Generalized polar coordinates, how to switch form cartesian to polar I was solving a problem: $$\left(\frac xa + \frac yb\right)^ 4=4xy,\quad a>0 , b>0 $$
Find the are bounded by curve.
It is not hard to see that x and y must have the same sign, that function exist only in first and third quadrant, and that the function is symetrical to ( 0,0).
So i want to transform this to polar coordinates. The regular transformation is $$ x=r\cos\phi \\ y=r\sin\phi $$ 
I have noticed that these are only used for circle with centre in (0,0). Whenever the function is  more complicated the transformation is different. Why ?
The book solution says that we will moved generalized coordinates use  moved generalized coordinates : 
$$ 
x-x_0 = ar\beta \cos \alpha \phi\\
y-y_0 = br\beta \sin \alpha \phi
$$
 Then it says $$ x_0=y_0=0 $$ $$ \alpha=2, \beta=1 $$ It says we did this to make the curve equation simpler. I see why the first equation is true but the second with $\alpha$ and $\beta$, I don't understand. Why are $\alpha =2, \beta = 1$, and where does this generalized polar coordinates equation come from ?
Also, how to think when switching to polar coordinates when dealing with ˝more complicated curves˝ ( for me they are complicated)
Edit : also i have seen a different transformation used in this problem $$ x=ar\cos^ 2\phi \\ y=br\sin^ 2\phi $$ then this was transformed with double integral identity.
 A: First of all, you can use "ordinary" polar coordinates for lots of curves other than circles centered at the origin. One of my particular favorites is a circle passing through the origin and centered on the $x$-axis. Its polar equation will be $r=2a\cos\theta$ (where $a$ is its radius). 
If you have an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, then it is natural to think of it as coming from the unit circle by stretching the $x$-axis by a factor of $a$ and the $y$-axis by a factor of $b$. So this would naturally be parametrized by $x=a\cos\theta$, $y=b\sin\theta$, and you get a modified polar coordinate system by "filling it in" with $x=ar\cos\theta$, $y=ar\sin\theta$. (Note that $\theta$ is no longer the angle between the $x$-axis and the ray to the point.)
Suppose we consider a slightly easier version of your question, since we've already covered the effect of stretching with $a$ and $b$. Let's look at the equation 
$$(x+y)^4 = 4xy.$$
I'm not sure I get the point of introducing $\alpha$. The equation of this curve in usual polar coordinates would be
$$r^2 = \frac{2\sin 2\theta}{(\cos\theta+\sin\theta)^4},$$
with $0\le\theta\le\pi/2$ or $\pi\le\theta\le 3\pi/2$ (to make $\sin 2\theta\ge 0$).
If we use $\alpha = 2$, with $x=r\cos 2\theta$ and $y=r\sin 2\theta$, this equation becomes
$$r^2 = \frac{2\sin 4\theta}{(\cos 2\theta + \sin 2\theta)^4},$$
with $0\le\theta\le \pi/4$ or $\pi/2\le\theta\le 3\pi/4$.
Although it's tempting to see that the difference in degree ($4$ on the LHS and $2$ on the RHS) should suggest some $\alpha$, I certainly don't see how to simplify the second equation any more easily. (It's tempting because of things like deMoivre's formula, but I don't see where it goes.)
So, how would we try to simplify either expression? We note that
$$\cos\theta+\sin\theta = \sqrt2\sin\big(\theta+\frac{\pi}4\big),$$
so we want to substitute $\psi = \theta+\frac{\pi}4$. Then
and then $\sin 2\theta = \sin 2(\psi-\frac{\pi}4)=\sin(2\psi - \frac{\pi}2) = -\cos 2\psi$. Then we end up with
$$r^2 = \frac{-\cos 2\psi}{2\sin^4\psi}, \quad \pi/4\le\psi\le 3\pi/4 \quad\text{or}\quad 5\pi/4\le\psi\le 7\pi/4.$$
Now let's try this with the second formulation. Analogously, we have
$$\cos 2\theta+\sin 2\theta = \sqrt2\sin\big(2\theta+\frac{\pi}4\big) = \sqrt2\sin 2\big(\theta+\frac{\pi}8\big).$$
This seems very unhelpful. Perhaps we should try scaling with $\alpha = 1/2$?
Then we'd have 
$$\sin\frac{\theta}2 + \cos\frac{\theta}2 = \sqrt2\sin\big(\frac{\theta}2 +\frac{\pi}4\big) = \sqrt2\sin\frac12\big(\theta+\frac{\pi}2\big).$$
Does that seem a bit more promising? 
At any rate, I am not sure I believe your book is right. A translation of $\theta$ is clearly called for before a rescaling of $\theta$.
