Compute
$$ \iint \arctan{\frac{y}{x}}\,\mathrm dx\,\mathrm dy,$$ where $1 \le x^2 + y^2 \le 9$, $y \ge \frac{x}{\sqrt{3}}$, $y \ge x\sqrt{3}$.
Introducing polar coordinates, I get that $1 \le r \le 3$ which is fine. However, I have trouble determining the boundaries for the angle $\phi$
From the conditions:
$y \ge \frac{x}{\sqrt{3}}$ we have:
$$\sin{\phi} \ge \frac{\cos{\phi}}{\sqrt{3}}$$
$$\tan{\phi} \ge \frac{\sqrt{3}}{3}$$
with $\phi \ne \frac{\pi}{2} + 2n\pi$
Applying $\arctan$ to both sides we get
$$\phi \in [\frac{\pi}{6}, \frac{\pi}{2})$$
Similarly, from the second condition we have the boundaries
$$\phi \in [\frac{\pi}{3}, \frac{\pi}{2})$$
Now, the boundaries for the angle are
$$\phi \in [\frac{\pi}{6}, \frac{\pi}{2})$$
But my workbook states that the boundaries are
$$\phi \in [\frac{\pi}{6}, \frac{7\pi}{6})$$
Can anyone help me see where I went wrong? Thanks!
Best Answer
In the first quadrant you know that for both $y \geq \dfrac{x}{\sqrt3}$ and $y \geq \sqrt3 x$ to be true, we must have $y \geq x \sqrt3 \implies \theta \in [\dfrac{\pi}{3}, \dfrac{\pi}{2}]$.
In second quadrant, $x \leq 0$ and so the inequality holds in the second quadrant $\theta \in [\dfrac{\pi}{2}, \pi]$.
In the third quadrant, both $x$ and $y$ are negative and the inequality reverses so for both $y \geq \dfrac{x}{\sqrt3}$ and $y \geq \sqrt3 x$ to be true, we must have $y \geq \dfrac{x}{\sqrt3} \implies \theta \in [\pi, \dfrac{7\pi}{6}]$
Here is a diagram that may help explain further.