# Find the area of a surface of revolution

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?

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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

\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.

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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.

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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:

http://tutorial.math.lamar.edu/Classes/CalcII/PolarSurfaceArea.aspx

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.

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