# 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 $D=\{x^2+y^2\le1,x+y\ge1\}$

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!

• Commented Jun 10 at 9:13

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 $$g(\theta)=\frac1{\sin \theta+\cos \theta}$$.