Let $\displaystyle\Omega$ a domain of $\displaystyle\mathbb{R}^{2}$ defined by $$ \Omega = \left\{\left(x,y\right) \in \mathbb{R}^{2},\ y > 1,\ y < \,\sqrt{\,4 - x^{2}\,}\,\right\} $$ I want to compute an integral over $\displaystyle\Omega$ but I can not defined the integral's bounds with polar coordinates.
Any suggestions ?. Thanks.
Since $x=\rho\cos\theta$ and $y=\rho\sin\theta$, $y>1\iff\rho\sin\theta>1(\implies\sin\theta>0)$ and\begin{align}y<\sqrt{4-x^2}&\iff\rho\sin\theta<\sqrt{4-\rho^2\cos^2\theta}\\&\iff\rho^2\sin^2\theta<4-\rho^2\cos^2\theta\\&\iff\rho<2.\end{align}Now note that since $\rho>\frac1{\sin\theta}$ and $\rho<2$, we have $\frac1{\sin\theta}<2$, which is equivalent to $\sin\theta>\frac12$. Therefore, if $f$ is your function, the integral in polar oordinates is$$\int_{\frac\pi6}^{\frac{5\pi}6}\int_{\frac1{\sin\theta}}^2f(\rho\cos\theta,\rho\sin\theta)\rho\,\mathrm d\rho\,\mathrm d\theta.$$