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Question Number 132838 by Ahmed1hamouda last updated on 16/Feb/21

	Answered by TheSupreme last updated on 17/Feb/21
	$∫∫(x+y)^2 sin^2 (x−y)dA R={(x,y)∣−1<x−y<1, 1<x+y<3} set u=x+y, v=x−y R′={(u,v)∣−1<v<1,1<u<3} { ((x=((u+v)/2))),((y=((u−v)/2))) :} J= determinant (((1/2),(1/2)),((1/2),(−(1/2))))=(1/2) dA=dxdy=−(1/2)dudv (1/2)∫_1 ^3 u^2 du∫_(−1) ^1 sin^2 (v)dv (1/2)[(u^3 /3)]_1 ^3 [(v/2)+((sin(v)cos(v))/2)]_(−1) ^1 (1/2)[9−(1/3)][1+sin(1)cos(1)]] ((13)/3)(1+((sin(2))/2))$
	$\int \int {(x + y)}^{2} {s i n}^{2} (x - y) d A$ $R = {(x, y) ∣ - 1 < x - y < 1, 1 < x + y < 3}$ $s e t u = x + y, v = x - y$ $R^{'} = {(u, v) ∣ - 1 < v < 1, 1 < u < 3}$ ${\begin{cases} x = \frac{u + v}{2} \\ y = \frac{u - v}{2} \end{cases}$ $J = \| \begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & - \frac{1}{2} \end{matrix} \| = \frac{1}{2}$ $d A = d x d y = - \frac{1}{2} d u d v$ $\frac{1}{2} \int_{1}^{3} u^{2} d u \int_{- 1}^{1} {s i n}^{2} (v) d v$ $\frac{1}{2} {[\frac{u^{3}}{3}]}_{1}^{3} {[\frac{v}{2} + \frac{s i n (v) c o s (v)}{2}]}_{- 1}^{1}$ $\frac{1}{2} [9 - \frac{1}{3}] [1 + s i n (1) c o s (1)]]$ $\frac{13}{3} (1 + \frac{s i n (2)}{2})$
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