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.

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}
