limits of Surface area of revolution in polar co-ordinates.

Question

My Question is Find the area of the surface generated by revolving the right-hand loop of the lemniscate $\;r^2=\cos2\theta\;$ about the vertical line through the origin (y-axis). I know the formula

$$S=2\pi \int_\alpha^\beta r \cos\theta\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta$$ But I cant figure out the limits $\alpha$ and $\beta$. in Cartesian co-ordinate I can easily picture a function segment from a to b and imagine it rotating around any axis, but In polar I am really having difficulty in capturing the limits and than imaging a line segment then rotating the segment around y-axis.I just cant picture it in my mind. any help? thanks in advance.

Answer 1

To generate the right-hand loop of the lemniscate $r^2=\cos(2\theta)$, you need to let $\theta$ vary from $-\frac{\pi}{4}$ to $\frac{\pi}{4}$; i.e. $\theta \in [-\frac{\pi}{4},\frac{\pi}{4}] $. To see this, we have

$$ r = \bar{+} \sqrt{\cos(2 \theta) } \implies \cos(2\theta)\geq 0 \implies -\frac{\pi}{2}\leq2\theta \leq \frac{\pi}{2}\implies -\frac{\pi}{4}\leq \theta \leq \frac{\pi}{4}. $$

Here is the plot

Here is the full graph

Answer 2

That branch of the lemniscate is generated by setting $\alpha = -\pi/4$ and $\beta = \pi/4$. You can see this by following the curve from $\theta=0$ at $r=1$ to where $r=0$ at $\theta=\pi/4$, and realizing the symmetry about the $x$ axis.

It turns out that the integral works out quite nicely,although as you have not asked for help in evaluating it, I will leave it to you to see how it all works out.