Question Number 201044 by mnjuly1970 last updated on 28/Nov/23
$$ \\ $$$$\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \left({x}−{y}\:\right)^{\mathrm{2}} {sin}^{\:\mathrm{2}} \:\left(\:{x}+{y}\:\right){dxdy}=? \\ $$
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](Q201069.png)
$$\Phi=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\left({x}−{y}\right)^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}+{y}\right){dxdy} \\ $$$${we}\:{use}\:{the}\:{diffeomorphisme} \\ $$$$\left({x},{y}\right)\rightarrow\left({f}_{\mathrm{1}} \left({u},{v}\right),{f}_{\mathrm{2}} \left({u},{v}\right)\right)={f}\left({u},{v}\right) \\ $$$$=\left({u},{v}\right)\:{with}\:{u}={x}−{y}\:{andv}={x}+{y} \\ $$$$\Rightarrow{x}=\frac{{u}+{v}}{\mathrm{2}}={f}_{\mathrm{1}} \:{and}\:{y}=\frac{−{u}+{v}}{\mathrm{2}}={f}_{\mathrm{2}} \\ $$$$\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:{f}\left({x},{y}\right){dxdy} \\ $$$$=\int_{{w}} \Phi{of}\:\mid{j}_{\Phi} \mid{dudv} \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\:\Rightarrow−\mathrm{1}\leqslant−{y}\leqslant\mathrm{0} \\ $$$$\mathrm{0}\leqslant{x}+{y}\leqslant\mathrm{2}\:{and}\:\:−\mathrm{1}\leqslant{x}−{y}\leqslant\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\mathrm{0}\leqslant{v}\leqslant\mathrm{2}\:{and}\:\:−\mathrm{1}\leqslant{u}\leqslant\mathrm{1} \\ $$$${m}_{{j}} \Phi=\begin{pmatrix}{\frac{\partial{f}_{\mathrm{1}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial{f}_{\mathrm{1}} }{\partial{v}}}\\{\frac{\partial{f}_{\mathrm{2}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial{f}_{\mathrm{2}} }{\partial{v}}}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mid{m}_{{j}} \Phi\mid=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{I}=\int\int_{−\mathrm{1}\leqslant{u}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant{v}\leqslant\mathrm{2}} \:{u}^{\mathrm{2}} {sin}^{\mathrm{2}} {v}\frac{\mathrm{1}}{\mathrm{2}}{du}\:{dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}} ^{\mathrm{1}} {u}^{\mathrm{2}} {du}.\int_{\mathrm{0}} ^{\mathrm{2}} {sin}^{\mathrm{2}} {v}\:{dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\right]_{−\mathrm{1}} ^{\mathrm{1}} .\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}−{cos}\left(\mathrm{2}{v}\right)}{\mathrm{2}}{dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}×\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{v}\right)\right){dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[{v}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{v}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\left\{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{4}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{12}}{sin}\mathrm{4} \\ $$$$ \\ $$