Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm a calculus II student and I'm completely stuck on one question:

Find the area of the surface generated by revolving the right-hand loop of the lemniscate $r^2 = \cos2 θ$ about the vertical line through the origin (y-axis).

Can anyone help me out?

Thanks in advance

share|cite|improve this question
any hints on how to go about this problem or a full explanation would be much appreciated. Thanks – Logan Besecker Apr 24 '12 at 22:59
Just to clarify - you definitely want surface area and not volume? – user29743 Apr 24 '12 at 23:53
this may help. – David Mitra Apr 25 '12 at 0:48
up vote 1 down vote accepted

\begin{aligned} ds^2 &= dr^2+r^2 d\theta^2\\ &=\left(\frac{4 \sin^2 2\theta}{\cos 2 \theta}+\cos 2\theta\right)d\theta^2\\ ds &= \sqrt{\frac{1+3 \sin^2 2\theta}{{\cos 2\theta}} }d\theta \\ A &=2\int_0^{\pi/4}2\pi r \cos \theta ds\\ &=4\pi \int_0^{\pi/4}\sqrt{1+3 \sin^2 2\theta} cos \theta d\theta\\ &=\int_0^{\frac{1}{\sqrt 2}} \sqrt{1+12 t^2 (1-t^2)} dt \end{aligned}

This is as simplified I was able to make it.

share|cite|improve this answer

Note some useful relationships and identities:

$r^2 = x^2 + y^2$

${cos 2\theta} = cos^2\theta - sin^2\theta$

${sin \theta} = {y\over r} = {y\over{\sqrt{x^2 + y^2}}}$

${sin^2 \theta} = {y^2\over {x^2 + y^2}}$

These hint at the possibility of doing this in Cartesian coordinates.

share|cite|improve this answer
I tried this but ran into a problem that the bounds of the integral were not possible to compute. The problem I ran into is that it's not easy to solve for x in terms of y (or vice versa) on the lemniscate. – user29743 Apr 25 '12 at 2:23

I stole the formula from a website for surfaces of revolution that was linked in the comment above:

They prove it more generally for parametric surfaces. I am not sure what you are allowed to assume in your calculus two course; I was unsuccessful in getting a correct formula from a direct polar slicing argument.

In what follows, I am going to be sloppy about whether I write $r$ as a variable or as a function $r(\theta)$ of $\theta$.

Since the right half of the lemniscate is traced out between $-\pi/4$ and $\pi/4$, the integral you want is $$ 2\pi\int_{-\pi/4}^{\pi/4} r(\theta)\cos\theta\sqrt{r(\theta)^2 + r'(\theta)^2}d\theta $$ We have $$ r^2 = \cos(2\theta) $$ and so $$ 2rr' = -2\sin(2\theta), $$ so \begin{align*} (r')^2 &= \frac{\sin^2(2\theta)}{r^2}\\ &=\tan^2(2\theta). \end{align*} The integrand isn't very pretty. I used wolfram alpha and it numerically approximated the integral (without the 2 $\pi$) to be 2.1028, which seems geometrically reasonable to me. I'm sorry for the unsatisfying conclusion.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.