2+\sqrt{a^4+4a^2u^2}} \,du
$$
substituting $u=x-R$. $u\mapsto u\sqrt{-a^2-u^2+\sqrt{a^4+4a^2u^2}}$ is odd, so the volume is equal to:
$$
4\pi R\int_{-\sqrt2a}^{\sqrt2a} \sqrt{-a^2-u^2+\sqrt{a^4+4a^2u^2}} \,du
$$
$$
= 4\pi Ra^2\int_{-\sqrt2}^{\sqrt2} \sqrt{-1-v^2+\sqrt{1+4v^2}} \,dv
$$
$$
= 4\pi Ra^2 \cdot \big(\textrm{area of upper half of lemniscate with }a=1\big)
$$
$$
= 4\pi Ra^2
$$
where we substituted $u=av$. I haven't looked into how the area of the lemniscate is found, but surely that can be looked up. I'm pretty sure it's a good idea to shift to polar coordinates here. 
It is interesting to note that neither wolframalpha nor maple can work out the integral exactly, even though the answer is just $1$. You can get like five lines of decimal places on wolframalpha though: $1.0000000000$... 
EDIT: For completeness: The right half of the lemniscate with $a=1$, i.e. $r^2=2\cos(2\theta)$, has area:
$$
\int_{-\pi/4}^{\pi/4} \int_0^{\sqrt{2\cos(2\theta)}} r \,dr\,d\theta
= \frac12 \int_{-\pi/4}^{\pi/4} 2\cos(2\theta) \,d\theta
= \frac12 \left(\sin\frac\pi2 - \sin\left(-\frac\pi2\right)\right)
= 1
$$
EDIT 2: It's no coincidence that the derivation was so similar to the normal torus, so let's generalise. Assume that the cross-section shape $A$ of our "torus" ($A$ corresponds to the circle or lemniscate) is vertically symmetric and scales with a parameter $a$. The height at $x$ of $A(a)$ is then:
$$
h_a(x) = a h\left(\frac xa\right) \quad {-c}a\le x\le ca 
$$
where $h(x)$ is the height function for $a=1$, and $h$ is even. Now the same calculations work:
$$
\big(\textrm{Volume of "$A$-torus"}\big)
= 2\pi\int_{R-ca}^{R+ca}xah\left(\frac {x-R}a\right) \, dx
= 2\pi\int_{-ca}^{ca}(u+R)ah\left(\frac {u}a\right) \, du
$$
$$
= 2\pi R\int_{-ca}^{ca}ah\left(\frac {u}a\right) \, du
= 2\pi Ra^2\int_{-c}^{c}h\left(v\right) \, dv
$$
$$
= 2\pi Ra^2 \cdot\big(\textrm{Area of $A$ with $a=1$}\big)
$$
In short, if the shape is vertically symmetric, then all you need to know is the area of one instance of the shape.
