Consider the ellipse
$$x^2/a^2 + y^2/b^2 = 1$$
(as a curve $L$) and its inner part $B$.
So $B$ is defined as
$$x^2/a^2 + y^2/b^2 < 1$$
Here $a,b$ are positive constants.
Both $L, B \subseteq \mathbb{R}^2$
Then consider just the inner part of this ellipse $B$ (excluding the contour/curve $L$).
Then let's consider the mapping $F$ from $(r, \theta)$ to $(x,y)$ defined as follows
$x = a \cdot r \cdot \cos(\theta)$
$y = b \cdot r \cdot \sin(\theta)$
for
$ 0 < r < 1$
$ 0 \le \theta < 2\pi $
How do we prove this $F$ is bijective and its image is $B$ (or maybe OK, it's $B$ without the origin point $(x,y) = (0,0)$)? How do we prove this is a bijection?
I am asking this because I noticed they often use this substitution when solving integrals involving ellipses or ellipsoids, but they somehow assume this mapping is bijective (don't prove it rigorously).
I can help you with injectivity, not yet sure about surjectivity.
Consider $(r_1,\theta_1) \neq (r_2,\theta_2) $ and $ f(r_1,\theta_1) = f(r_2,\theta_2) $, or, $$\begin{align*} (a \cdot r_1 \cos \theta_1, b\cdot r_1 \sin \theta_1) &= (a\cdot r_2 \cos \theta_2, b \cdot r_2 \sin \theta_2) \end{align*}$$ This gives us the following two equations, $$\begin{align*} \frac { \cos \theta_1 }{\cos \theta_2} &= \frac { r_1 }{r_2} \tag{1} \\ \frac { \sin \theta_1 }{\sin \theta_2} &= \frac { r_1 }{r_2} \tag{2} \end{align*}$$
Equating (1) and (2), we get,
$$\begin{align*} \sin \theta_2 \cdot \cos \theta_1 &= \sin \theta_1 \cdot \cos \theta_2 \\ \frac{ \sin(\theta_2 + \theta_1) + \sin(\theta_2 - \theta_1)}{2} &= \frac{\sin(\theta_1 + \theta_2) + \sin(\theta_1 - \theta_2) }{2} \tag{$\sin A \cos B$} \\ \sin(\theta_2 - \theta_1) &= \sin(\theta_1 - \theta_2) \end{align*}$$
WLOG now assume $\theta_2 > \theta_1$. Then $\sin(\theta_2 - \theta_1) = - \sin(\theta_1 - \theta_2)$. Same goes for $\theta_2 < \theta_1$. Thus, $\theta_1 = \theta_2$ is the only valid solution.
Using this in (1), we get $r_1 = r_2$ which completes the proof for injectivity.
Edit: Perhaps an easy way to show surjectivity is to show that the image is exactly $B\backslash\{(0,0)\}$, then by definition, a function is surjective to its image and the result will follow. To do so, we can bound our set as follows,
$$\begin{align*} \frac{(a \cdot 0 \cdot \cos \theta)^2}{a^2} + \frac{(b \cdot 0 \cdot \sin \theta)^2}{b^2} < \frac{x^2}{a^2} &+ \frac{y^2}{b^2} < \frac{(a \cdot 1 \cdot \cos \theta)^2}{a^2} + \frac{(b \cdot 1 \cdot \sin \theta)^2}{b^2} \\ 0 < \frac{x^2}{a^2} &+ \frac{y^2}{b^2} < 1 \end{align*}$$ Now we can argue since it is a continuous function with image between $0$ and $1$, it will take every value in between (exactly once due to injectivity) which gives us exactly the set $B \backslash \{(0,0)\}$. Hence, bijection is shown.