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
Best Answer
\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.