Question Number 132838 by Ahmed1hamouda last updated on 16/Feb/21
Answered by TheSupreme last updated on 17/Feb/21
![∫∫(x+y)^2 sin^2 (x−y)dA R={(x,y)∣−1<x−y<1, 1<x+y<3} set u=x+y, v=x−y R′={(u,v)∣−1<v<1,1<u<3} { ((x=((u+v)/2))),((y=((u−v)/2))) :} J= determinant (((1/2),(1/2)),((1/2),(−(1/2))))=(1/2) dA=dxdy=−(1/2)dudv (1/2)∫_1 ^3 u^2 du∫_(−1) ^1 sin^2 (v)dv (1/2)[(u^3 /3)]_1 ^3 [(v/2)+((sin(v)cos(v))/2)]_(−1) ^1 (1/2)[9−(1/3)][1+sin(1)cos(1)]] ((13)/3)(1+((sin(2))/2))](https://www.tinkutara.com/question/Q132875.png)
$$\int\int\left({x}+{y}\right)^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}−{y}\right){dA} \\ $$$${R}=\left\{\left({x},{y}\right)\mid−\mathrm{1}<{x}−{y}<\mathrm{1},\:\mathrm{1}<{x}+{y}<\mathrm{3}\right\} \\ $$$${set}\:{u}={x}+{y},\:{v}={x}−{y} \\ $$$${R}'=\left\{\left({u},{v}\right)\mid−\mathrm{1}<{v}<\mathrm{1},\mathrm{1}<{u}<\mathrm{3}\right\} \\ $$$$\begin{cases}{{x}=\frac{{u}+{v}}{\mathrm{2}}}\\{{y}=\frac{{u}−{v}}{\mathrm{2}}}\end{cases} \\ $$$${J}=\begin{vmatrix}{\frac{\mathrm{1}}{\mathrm{2}}}&{\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}}&{−\frac{\mathrm{1}}{\mathrm{2}}}\end{vmatrix}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${dA}={dxdy}=−\frac{\mathrm{1}}{\mathrm{2}}{dudv} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{3}} {u}^{\mathrm{2}} {du}\int_{−\mathrm{1}} ^{\mathrm{1}} {sin}^{\mathrm{2}} \left({v}\right){dv} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{1}} ^{\mathrm{3}} \left[\frac{{v}}{\mathrm{2}}+\frac{{sin}\left({v}\right){cos}\left({v}\right)}{\mathrm{2}}\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$\left.\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{9}−\frac{\mathrm{1}}{\mathrm{3}}\right]\left[\mathrm{1}+{sin}\left(\mathrm{1}\right){cos}\left(\mathrm{1}\right)\right]\right]\:\: \\ $$$$\frac{\mathrm{13}}{\mathrm{3}}\left(\mathrm{1}+\frac{{sin}\left(\mathrm{2}\right)}{\mathrm{2}}\right) \\ $$$$ \\ $$