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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1} \int\limits_{0}^{\sqrt{3}x} \frac{ 1 }{\sqrt{x^{2}+y^{2}}} dy dx\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436659558096:dw

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1dw:1436659627501:dw\[ 0\le \theta \le \frac{\pi}{3}\\\]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1\[x=rcos (\theta)\\x=1 \rightarrow rcos \theta =1 \rightarrow r=\frac{1}{\cos \theta}=\sec theta \]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta} \]

amoodarya
 one year ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta} \frac{ 1 }{\sqrt{x^2+y^2} }dxdy=\\ =\int\limits_{0}^{\frac{\pi}{3}}\int\limits_{0}^{\sec \theta} \frac{ 1 }{ \sqrt{r^2} }*r dr d \theta\] it is simple to integrate
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