anonymous one year ago Medal ** use polar coordinates to evaluate the integral

1. anonymous

$\int\limits_{0}^{1} \int\limits_{0}^{\sqrt{3}x} \frac{ 1 }{\sqrt{x^{2}+y^{2}}} dy dx$

2. amoodarya

|dw:1436659558096:dw|

3. amoodarya

|dw:1436659627501:dw|$0\le \theta \le \frac{\pi}{3}\\$

4. amoodarya

$x=rcos (\theta)\\x=1 \rightarrow rcos \theta =1 \rightarrow r=\frac{1}{\cos \theta}=\sec theta$

5. amoodarya

$\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta}$

6. amoodarya

$\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta} \frac{ 1 }{\sqrt{x^2+y^2} }dxdy=\\ =\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta} \frac{ 1 }{ \sqrt{r^2} }*r dr d \theta$ it is simple to integrate