Integral in polar coordinates

Question

I was wondering if I had set up this integral correctly. If anyone could help me, I would greatly appreciate it. I am available in case there are any unclear things! $$\iint_D x^2+y^2dxdy, \quad D=\{1\leq x^2+y^2 \leq 4, y \geq \sqrt{3}\}$$$x=r\cos(\theta)$
$y=r\sin(\theta)$
$dxdy=r \ dr \ d\theta$
$x^2+y^2=r^2$

$1 \leq r^2 \leq 4 \Rightarrow 1 \leq r \leq 2 \ *$
$r\sin(\theta)\geq \sqrt{3} \Rightarrow \sin(\theta)\geq \frac{\sqrt{3}}{r}\Rightarrow \arcsin(\sin(\theta)) \geq \arcsin\left(\frac{\sqrt{3}}{r}\right) \Rightarrow \theta \geq \arcsin\left(\frac{\sqrt{3}}{r}\right)$ $r\sin(\pi-\theta)\leq \sqrt{3} \Rightarrow \sin(\pi-\theta)\leq \frac{\sqrt{3}}{r}\Rightarrow \arcsin(\sin(\pi-\theta)) \leq \arcsin\left(\frac{\sqrt{3}}{r}\right) \Rightarrow \pi-\theta \leq \arcsin\left(\frac{\sqrt{3}}{r}\right)\Rightarrow \theta \leq \pi-\arcsin\left(\frac{\sqrt{3}}{r}\right)$ $*\arcsin$ is only defined between $-1$ and $1$. So i have to consider $\frac{\sqrt{3}}{r} \leq 1 \Leftrightarrow r \geq \sqrt{3}$
This means my bounds for $r$ are $\sqrt{3}$ and $2$.

So the integral becomes:$$\int_{\sqrt{3}}^2 \int_{\arcsin\left(\frac{\sqrt{3}}{r}\right)}^{\pi-\arcsin\left(\frac{\sqrt{3}}{r}\right)} r^3 d\theta dr$$

This should be the domain graph:

Answer 1

I'd strongly recommend that you instead choose the reverse order of integration; that is, let $r$ be a function of $\theta$.

The region is correct, although it is a little strange that the question would specify $1 \le x^2 + y^2$ when the condition $y \ge \sqrt{3}$ automatically guarantees that $x^2 + y^2 > 1$.

The point $P$ in your diagram is given by the coordinate $(x,y) = (1,\sqrt{3})$, and similarly, $O = (-1,\sqrt{3})$. Hence for the polar transformation $$(x,y) = (r \cos \theta, r \sin \theta),$$ we have $P(r,\theta) = (2, \pi/3)$, and $O(r,\theta) = (2,2\pi/3)$. While it's clear that the arc $OQP$ corresponds to $r = 2$, the segment $OP$ is determined by setting $$r \sin \theta = y = \sqrt{3}$$ and then solving for $r$ as a function of $\theta$: $$r = \sqrt{3} \csc \theta.$$ It follows that $$\iint_D x^2 + y^2 \, dx \, dy = \int_{\theta = \pi/3}^{2\pi/3} \int_{r=\sqrt{3} \csc \theta}^2 r^3 \, dr \, d\theta.$$ The evaluation of this iterated integral is left as an exercise.

To be sure, your integral is also correct. You are effectively integrating along concentric arcs whose angle subtending the origin is an increasing function of their distance from the origin. My integral is integrating along selected portions of rays emanating from the origin, whose lengths are functions of the counterclockwise angle the ray makes with the positive $x$-axis.