Using polar coordinates, calculate the double integral

Question

$$ f(x,y) = \frac{1}{1+x^2+y^2} \qquad D:\{x^2 +y^2 \leq 1;~ x\leq 0;~ y \geq \sqrt{3}x \} $$ Can you help me please?

Answer 1

$f(x, y)=\frac{1}{1+x^{2}+y^{2}}$ transform your function to polar form ($f(r, \theta)$) by using $r=\sqrt{x^{2}+y^{2}}$ then the domain in polar form: \begin{equation} \begin{array}{l} \text { (using the given transformations for r) } \\ D=\left\{(r, \theta): 0 \leqslant r \leqslant 1, \frac{\pi}{2} \leqslant \theta \leq \frac{4 \pi}{3}\right\} \end{array} \end{equation} , integral will be $\int_{\pi / 2}^{4 \pi / 3} \int_{0}^{1} f(r, \theta) r\, d r\, d \theta$ \begin{equation} =\int_{\frac{\pi}{2}}^{\frac{4 \pi}{3}} \int_{0}^{1} \frac{r}{1+r^{2}}\, d r\, d \theta \end{equation} \begin{equation} \begin{aligned} =&\int_{\pi/2}^{4 \pi / 3}\left[\int_{0}^{1} \frac{r}{1+r^{2}} d r\right] d \theta\\ &=\int_{\pi/2}^{4\pi/3}\left[\frac{\ln \left(1+r^{2}\right)}{2}\right]_{0}^{1} d \theta \end{aligned} \end{equation} \begin{equation} \begin{aligned} =& \int_{\pi / 2}^{4 \pi / 3} \frac{\ln (2)}{2} d \theta \\ =& \frac{\ln (2)}{2}[\theta]_{\pi / 2}^{4 \pi / 3} \\ =& \frac{\ln (2)}{2}\left(\frac{4 \pi}{3}-\frac{\pi}{2}\right) \\ =& \frac{\ln (2)}{2}\left(\frac{5 \pi}{6}\right) \end{aligned} \end{equation}