Surface area generated by revolving $r = \sqrt {\cos 2\theta}$ I've been giving a good time trying to solve this problem, I do not find a clear way to solve appreciate your help.
\begin{array}{rcl}
r& =& \sqrt{\cos 2\theta } 
\end{array}
This Around to axis y and the limits of integration is 0 to π/4
P.d :
I would appreciate if anyone could recommend me some text or page in which you could learn to plot in polar and learn more about these problems.
 A: Parametric Breakdown
Start by dividing the function from polar form into a set of two parametric functions. That is, the horizontal and vertical component of the radial coordinate $r$:
$$
x =r \cos{\theta}\qquad\quad y=r\sin{\theta}\\
\implies \dfrac{dx}{d\theta}=\dfrac{dr}{d\theta}\cos{\theta}-r\sin{\theta} \qquad\quad \dfrac{dy}{d\theta} = \dfrac{dr}{d\theta}\sin{\theta}+r\cos{\theta}\\ 
$$
However the components alone don't really help, we need to find a differential element $ds$ that represents the hypotenuse formed by the two differential components (basically Pythagorean Theorem) :
\begin{align} 
(ds)^2 &= (\dfrac{dx}{d\theta})^2+(\dfrac{dy}{d\theta})^2\\
& = (\dfrac{dr}{d\theta}\cos{\theta}-r\sin{\theta})^2 + (\dfrac{dr}{d\theta}\sin{\theta}+r\cos{\theta})^2\\
& = \left[(\dfrac{dr}{d\theta})^2\cos^2\theta\color{red}{-(\dfrac{dr}{d\theta})2r\cos\theta\sin\theta} +r^2\sin^2\theta\right] \space\space\space + \left[(\dfrac{dr}{d\theta})^2\sin^2\theta \color{red}{+ (\dfrac{dr}{d\theta})2r\cos\theta\sin\theta} +r^2cos^2\theta \right]\\
& =(\dfrac{dr}{d\theta})^2(\cos^2\theta + \sin^2\theta) + r^2(\cos^2\theta + \sin^2\theta)\\
&=(\dfrac{dr}{d\theta})^2+r^2
\end{align}
Hence we conclude for this step that:
$$ds = \sqrt{(\dfrac{dr}{d\theta})^2+r^2} d\theta$$
Surface of Revolution Derivation
We should notice (through derivation using frustrums not shown here) that just as in moving from the arc length of a curve to its surface area in parametric coordinates requires multiplication by $2\pi * \left(x(t) \,\mathrm{or}\,y(t)\right)$, the same applies in polar coordinates:
$$ L = \int ds $$
Since we are rotating about the y-axis, the height of each frustrum would be the $x$ component of the polar equation:
$$A = 2\pi\int \operatorname{x}(\theta)\, ds\\
\boxed{A = 2\pi\int \operatorname{x}(\theta)\,\sqrt{(\dfrac{dr}{d\theta})^2+r^2} d\theta}
$$
If you wish to try it from here yourself, don't continue reading as I will propose the calculated solution!
Calculated Solution
Recollecting variables:
$r = \sqrt{\cos2\theta}\\
\operatorname{x}(\theta)= r\cos\theta = \cos\theta\sqrt{\cos2\theta}\\
\dfrac{dr}{d\theta}= \dfrac{-2\sin2\theta}{2\sqrt{\cos2\theta}} = \dfrac{-\sin2\theta}{\sqrt{\cos2\theta}}
$
Plugging in variables and simplifying:
\begin{align}
A &= 2\pi\int_0^{\pi/4} \operatorname{x}(\theta)\,\sqrt{(\dfrac{dr}{d\theta})^2+r^2}\, d\theta\\
& = 2\pi\int_0^{\pi/4}\cos\theta\sqrt{\cos2\theta}\,\sqrt{(\dfrac{-\sin2\theta}{\sqrt{\cos2\theta}})^2+(\sqrt{\cos2\theta})^2}\, d\theta\\
& = 2\pi\int_0^{\pi/4}\cos\theta\sqrt{\cos2\theta}\,\sqrt{\dfrac{\sin^22\theta}{\cos2\theta}+\cos2\theta}\, d\theta
\end{align}
Combining the square roots and using the fact that $\sin^2x+\cos^2x=1$:
\begin{align}
&= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{\cos2\theta \left(\dfrac{\sin^22\theta}{\cos2\theta}+\cos2\theta\right)}\, d\theta\\
&= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{\sin^22\theta + \cos^22\theta}\, d\theta\\
&= 2\pi\int_0^{\pi/4}\cos\theta\,\sqrt{(1)}\, d\theta\\
&= 2\pi\, \left.\sin\theta\right\rvert_0^{\pi/4}\\
&= 2\pi\, \left(\dfrac{\sqrt{2}}{2}\right)\\
& = \boxed{\pi\sqrt{2}}
\end{align}
