Question Number 141855 by rs4089 last updated on 24/May/21
Answered by TheSupreme last updated on 24/May/21
![E={(x,y)∣y<x<a ∨ 0<y<a} E^1 ={(ρ,θ)∣ 0<ρ<a, 0<θ<(π/4)} det∣J∣=ρ ∫_0 ^a ∫_0 ^(π/4) ((ρ^3 cos^2 (θ))/ρ^3 )dρdθ=a∫_0 ^(π/4) cos^2 (θ)dθ ∫cos^2 (θ)dθ=∫(1/2)+(1/2)(2cos^2 θ−1)dθ=(θ/2)+(1/2)∫cos(2θ)dθ=2θ+(1/4)sin(2θ) ...=a[(π/8)+(1/4)]](https://www.tinkutara.com/question/Q141869.png)
$${E}=\left\{\left({x},{y}\right)\mid{y}<{x}<{a}\:\vee\:\mathrm{0}<{y}<{a}\right\} \\ $$$${E}^{\mathrm{1}} =\left\{\left(\rho,\theta\right)\mid\:\mathrm{0}<\rho<{a},\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{4}}\right\} \\ $$$${det}\mid{J}\mid=\rho \\ $$$$\int_{\mathrm{0}} ^{{a}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\rho^{\mathrm{3}} {cos}^{\mathrm{2}} \left(\theta\right)}{\rho^{\mathrm{3}} }{d}\rho{d}\theta={a}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cos}^{\mathrm{2}} \left(\theta\right){d}\theta \\ $$$$\int{cos}^{\mathrm{2}} \left(\theta\right){d}\theta=\int\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{cos}^{\mathrm{2}} \theta−\mathrm{1}\right){d}\theta=\frac{\theta}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int{cos}\left(\mathrm{2}\theta\right){d}\theta=\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}\theta\right) \\ $$$$…={a}\left[\frac{\pi}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{4}}\right] \\ $$