Question Number 181857 by KINMATICS last updated on 01/Dec/22
Answered by hmr last updated on 01/Dec/22
![∫_(π/4) ^( (π/2)) ∫_0 ^( (√2)) ((rcos(θ))/r^2 ) r dr dθ = ∫_(π/4) ^( (π/2)) ∫_0 ^( (√2)) cos(θ) dr dθ = ∫_(π/4) ^( (π/2)) [r cos(θ)]_0 ^(√2) dθ = ∫_(π/4) ^( (π/2)) (√2) cos(θ) dθ = (√2) [sin(θ)]_(π/4) ^(π/2) = (√2) (1 − ((√2)/2)) = (√2) −1](https://www.tinkutara.com/question/Q181865.png)
$$\int_{\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\sqrt{\mathrm{2}}} \:\frac{{rcos}\left(\theta\right)}{{r}^{\mathrm{2}} }\:\:{r}\:{dr}\:{d}\theta \\ $$$$=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\:\sqrt{\mathrm{2}}} \:{cos}\left(\theta\right)\:{dr}\:{d}\theta \\ $$$$=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{2}}} \left[{r}\:{cos}\left(\theta\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:{d}\theta \\ $$$$=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\:\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{2}}\:{cos}\left(\theta\right)\:{d}\theta \\ $$$$=\:\sqrt{\mathrm{2}}\:\left[{sin}\left(\theta\right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\:\sqrt{\mathrm{2}}\:\left(\mathrm{1}\:−\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$=\:\sqrt{\mathrm{2}}\:−\mathrm{1} \\ $$