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.
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 $\theta$.