Integration Questions

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

$$\Phi=\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\left({x}−{y}\right)^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}+{y}\right){dxdy} \\$$$${we}\:{use}\:{the}\:{diffeomorphisme} \\$$$$\left({x},{y}\right)\rightarrow\left({f}_{\mathrm{1}} \left({u},{v}\right),{f}_{\mathrm{2}} \left({u},{v}\right)\right)={f}\left({u},{v}\right) \\$$$$=\left({u},{v}\right)\:{with}\:{u}={x}−{y}\:{andv}={x}+{y} \\$$$$\Rightarrow{x}=\frac{{u}+{v}}{\mathrm{2}}={f}_{\mathrm{1}} \:{and}\:{y}=\frac{−{u}+{v}}{\mathrm{2}}={f}_{\mathrm{2}} \\$$$$\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:{f}\left({x},{y}\right){dxdy} \\$$$$=\int_{{w}} \Phi{of}\:\mid{j}_{\Phi} \mid{dudv} \\$$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}\:\Rightarrow−\mathrm{1}\leqslant−{y}\leqslant\mathrm{0} \\$$$$\mathrm{0}\leqslant{x}+{y}\leqslant\mathrm{2}\:{and}\:\:−\mathrm{1}\leqslant{x}−{y}\leqslant\mathrm{1}\:\Rightarrow \\$$$$\Rightarrow\mathrm{0}\leqslant{v}\leqslant\mathrm{2}\:{and}\:\:−\mathrm{1}\leqslant{u}\leqslant\mathrm{1} \\$$$${m}_{{j}} \Phi=\begin{pmatrix}{\frac{\partial{f}_{\mathrm{1}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial{f}_{\mathrm{1}} }{\partial{v}}}\\{\frac{\partial{f}_{\mathrm{2}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial{f}_{\mathrm{2}} }{\partial{v}}}\end{pmatrix} \\$$$$=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\$$$$\mid{m}_{{j}} \Phi\mid=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}} \\$$$$\Rightarrow{I}=\int\int_{−\mathrm{1}\leqslant{u}\leqslant\mathrm{1}\:{and}\:\mathrm{0}\leqslant{v}\leqslant\mathrm{2}} \:{u}^{\mathrm{2}} {sin}^{\mathrm{2}} {v}\frac{\mathrm{1}}{\mathrm{2}}{du}\:{dv} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\mathrm{1}} ^{\mathrm{1}} {u}^{\mathrm{2}} {du}.\int_{\mathrm{0}} ^{\mathrm{2}} {sin}^{\mathrm{2}} {v}\:{dv} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\right]_{−\mathrm{1}} ^{\mathrm{1}} .\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}−{cos}\left(\mathrm{2}{v}\right)}{\mathrm{2}}{dv} \\$$$$=\frac{\mathrm{1}}{\mathrm{12}}×\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{v}\right)\right){dv} \\$$$$=\frac{\mathrm{1}}{\mathrm{6}}\left[{v}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{v}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \\$$$$=\frac{\mathrm{1}}{\mathrm{6}}\left\{\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{4}\right\} \\$$$$=\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{12}}{sin}\mathrm{4} \\$$$$\\$$