Volume of Lemniscate rotated using High School methods

Question

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

Answer 1

Part 1: Volume of Lemniscate by integrating with respect to $r$

Given the rotational symmetry around the x-axis the volume of a Lemniscate [ defined by $r^2=a^2\cos(2\theta)$ ] $V$ can be determined using the integral below $$V=2\pi \int_0^a y^2\;dx$$ which may be expanded as $$V=V_1-V_2= \left(2\pi \int_0^a r^2\;dx \right)-\left(2\pi \int_0^a x^2\;dx \right)$$ Finding $V_2$ is straight forward the answer being $V_2=\frac{2\pi a^3}{3}$

The $V_1$ integral can (with a lot of work) be transformed to an integral w.r.t. $r$. Since $x=r\cos(\theta)$ we first need to rewrite $\cos(\theta)$ as a function of $r$. This can be done using the defining equation $r^2=a^2\cos(2\theta)$ and the identity $\cos(2\theta)=2\cos^2(\theta)-1$ giving

$$x=\frac{r}{\sqrt{2}\;a}\sqrt{a^2+r^2}$$

which can be differentiated w.t.r. $r$ giving $$dx= \frac{1}{\sqrt{2}\;a} \left( \frac{r^2}{\sqrt{a^2+r^2}}+\sqrt{a^2+r^2}\right) dr$$

$V_1$ can now be rewritten in terms of $a$ and $r$ only $$V_1=\frac{2\pi}{\sqrt{2}\; a} \int_0^a \left( \frac{r^2}{\sqrt{a^2+r^2}}+\sqrt{a^2+r^2}\right) dr $$ and integrated $$V_1=\frac{\pi}{2\sqrt{2}\; a}\left[ \left( 2r^3-a^2r\right) \sqrt{a^2+r^2}+a^4 \log(r+\sqrt{a^2+r^2})\right]_0^a$$ with the final result for $V_1$ $$V_1=\frac{\pi a^3}{2}+\frac{\pi a^3}{2 \sqrt{2}} \log(1+\sqrt{2})$$ Therefore $$ V=V_1-V_2= \frac{\pi a^3}{2}+\frac{\pi a^3}{2 \sqrt{2}} \log(1+\sqrt{2}) - \frac{2\pi a^3}{3} $$ $$ V= -\frac{\pi a^3}{6}+\frac{\pi a^3}{2 \sqrt{2}} \log(1+\sqrt{2})$$

In agreement with B. Goddard's result.

Part 2: Is there a simpler route?

In messing about with this problem whilst getting the derivation profoundly wrong I found a simpler integral w.r.t. r which leads to the same result as above

$$ V=\frac{4\pi a^3}{3}-\frac{4 \pi}{a \sqrt{2}} \int_0^a \left( r^2 \sqrt{a^2+r^2}\right) dr$$

I have to be honest I don't understand this but seems to arise from the integral

$$V=V_1-V_2=\left(4\pi \int_0^a x^2\;dx \right)-\left(4\pi \int_0^a r^2 \cos(\theta)\;dr \right)$$ with $\cos(\theta)$ defined as above. How one correctly derives and interprets this last formula I do not know.

Answer 2

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.

Answer 3

There seems to be a lot of teeth gnashing here so I wanted to demonstrate how relatively simple this is in complex plane. We will, however, start with the well known Pappus's $2^{nd}$ centroid theorem: the volume of a planar area of revolution is the product of the area $A$ and the length of the path traced by its centroid $R$, i.e., $2πR$. The bottom line is that the volume is given simply by $V=2πRA$.

For an closed area in the complex plane, say $z$, the area and centroid are given by

$$ A=\frac{1}{2}\int \Im\{z^*\dot z\}~du\\ R=\frac{1}{3A}\int z~\Im\{z^*\dot z\}~du\\ RA=\frac{1}{3}\int z~\Im\{z^*\dot z\}~du $$

Now, for the present problem, we must first put the lemniscate in the proper form. First, we express it as a complex variable, the we rotate it so that the axis of rotation becomes the $x$-axis, thus,

$$w=re^{i\theta},\quad \theta\in[0,\pi]\\ z=we^{-i\pi/4}\\ $$

The area that will be rotating is that under the curve $z\in[0,\pi/4]$, that's the lemniscate in one quadrant, so we'll need to double the volume of that rotation.

Okay, we are ready to start constructing the volume given by

$$V=2\cdot2\pi\frac{1}{3}\int_0^{\pi/4} z~\Im\{z^*\dot z\}~d\theta$$

Without any loss in generality we can take $a=1$ and scale the result by $a^3$ otherwise. $$ z=re^{i\theta}e^{-i\pi/4}=r(\cos\theta+i\sin\theta)\frac{\sqrt{2}}{2}(1-i)\\ \dot z=(\dot r +ir)e^{i\theta}e^{-i\pi/4}\\ z^*=re^{-i\theta}e^{i\pi/4}\\ z^*\dot z=r(\dot r +ir)\\ \Im\{z^*\dot z\}=r^2=\sin2\theta\\ $$

For the volume of the area rotated about the $x$-axis we obtain

$$ \begin{align} V &=\frac{4\pi}{3}\frac{\sqrt{2}}{2}\int_0^{\pi/4}[\sin2\theta]^{3/2}(\cos\theta-\sin\theta)\\ &=\frac{4\pi}{3}\frac{\sqrt{2}}{2}\frac{1}{8}\left(3\sinh^{-1}(1)-\sqrt{2} \right)\\ &=\frac{\pi\sqrt{2}}{12}\left(3\sinh^{-1}(1)-\sqrt{2} \right) \end{align} $$

I have verified this result numerically.

Answer 4

thanks for all of your answers, not only were they insightful in seeing how many different ways a math problem can be tackled, I was able to find my own solution drawing from the manipulations you presented -

1. Taking slices perpendicular to the x-axis, we have $\displaystyle V=2\pi\int_{0}^{a}y^{2}dx=-\frac{2\pi}{3}a^{3}+2\pi\int_{0}^{a}r^{2}dx$ (using the fact that $y^{2}=r^{2}-x^{2}$.
2. Changing the subject of our polar equation $r^{2}=a^{2}\cos2\theta$, we get $\cos\theta=\frac{1}{a\sqrt{2}}\sqrt{a^{2}+r^{2}}$. Now, $x=r\cos\theta$, so substituting our result for $\cos\theta$ gives $x$ in terms of $r$, which can be rearranged to the quadratic $r^{4}+a^{2}r^{2}-2a^{2}x^{2}=0$
3. Using the quadratic formula gives us $\displaystyle r^{2}=\displaystyle\frac{a^{2}}{2}\left(-1+\sqrt{a^{2}+8x^{2}}\right)$ which is easily integrated by parts