Switching to polar coordinates with $\Big\{(x, y) \in \Bbb R^2,$ $x^2 \over a^2$ $+$ $y^2 \over b^2$ $\le 1\Big\}$

Question

In one solution of an exercise, there is the following equality:

$\Big\{(x, y) \in \Bbb R^2,$ $x^2 \over a^2$ $+$ $y^2 \over b^2$ $\le 1\Big\} =$ $\left\{ \begin{pmatrix} r a \cos\phi \\ rb \sin\phi \\ \end{pmatrix} : 0 \le r \le 1, \ 0 \le \phi \le 2\pi\right\}$

This was done by switching the coordinates to polar coordinates, so $(x, y) = (r \cos \phi, r \sin \phi).$ I don't see how this was done though. How does one have to work with the $a^2, b^2$ in this case?

Answer 1

The change in coordinates is: $$ x=ar\cos{\phi}\\ y=br\sin{\phi} $$ with $r\in [0,1]$, $\phi\in [0,2\pi]$.

Substituting in the inequality you have gives: $r^2 \cos^2{\phi}+r^2 \sin^2{\phi}=r^2\leq 1$ which is trivially satisfied since $r\in [0,1]$.

So the set you have on left-hand-side (which is an ellipse) can be re-written as the set you have on the right-hand-side.

---

Polar coordinates are used when the domainyou have i.e. when you have two variables whose relation determine a circle or disk or a portion of a circle or disk. Then you use the transformation as you wrote it that goes from the domain you originally have to a new one that is a rectangle. The values of $r,\phi$ are determined accordingly.

---

Now the original domain you have is an ellipse. Hence in order to transform it to a square which is simpler to work with you apply two transformations (two change of coordinates). The first one sends the ellipse to a disk using $x=ax', y=ay'$ Then the inequality you have becomes $x'^2+y'^2\leq 1$.

The second one sends the disk to a rectangle using $x'=r \cos{\phi}$, $y'=r\sin{\phi}$.

If you take the composition of the two transformations you have $$ x=ar\cos{\phi}\\ y=br\sin{\phi} $$ that sends the ellipse to a rectangle.

Answer 2

Instead of switching to polar coordinates, it may be easier to switch to slightly different coordinates (I am assuming $a,b>0$, if not replace $a$ by $|a|$ and same for $b$ in the rest). We set

$x=ra\cos{\phi}$
$y=rb\sin\phi$

with $0\leq r$ and $0\leq\phi<2\pi$. Then

$\{(x,y) : \frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1\} = \left\{ \begin{pmatrix} r a \cos\phi \\ rb \sin\phi \\ \end{pmatrix} : r^2(\cos^2\phi+sin^2\phi)\leq 1, r\geq 0, 0\leq\phi<2\pi\right\}$

Using $\cos^2\phi+\sin^2\phi = 1$, we find that

$r^2(\cos^2\phi+\sin^2\phi)\leq 1 \Leftrightarrow -1\leq r\leq 1$.

Together with the other conditions ($r\geq 0$ and $0\leq\phi<2\pi$) we end up with

$\left\{ \begin{pmatrix} r a \cos\phi \\ rb \sin\phi \\ \end{pmatrix} : 0\leq r\leq 1, 0\leq\phi<2\pi\right\}$

which is what we were looking for.

If your question was in particular to understand how it is done using polar coordinates then let me know and I'll remove this answer.