Ω = ∫_0 ^( 1) ∫_0 ^( 1) (x−y )^2 sin^( 2) ( x+y )dxdy=?


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Question Number 201044 by mnjuly1970 last updated on 28/Nov/23

$Ω = \int_{0}^{1} \int_{0}^{1} {(x - y)}^{2} {s i n}^{2} (x + y) d x d y = ?$
	Answered by mathematicsmagic last updated on 29/Nov/23

	$u = x - y, v = x + y$
	Answered by Mathspace last updated on 29/Nov/23
	$Φ=∫∫_([0,1]^2 ) (x−y)^2 sin^2 (x+y)dxdy we use the diffeomorphisme (x,y)→(f_1 (u,v),f_2 (u,v))=f(u,v) =(u,v) with u=x−y andv=x+y ⇒x=((u+v)/2)=f_1 and y=((−u+v)/2)=f_2 ∫_([0,1]^2 ) f(x,y)dxdy =∫_w Φof ∣j_Φ ∣dudv 0≤x≤1 and 0≤y≤1 ⇒−1≤−y≤0 0≤x+y≤2 and −1≤x−y≤1 ⇒ ⇒0≤v≤2 and −1≤u≤1 m_j Φ= ((((∂f_1 /∂u) (∂f_1 /∂v))),(((∂f_2 /∂u) (∂f_2 /∂v))) ) = ((((1/2) (1/2))),((−(1/2) (1/2))) ) ∣m_j Φ∣=(1/4)+(1/4)=(1/2) ⇒I=∫∫_(−1≤u≤1 and 0≤v≤2) u^2 sin^2 v(1/2)du dv =(1/2)∫_(−1) ^1 u^2 du.∫_0 ^2 sin^2 v dv =(1/2)[(u^3 /3)]_(−1) ^1 .∫_0 ^2 ((1−cos(2v))/2)dv =(1/(12))×2∫_0 ^2 (1−cos(2v))dv =(1/6)[v−(1/2)sin(2v)]_0 ^2 =(1/6){2−(1/2)sin4} =(1/3)−(1/(12))sin4$
	$Φ = \int \int_{{[0, 1]}^{2}} {(x - y)}^{2} {s i n}^{2} (x + y) d x d y$ $w e u s e t h e d i f f e o m o r p h i s m e$ $(x, y) \to (f_{1} (u, v), f_{2} (u, v)) = f (u, v)$ $= (u, v) w i t h u = x - y a n d v = x + y$ $\Rightarrow x = \frac{u + v}{2} = f_{1} a n d y = \frac{- u + v}{2} = f_{2}$ $\int_{{[0, 1]}^{2}} f (x, y) d x d y$ $= \int_{w} Φ o f ∣ j_{Φ} ∣ d u d v$ $0 ⩽ x ⩽ 1 a n d 0 ⩽ y ⩽ 1 \Rightarrow - 1 ⩽ - y ⩽ 0$ $0 ⩽ x + y ⩽ 2 a n d - 1 ⩽ x - y ⩽ 1 \Rightarrow$ $\Rightarrow 0 ⩽ v ⩽ 2 a n d - 1 ⩽ u ⩽ 1$ $m_{j} Φ = (\begin{matrix} \frac{\partial f_{1}}{\partial u} \frac{\partial f_{1}}{\partial v} \\ \frac{\partial f_{2}}{\partial u} \frac{\partial f_{2}}{\partial v} \end{matrix})$ $= (\begin{matrix} \frac{1}{2} \frac{1}{2} \\ - \frac{1}{2} \frac{1}{2} \end{matrix})$ $∣ m_{j} Φ ∣= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ $\Rightarrow I = \int \int_{- 1 ⩽ u ⩽ 1 a n d 0 ⩽ v ⩽ 2} u^{2} {s i n}^{2} v \frac{1}{2} d u d v$ $= \frac{1}{2} \int_{- 1}^{1} u^{2} d u . \int_{0}^{2} {s i n}^{2} v d v$ $= \frac{1}{2} {[\frac{u^{3}}{3}]}_{- 1}^{1} . \int_{0}^{2} \frac{1 - c o s (2 v)}{2} d v$ $= \frac{1}{12} \times 2 \int_{0}^{2} (1 - c o s (2 v)) d v$ $= \frac{1}{6} {[v - \frac{1}{2} s i n (2 v)]}_{0}^{2}$ $= \frac{1}{6} {2 - \frac{1}{2} s i n 4}$ $= \frac{1}{3} - \frac{1}{12} s i n 4$
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