I have an integral $$ \int_{1}^{2} \int_{0}^{2 - y} f\left(x, y\right) \mathrm{d}x \: \mathrm{d}y$$ and I need to change into polar coordinates. The area looks like this:
Since there's no easy way of saying which bounds $r$ will have, we need to use a substitution into one of the following inequalities:
$$ 1 \leq y \leq 2, $$ $$ 0 \leq x \leq 2 - y. $$
By substituting into $x$ inequality I get: $$ x \geq 0 \implies r \cdot \cos \varphi \geq 0 \implies r \geq 0 $$ $$ x \leq 2 - y \implies r \cdot \cos \varphi \leq 2 - r \cdot \sin \varphi \implies r \leq \frac{2}{\sin \varphi + \cos \varphi},$$ so the integral would look like this $$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{0}^{\frac{2}{\sin \varphi + \cos \varphi}} f \left( r \cos \varphi, r \sin \varphi \right) \mathrm{d}r \: \mathrm{d}\varphi $$ But when doing the same for the $y$ inequality, the integral would be different:
$$ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_{\frac{1}{\sin \varphi}}^{\frac{2}{\sin \varphi}} f \left( r \cos \varphi, r \sin \varphi \right) \mathrm{d}r \: \mathrm{d}\varphi $$
Which one is right and how to choose where to subtitute?