Question Number 128620 by Ahmed1hamouda last updated on 08/Jan/21
Answered by TheSupreme last updated on 09/Jan/21
![u=x+y v=x−y x=((u+v)/2) y=((u−v)/2) J= [(((1/2) (1/2))),(((1/2) −(1/2))) ] det(J)=−(1/2) R={(u,v)∣−1≤v≤1,1≤u≤3} ∫∫u^2 sin(v)(−(1/2))dudv=(u^3 /6)∣_1 ^3 cos(v)∣_(−1) ^1 =((26)/6)2cos(1)=((26)/3)cos(1)](https://www.tinkutara.com/question/Q128627.png)
$${u}={x}+{y} \\ $$$${v}={x}−{y} \\ $$$${x}=\frac{{u}+{v}}{\mathrm{2}} \\ $$$${y}=\frac{{u}−{v}}{\mathrm{2}} \\ $$$${J}=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}}}\end{bmatrix} \\ $$$${det}\left({J}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}=\left\{\left({u},{v}\right)\mid−\mathrm{1}\leqslant{v}\leqslant\mathrm{1},\mathrm{1}\leqslant{u}\leqslant\mathrm{3}\right\} \\ $$$$\int\int{u}^{\mathrm{2}} {sin}\left({v}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\right){dudv}=\frac{{u}^{\mathrm{3}} }{\mathrm{6}}\mid_{\mathrm{1}} ^{\mathrm{3}} {cos}\left({v}\right)\mid_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{26}}{\mathrm{6}}\mathrm{2}{cos}\left(\mathrm{1}\right)=\frac{\mathrm{26}}{\mathrm{3}}{cos}\left(\mathrm{1}\right) \\ $$