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Question Number 62201 by maxmathsup by imad last updated on 17/Jun/19

calculate ∫∫_W e^(x−2y) sin(x+2y) dxdy W ={(x,y)^2 / 0≤x≤1 and 2≤y≤(√5)}

$${calculate}\:\int\int_{{W}} \:\:{e}^{{x}−\mathrm{2}{y}} {sin}\left({x}+\mathrm{2}{y}\right)\:{dxdy} \\ $$$${W}\:=\left\{\left({x},{y}\right)^{\mathrm{2}} /\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:\:{and}\:\:\mathrm{2}\leqslant{y}\leqslant\sqrt{\mathrm{5}}\right\} \\ $$

Commented by mathmax by abdo last updated on 04/Jul/19

let use the diffeomorphism (u,v)→ϕ(u,v)=(x,y) /x−2y =u and x+2y =v ⇒ x =((u+v)/2) and y =((−u+v)/4) ⇒ϕ_1 (u,v)=((u+v)/2) and ϕ_2 (u,v)=((−u+v)/4) we have 0≤x≤1 and −2(√5)≤−2y≤−4 ⇒−2(√5)≤x−2y≤−3 ⇒−2(√5)≤u≤−3 4≤2y≤2(√5) ⇒ 4≤x+2y≤ 1+2(√5) ⇒ 4≤v≤1+2(√5) ∫∫_W f(x,ydxdy =∫∫_w foϕ(u,v)∣j_ϕ ∣ du dv M_j (ϕ) = ((((∂ϕ_1 /∂u) (∂ϕ_1 /∂v))),(((∂ϕ_2 /∂u) (∂ϕ_2 /∂v))) ) = ((((1/2) (1/2))),((−(1/4) (1/4))) ) det Mj(ϕ) =(2/8) =(1/4)⇒ I =∫∫_(−2(√5)≤u≤−3 and 4≤v≤1+2(√5)) e^u sin(v) (1/4) du dv =(1/4) ∫_(−2(√5)) ^(−3) e^u du . ∫_4 ^(1+2(√5)) sin(v)dv =(1/4)( e^(−3) −e^(−2(√5)) )( cos(4)−cos(1+2(√5))) .

$${let}\:{use}\:{the}\:{diffeomorphism}\:\left({u},{v}\right)\rightarrow\varphi\left({u},{v}\right)=\left({x},{y}\right)\:/{x}−\mathrm{2}{y}\:={u}\:{and}\:{x}+\mathrm{2}{y}\:={v}\:\Rightarrow \\ $$$${x}\:=\frac{{u}+{v}}{\mathrm{2}}\:{and}\:\:{y}\:=\frac{−{u}+{v}}{\mathrm{4}}\:\Rightarrow\varphi_{\mathrm{1}} \left({u},{v}\right)=\frac{{u}+{v}}{\mathrm{2}}\:{and}\:\varphi_{\mathrm{2}} \left({u},{v}\right)=\frac{−{u}+{v}}{\mathrm{4}} \\ $$$$\:{we}\:{have}\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\:−\mathrm{2}\sqrt{\mathrm{5}}\leqslant−\mathrm{2}{y}\leqslant−\mathrm{4}\:\Rightarrow−\mathrm{2}\sqrt{\mathrm{5}}\leqslant{x}−\mathrm{2}{y}\leqslant−\mathrm{3}\:\Rightarrow−\mathrm{2}\sqrt{\mathrm{5}}\leqslant{u}\leqslant−\mathrm{3} \\ $$$$\mathrm{4}\leqslant\mathrm{2}{y}\leqslant\mathrm{2}\sqrt{\mathrm{5}}\:\Rightarrow\:\mathrm{4}\leqslant{x}+\mathrm{2}{y}\leqslant\:\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}\:\Rightarrow\:\mathrm{4}\leqslant{v}\leqslant\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\int\int_{{W}} \:\:\:{f}\left({x},{ydxdy}\:=\int\int_{{w}} {fo}\varphi\left({u},{v}\right)\mid{j}_{\varphi} \mid\:{du}\:{dv}\right. \\ $$$${M}_{{j}} \left(\varphi\right)\:=\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial{v}}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial{v}}}\end{pmatrix}\:\:\:\:\:\:=\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}}\end{pmatrix} \\ $$$${det}\:{Mj}\left(\varphi\right)\:=\frac{\mathrm{2}}{\mathrm{8}}\:=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow \\ $$$${I}\:=\int\int_{−\mathrm{2}\sqrt{\mathrm{5}}\leqslant{u}\leqslant−\mathrm{3}\:{and}\:\:\mathrm{4}\leqslant{v}\leqslant\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}} \:\:\:\:\:{e}^{{u}} {sin}\left({v}\right)\:\frac{\mathrm{1}}{\mathrm{4}}\:{du}\:{dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{−\mathrm{2}\sqrt{\mathrm{5}}} ^{−\mathrm{3}} \:{e}^{{u}} {du}\:.\:\int_{\mathrm{4}} ^{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}} \:\:\:{sin}\left({v}\right){dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\:{e}^{−\mathrm{3}} \:−{e}^{−\mathrm{2}\sqrt{\mathrm{5}}} \right)\left(\:\:{cos}\left(\mathrm{4}\right)−{cos}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{5}}\right)\right)\:. \\ $$

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