how to write this region $D$ in relation to $r,\theta$ in this $\iint_Df(x,y)dxdy$ where $D=\{x^2+y^2 \le1,x+y\le 1\}$ and $D=\{x^2+y^2\le1,x+y\ge1\}$

Question

I have attached two photos showing the integration bounds and I find it tricky how to express $r$ and $\theta$ in those two, if $x=r \cos{\theta}$ and $y=r\sin{\theta}$, so any help is very much appreciated!

in the first graph I can say that $0 \le r \le1$ from $x^2+y^2 \le1$ and from the second one $(x+y \le1 )$ what can I derive? Should I derive it with respect to $r$ or $\theta$? I can write it as $r\le\frac1{\cosθ+\sinθ}=\frac1{\sqrt{2}\sin(θ+\pi/4)}$

and after that I can write it as $r\sqrt2\sin(θ+π/4) \le1$. So then do I solve as where the line intersects the circle, so when $r=1$ and thus $θ+π/4= kπ +π/4$ which gives 2 solutions in $(0,2π)$ which is $0$ and $π/2$. So is it $0 \le θ\le π/2$ and $\frac1{\cosθ+\sinθ}\le r \le 1$ ?

And with the same logic the second image is $π/2 \le θ\le 0$ and $0\le r \le\frac1{\cosθ+\sinθ}$.

Can someone tell me if I am right? Or is the right answer for the second one a sum of two integrals: the first one integral with bounds $0 \le θ\le 2π$ and $0\le r \le\frac1{\cosθ+\sinθ}$ plus the second integral with bounds $π/2 \le θ\le 2π$ and $0\le r \le 1$? A confirmation or a correction would be much appreciated! Also if there is a simpler solution I am all ears!

Answer 1

In the case $D=\{x^2+y^2 \le 1,\, x+y\ge 1\}$, you have $0\le r\le1$ and $r(\cos \theta+\sin \theta)\ge 1$. The angle $\theta$ varies between $0$ and $\pi/2$, so the integral becomes: $$\iint_D f(x,y)dxdy=\int_0^{\pi/2}\int_{g(\theta)}^1 f(r\cos \theta,r\sin \theta)rdrd\theta $$ where $g(\theta)=\frac1{\sin \theta+\cos \theta}$.