I'd like to solve $\iint_R \tan^{-1} \frac{y}{x}dA$, where $R=\{ (x,y) : x\ge0, y\ge 0, x^2+y^2 \le 4\}.$
I'm wondering if the following calculation is true: $$ \begin{split} \iint_R \tan^{-1} \frac{y}{x}dA &=\int_0^{\pi/2}\int_0^2 \tan^{-1} \left(\frac{r \sin \theta}{r\cos\theta}\right)rdrd\theta\\ &=\int_0^{\pi/2}\int_0^2 \tan^{-1}(\tan\theta)rdrd\theta\\ &=\int_0^{\pi/2}\int_0^2\theta rdrd\theta\\ &=\frac{\pi^2}{4} \end{split} $$
It might be true, but I don't think it is a rigid proof. Apparently, $\tan^{-1} \frac{y}{x}$ is not defined for $x=0,$ but since $\lim_{x \to +0} \frac{y}{x} \to \infty$, we may extend or regard $\tan^{-1} \frac{y}{x}=\frac{\pi}{2}$ for $x=0.$
Please let me know if you have any comments for this problem. Thanks in advance!
You're noticing is that the integral is improper when $x=0$. This translates to it being improper at $\theta=\pi/2$ in polar coordinates. So it's slightly better to include the limits.
$$ \begin{split} \iint_R \tan^{-1} \frac{y}{x}dA &=\int_0^2\int_0^\sqrt{4-x^2}\tan^{-1}\frac{y}{x}\,dy\,dx\\ &=\lim_{a\to 0^+}\int_a^2\int_0^\sqrt{4-x^2}\tan^{-1} \frac{y}{x}\,dy\,dx\\ &=\lim_{a\to 0^+}\int_0^{\alpha(a)}\int_{a^2\sec\theta}^2\tan^{-1} \frac{r\sin\theta}{r\cos\theta}\,r\,dr\,d\theta\\ &=\lim_{a\to 0^+}\int_0^{\alpha(a)}\int_{a\sec\theta}^2 \theta r\,dr\,d\theta\\\ &=\lim_{a\to 0^+}\int_0^{\alpha(a)} 2\theta-\frac{a^2}{2}\theta \sec^2(\theta)\, d\theta\\\ &=\lim_{a\to 0^+} \theta^2 |_0^{\alpha(a)}-\frac{a^2}{2}\left[\ln\left|\cos\theta\right|-\theta\tan\theta\right]_0^{\alpha(a)}\\ &=\frac{\pi^2}{4} \end{split} $$
Note $x=a$ under the change of coordinates is $r\cos \theta=a$ so $r=a\sec\theta$. Also define the angle to the point $(a,\sqrt{4-a^2})$ to be $\alpha(a)$ for all $a>0$ that is $\alpha(a)=\tan^{-1}\sqrt{4/a^2-1}$.
First part simply becomes $\lim_{a\to 0^+} (\tan^{-1}\sqrt{4/a^2-1})^2=(\lim_{x\to\infty}\tan^{-1}(x))^2=\left(\frac{\pi}{2}\right)^2$ the right hand part is zero. Here are some notes on how to see that
$$\begin{split} \lim_{a\to 0^+}a^2\alpha(a)\tan(\alpha(a)) &= \lim_{a\to 0^+}a^2\sqrt{4/a^2-1}\tan^{-1}\sqrt{4/a^2-1}\\ &=\lim_{a\to 0^+}a\sqrt{4-a^2}\tan^{-1}\sqrt{4/a^2-1}\\ &=(0)(2)\left(\frac{\pi}{2}\right)=0 \end{split}$$
$$\begin{split} \lim_{a\to 0^+}a^2\ln\left|\cos\alpha(a)\right|&= \lim_{a\to 0^+}a^2\ln\left|\frac{a^2}{4}\right|\\ &=\lim_{a\to 0^+}\frac{\ln\left|\frac{a^2}{4}\right|}{a^{-2}}\\ &=\lim_{a\to 0^+}-a^2=0 \end{split}$$
where the last step is by L'Hopital's rule