Question Number 129370 by bramlexs22 last updated on 15/Jan/21
Answered by Ar Brandon last updated on 15/Jan/21
![Θ=∫_1 ^2 ∫_0 ^y ((dxdy)/(x^2 +y^2 ))=∫_1 ^2 (1/y)[tan^(−1) ((x/y))]_0 ^y dy =∫_1 ^2 (π/(4y))dy=[((πlny)/4)]_1 ^2 =((πln2)/4)](Q129372.png)
$$\Theta=\int_{\mathrm{1}} ^{\mathrm{2}} \int_{\mathrm{0}} ^{\mathrm{y}} \frac{\mathrm{dxdy}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{y}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)\right]_{\mathrm{0}} ^{\mathrm{y}} \mathrm{dy} \\ $$$$\:\:\:\:=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\pi}{\mathrm{4y}}\mathrm{dy}=\left[\frac{\pi\mathrm{lny}}{\mathrm{4}}\right]_{\mathrm{1}} ^{\mathrm{2}} =\frac{\pi\mathrm{ln2}}{\mathrm{4}} \\ $$