Double integrals in polar coordinates with unexpected result

Question

I have been working on an exercise where I need to convert to polar coordinates :

$$\iint_R (x+y)\,dA\,,$$

where $$R=\left\{(x,y)\mid 1\le x^2+y^2\le 4\,,\, x\le 0\right\}$$

To me, it makes perfect sense that $ (x + y)\,dx\,dy $ became : $(r\cos(\theta) + r\sin(\theta))r\,dr\,d\theta$

What I don't understand is that, I was expecting to work between $r = -1$ and $r = -2$ ; $\theta = - \pi/2$ and $\theta = \pi/2$

Instead, the solution puts $r = 1$ and $r = 2$ ; $\theta = \pi/2$ and $\theta = 3\pi/2$

Could you tell me why ? (I am trying to grasp this new concept)

Answer 1

Perhaps this drawing can help you in understanding the solution

To integrate the purlple area you have to chose a radius in

$$1<r<2$$

and a corresponding angle in

$$\frac{\pi}{2}<\theta<\frac{\pi}{2}+\pi=\frac{3}{2}\pi$$

Thus rotating this positive radius in the angle's interval you get exactly the desired area to integrate

Answer 2

Keep in mind that $r = \sqrt{x^2+y^2}$ cannot be negative... It represents the distance to the origin. The condition $1\leq x^2+y^2 \leq 4$ puts our domain between the circles with radius 1 and 2, and so $r \in[1,2]$. The condition $x \leq 0$ puts our domain in the second and third quadrants, for which $\theta \in [\frac{\pi}{2}, \frac{3\pi}{2}]$