I am trying to find the volume of a loop of the lemniscate $r^{2}=a^{2}\sin2\theta$ when rotated about the line $y=x$. Is it possible to do this using only high school methods? (i.e taking washers/slices or cylindrical shells).
My first thought was to take slices perpendicular to the line $y=x$ of radius $\left|\frac{x-y}{\sqrt{2}}\right|$ (perpendicular distance of point to line y=x). However the cartesian equation of the lemniscate is $(x^{2}+y^{2})^{2}=a^{2}(y^{2}-x^{2})$, which I can't seem to express nicely in terms of $y$ so that I can calculate the volume using an integral with respect to $x$.
I then thought of rotating the lemniscate $\frac{\pi}{4}$ clockwise to the curve $r^{2}=a^{2}\cos2\theta$ and using cylindrical shells or slices but again I ran into the problem of trying to express $y$ in terms of $x$ (using the quadratic formula gave me the square root of a quartic which doesn't seem integrable). I know there are approximations I can use, but I just want to know if this problem is possible with only these basic methods (washers/slices and cylindrical shells) and not Pappus's centroid theorem which I have learnt yet.
Thanks
Best Answer
This seems to work, although at first I was very unsure:
First, I changed to $r^2=a^2\cos 2\theta$. Then for convenience, I set $a=1$ and $t=\theta.$ So I rotate the $1/4$ lemniscate as $t=0$ to $\pi/4$ about the $x$-axis, and try to do method of disks.
A point on the lemniscate has Cartesian coordinates $(r\cos t, r\sin t) =$ $ (\sqrt{\cos 2t}\cos t, \sqrt{\cos 2t}\sin t).$ Then we have the area of the disk
$$\pi y^2 = \pi \cos 2t \sin^2 t. $$
And we have
$$dx =\sqrt{\cos 2t}(-\sin t) + \frac{-2\sin 2t}{\sqrt{\cos 2t}}\cos t \; dt.$$
So the volume of one lobe should be
$$\int_{x=0}^{x=1} \pi y^2 \; dx = \int_{x=0}^{x=1} \pi\cos 2t \sin^2 t\left(\sqrt{\cos 2t}(-\sin t) + \frac{-2\sin 2t}{\sqrt{\cos 2t}}\cos t\right) \; dt$$
$$=-\pi \int_{t=\pi/4}^{t=0} \cos^{3/2}2t\sin^3 t +\sqrt{\cos2t}\sin2t\sin^2 t\cos t \; dt. $$
Flip the integral to get rid of the minus sign. I put this in Maple and it gave me
$$-\frac{\pi}{12} +\frac{\pi\sqrt{2}}{8}\ln(1+\sqrt{2}) \approx 0.228.$$
We'd have to double that for the total volume. The antiderivative Maple gives for the integral above is not that crazy, so I think a very stubborn high school calculus student could do it by hand.